Chemical kinetics

Class 11 Chemistry, unit 7, book pages 144–165

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Chemical kinetics studies how fast chemical reactions go, what controls their speed and the step-by-step path (mechanism) by which reactants become products. The unit defines the rate of reaction, the rate law, the order and the rate constant, and shows how to find them from experimental data and from the half-life. It then explains, with collision theory, the Maxwell-Boltzmann curve, the Arrhenius equation and energy profiles, how temperature, activation energy and catalysts change the rate, and ends with the feasibility of a reaction from the Gibbs free energy change.

Learning outcomes (11)

Explain the rate of reaction, and rate constant.
Use experimental data to calculate the rate of a chemical reaction.
Explain the concept of activation energy and its role in chemical reactions.
Use the Boltzmann distribution curve to explain the effect of temperature on the rate of a reaction.
Explain the concept of catalyst and how they increase the rate of a reaction by lowering the activation energy.
Interpret reaction pathway diagrams, including in the presence and absence of catalysts.
Explain the relationship between Gibbs free energy change, $Δ G^{\circ}$ and the feasibility of a reaction.
Use rate equations, including orders of reaction and rate constant.
Suggest a reaction mechanism that is consistent with a given rate equation and rate-determining step.
Calculate the numerical value of a rate constant using the initial rates and half-life method.
Describe the effect of temperature change on the rate constant and rate of reaction.

What is chemical kinetics?

Chemical reactions are dynamic processes in which matter and energy are constantly changing. They occur at very different rates, from very fast to very slow. Fermentation is a slow reaction that can take several weeks to produce enough product. Acid-base neutralisation reactions, on the other hand, are complete in microseconds. Some reactions go at a reasonable rate, for example the reactions that contract muscles and transmit impulses along nerves, and the reactions that record photographic images.

In industry it is important to know the conditions under which a reaction goes most favourably. Rate information is also the most important information used to infer the mechanism of a reaction.

Chemical kinetics: The study of the rates, mechanisms and factors that affect the rate of chemical reactions.
Mechanism: The sequence of all the chemical steps leading from the starting materials to the products.

7.1Rates of reactionsOften examined

The rate of reaction is the change in concentration of reactants or products per unit time. Mathematically,

Rate = \frac{change in concentration of a substance}{time taken for the change}

The concentration of the reactants decreases and the concentration of the products increases with the passage of time. A reaction rate can therefore also be defined as the decrease in concentration of reactants per unit time or the increase in concentration of products per unit time. The unit of concentration is $mol d m^{- 3}$ and the unit of time is the second, so the unit of the reaction rate is

mol d m^{- 3} s^{- 1}

The change in concentration of the reactant and the product of a general reaction $A B$ can be shown on a graph (Figure 7.1).

100

Time

Conc.

A molecules

B molecules

Figure 7.1: Change in concentration of reactants and products with passage of time (for

A B

). The curves show the shape only: A falls and B rises, steeply at first and then more slowly.

The slope of the curve, for both the starting material and the product, is steeper at the beginning than at later stages. This means a rapid decrease of the reactant, or a rapid increase of the product, at the start. As the reaction goes on, the slope becomes less steep, which shows that the reaction rate is falling. Finally the graph becomes horizontal and the reaction stops. So the reaction rate is never constant: the reactant concentration keeps falling and the product concentration keeps rising, and therefore the reaction rate keeps falling.

Rate of reaction as a derivative

If $d x$ is a very small change in concentration of a product in a very small time interval $d t$ , the rate of reaction is expressed as

Rate of reaction = \frac{d x}{d t}

For the general reaction $A B$ the rate can be expressed either as the rate of disappearance of the reactant A or as the rate of appearance of the product B:

\frac{d x}{d t} = - \frac{d [ A ]}{d t} \frac{d x}{d t} = + \frac{d [ B ]}{d t}

Here $d [A]$ and $d [B]$ are the changes in the concentrations of A and B. The negative sign shows a decrease in the concentration of the reactant A, and the positive sign shows an increase in the concentration of the product B. The rate is therefore always a positive number. With these relationships you can find the rate of reaction between any two time intervals.

Example 7.1: Rate of the reaction A + B giving C

Determine the rate of the reaction

A + B C

between

0 s

and

20 s

, from the concentration of the product C measured at different times.

Sr. No.	Time (s)	Concentration of C ( $mol d m^{- 3}$ )
1	0.0	0.0
2	10	0.20
3	20	0.38
4	30	0.45
5	40	0.60

Rate between two time intervals is found by following the change in concentration of C. Take the interval $0.0 s$ to $20 s$ and find the change in time.
$d t = 20 s - 0.0 s = 20 s$
Find the change in the concentration of C over the same interval (the table gives $0.0$ at $0 s$ and $0.38$ at $20 s$ ).
$d [C] = 0.38 - 0.0 = 0.38 mol d m^{- 3}$
C is a product, so the rate is $+ d [C] / d t$ .
$\frac{d x}{d t} = + \frac{d [ C ]}{d t} = \frac{0.38}{20} = 0.019 mol d m^{- 3} s^{- 1}$

Rate = 0.019 mol d m^{- 3} s^{- 1}

7.2Rate lawOften examined

The rate of a reaction is the instantaneous change in concentration of a reactant or product at a given time. Experimental studies of reaction rates show that the rate of a chemical reaction is proportional to the molar concentration of each reactant raised to a power, and the value of that power is found by experiment. For a general reaction $A products$ :

Rate \propto [A]^{x}

Rate = k [A]^{x}

Rate constant ( $k$ ): The proportionality constant in the rate law. It is the rate of reaction when the molar concentration of each reactant is unity.
Rate law (rate equation): The expression for the rate of the reaction in terms of $k$ and the concentrations of the reactants, each raised to a power found by experiment.
Order with respect to A: The exponent $x$ of the concentration of A in the rate equation.

When $[A] = 1 M$ , the rate law gives

Rate = k

So the rate constant is the rate of reaction when the molar concentration of each reactant is one. It links the concentration of the reactants to the reaction rate. Every reaction has its own specific rate constant, which does not depend on concentration or on time. But the value of the rate constant changes with temperature (see section 7.3).

7.2.1Order of reaction and rate equationOften examined

Order of reaction (book definition): The number of molecules of reactants participating in the rate-determining step.
Order of reaction (as the sum of exponents): The sum of all the exponents to which the molar concentration terms in the rate equation are raised.

Consider a general reaction between $a$ moles of A and $b$ moles of B to give $c$ moles of C and $d$ moles of D:

a A + b B ⟶ c C + d D

The rate equation can be written as

Rate \propto [A]^{x} [B]^{y}

Rate = k [A]^{x} [B]^{y}

The exponent $x$ is the order of reaction with respect to species A and the exponent $y$ is the order of reaction with respect to species B. The order of the reaction expresses the effect of concentration on the rate of reaction. The sum $x + y$ is the overall order of the reaction, or simply the order of the reaction.

The values of $x$ and $y$ may or may not be the same as $a$ and $b$ . The order of a reaction for a particular species cannot be predicted from the balanced chemical equation. It can be determined only by experiment. For example, for the reaction

2 NO X_{2} + O X_{3} ⟶ N X_{2} O X_{5} + O X_{2}

experiment shows that

Rate = k [NO X_{2}] [O X_{3}]

The order with respect to $NO X_{2}$ is one, whereas its stoichiometric coefficient is two. An expression which shows how the reaction rate is related to the concentration of the reactants is the rate law or rate equation. The order of a reaction may be a whole number, zero or a fraction, and it helps in finding the mechanism of a reaction.

Types of order of reactions

a) Zero order reaction

A reaction that is independent of the concentration of the reactant molecules is called a zero order reaction. An example is the decomposition of ammonia on heated tungsten.

2 NH X_{3} (g) ⟶ N X_{2} (g) + 3 H X_{2} (g) Rate = k [NH X_{3}]^{0}

The concentration of ammonia decreases continuously until it reaches zero, but the rate stays the same. The combination of $H X_{2}$ and $Cl X_{2}$ in the presence of sunlight is also a zero order reaction. Reactions catalysed by enzymes also follow zero order kinetics.

b) First order reactions

A reaction whose rate is directly proportional to the first power of the concentration of a single reactant molecule is called a first order reaction: $Rate \propto [A]^{1}$ .

Thermal decomposition of $N X_{2} O X_{5}$ : $2 N X_{2} O X_{5} (g) \to 2 N X_{2} O X_{4} (g) + O X_{2} (g)$ , with $Rate = k [N X_{2} O X_{5}]$ .
Decomposition of ammonium nitrite in aqueous solution: $NH X_{4} NO X_{2} N X_{2} + 2 H X_{2} O$ , with $Rate = k [NH X_{4} NO X_{2}]$ .

c) Second order reactions

A reaction for which the sum of the exponents in the rate equation is two is called a second order reaction: $Rate \propto [A]^{2}$ or $Rate \propto [A]^{1} [B]^{1}$ .

Decomposition of nitrogen dioxide: $2 NO X_{2} (g) \to 2 NO (g) + O X_{2} (g)$ , with $Rate = k [NO X_{2}]^{2}$ .
$NO (g) + O X_{3} (g) \to NO X_{2} (g) + O X_{2} (g)$ , with $Rate = k [NO] [O X_{3}]$ . Order of reaction $= 1 + 1 = 2$ .
$2 NO X_{2} (g) + O X_{3} (g) \to N X_{2} O X_{5} (g) + O X_{2} (g)$ , with $Rate = k [NO X_{2}] [O X_{3}]$ .

d) Third order reactions

A reaction for which the sum of the exponents in the rate equation is three is called a third order reaction: $Rate \propto [A]^{3}$ , or $[A]^{2} [B]^{1}$ , or $[A]^{1} [B]^{1} [C]^{1}$ .

The oxidation of $NO$ by $O X_{2}$ : $2 NO (g) + O X_{2} (g) \to 2 NO X_{2} (g)$ , with $Rate = k [NO]^{2} [O X_{2}]$ .
$2 FeCl X_{3} (a q) + 6 KI (a q) \to 2 FeI X_{2} (a q) + 6 KCl (a q) + I X_{2} (s)$ , with $Rate = k [FeCl X_{3}] [KI]^{2}$ .

e) Fractional order reactions

A reaction for which the sum of the exponents in the rate equation is a fraction is called a fractional order reaction.

The reaction between $H X_{2}$ and $Br X_{2}$ to give $HBr$ is first order in $H X_{2}$ and half order in $Br X_{2}$ : $H X_{2} (g) + Br X_{2} (g) \to 2 HBr (g)$ , with $Rate = k [H X_{2}] [Br X_{2}]^{1/2}$ . Order of reaction $= 1 + 0.5 = 1.5$ .
$CHCl X_{3} (l) + Cl X_{2} (g) \to CCl X_{4} (l) + HCl (g)$ , with $Rate = k [CHCl X_{3}] [Cl X_{2}]^{1/2}$ .

f) Pseudo first order reaction

A bimolecular reaction in which the solvent is in excess, so that its concentration stays constant and it does not take part in the rate-determining step, is called a pseudo first order reaction. Two reactants take part, but only one concentration changes, so the reaction behaves as first order.

(CH X_{3})_{3} C - Br (l) + H X_{2} O (excess) ⟶ (CH X_{3})_{3} C - OH (l) + HBr (l) Rate = k [(CH X_{3})_{3} C - Br]

tert-butyl bromide

tert-butyl alcohol

Summary of the orders of reaction in the book
Order	Rate law (example)	Reaction
Zero	$Rate = k [NH X_{3}]^{0}$	$2 NH X_{3} \to N X_{2} + 3 H X_{2}$ on heated tungsten
First	$Rate = k [N X_{2} O X_{5}]$	$2 N X_{2} O X_{5} \to 2 N X_{2} O X_{4} + O X_{2}$
Second	$Rate = k [NO X_{2}]^{2}$	$2 NO X_{2} \to 2 NO + O X_{2}$
Third	$Rate = k [NO]^{2} [O X_{2}]$	$2 NO + O X_{2} \to 2 NO X_{2}$
Fractional ( $1.5$ )	$Rate = k [H X_{2}] [Br X_{2}]^{1/2}$	$H X_{2} + Br X_{2} \to 2 HBr$
Pseudo first	$Rate = k [(CH X_{3})_{3} CBr]$	$(CH X_{3})_{3} CBr + H X_{2} O \to (CH X_{3})_{3} COH + HBr$ (water in excess)

7.2.2Determination of reaction order, rate law and rate constantOften examined

The effect of a change in reactant concentration on the reaction rate cannot be derived from the chemical equation. It can only be found by experiment, by determining the order of the reaction. For this, the method of initial rates is a simpler way to find the order. An experiment is designed in which the concentration of one reactant is changed while everything else is kept constant. The concentration of one reactant is varied systematically and the initial reaction rate is found for each change.

Method of initial rates

Keep the concentration of all reactants but one constant, and change that one.
Compare the initial rates of the two experiments. If doubling a concentration doubles the rate, the order in that reactant is 1. If it quadruples the rate, the order is 2. If the rate does not change, the order is 0.
Repeat for each reactant, then combine the orders into the rate law \text{Rate}=k[\ce{A}]^{x}[\ce{B}]^{y}.
Put the data of any one experiment into the rate law to find k.

Example 7.2: Rate law and rate constant for NO + O3

Jet engines release nitrogen(II) oxide in the upper atmosphere. In the ozone layer nitrogen(II) oxide reacts with ozone to form nitrogen(IV) oxide and oxygen: \ce{NO}(g)+\ce{O3}(g)\to\ce{NO2}(g)+\ce{O2}(g). The following data were obtained at 25^\circ\mathrm{C}. Use the data to determine the rate law for the reaction.

Experiment	Initial [\ce{NO}]	Initial [\ce{O3}]	Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})
1	1.00\times10^{-6}	9.00\times10^{-6}	1.98\times10^{-4}
2	2.00\times10^{-6}	9.00\times10^{-6}	3.96\times10^{-4}
3	1.00\times10^{-6}	3.00\times10^{-6}	6.60\times10^{-5}

To find the order with respect to \ce{NO}, compare experiments 1 and 2. [\ce{O3}] is constant at 9.00\times10^{-6}\,\mathrm{M} while [\ce{NO}] doubles from 1.00\times10^{-6} to 2.00\times10^{-6}\,\mathrm{M}. The rate goes from 1.98\times10^{-4} to 3.96\times10^{-4}.
\dfrac{1.98\times10^{-4}}{1.98\times10^{-4}}:\dfrac{3.96\times10^{-4}}{1.98\times10^{-4}}=1:2
The initial rate doubles when [\ce{NO}] doubles, so the rate is directly proportional to the first power of [\ce{NO}].
\text{Rate}\propto[\ce{NO}]
To find the order with respect to \ce{O3}, compare experiments 1 and 3. [\ce{NO}] is constant at 1.00\times10^{-6}\,\mathrm{M} while [\ce{O3}] falls to one third, from 9.00\times10^{-6} to 3.00\times10^{-6}\,\mathrm{M}. The rate goes from 1.98\times10^{-4} to 6.60\times10^{-5}.
\dfrac{1.98\times10^{-4}}{1.98\times10^{-4}}:\dfrac{6.60\times10^{-5}}{1.98\times10^{-4}}=1:\dfrac13
The rate also falls to one third, so the rate is directly proportional to the first power of [\ce{O3}].
\text{Rate}\propto[\ce{O3}]
Combine the two results to get the rate law. The overall order is 1+1=2, so this is a second order reaction.
\text{Rate}=k\,[\ce{NO}]\,[\ce{O3}]
To find k, put the data of experiment 1 into the rate law.
1.98\times10^{-4}=k\,(1\times10^{-6})(9\times10^{-6})=k\,(9\times10^{-12})
Solve for k.
k=\dfrac{1.98\times10^{-4}}{9\times10^{-12}}=2.2\times10^{7}\,\mathrm{dm^{3}\,mol^{-1}\,s^{-1}}
Check with another experiment, as the book suggests. Experiment 2 gives k=\dfrac{3.96\times10^{-4}}{(2.00\times10^{-6})(9.00\times10^{-6})}=2.2\times10^{7} and experiment 3 gives k=\dfrac{6.60\times10^{-5}}{(1.00\times10^{-6})(3.00\times10^{-6})}=2.2\times10^{7}. The same value is obtained.

\text{Rate}=k\,[\ce{NO}][\ce{O3}],\qquad k=2.2\times10^{7}\,\mathrm{dm^{3}\,mol^{-1}\,s^{-1}}

Example 7.3: Writing a rate law from the orders

The reaction \ce{H2 + Br2 -> 2HBr} is first order in \ce{H2} and half order in \ce{Br2}. Write the rate law for the reaction.

The first order in \ce{H2} means the rate is proportional to the first power of [\ce{H2}].
\text{Rate}\propto[\ce{H2}]\qquad\text{(i)}
The half order in \ce{Br2} means the rate is proportional to the square root of [\ce{Br2}].
\text{Rate}\propto[\ce{Br2}]^{1/2}\qquad\text{(ii)}
Combine (i) and (ii) and introduce the rate constant.
\text{Rate}\propto[\ce{H2}][\ce{Br2}]^{1/2}\;\Rightarrow\;\text{Rate}=k\,[\ce{H2}][\ce{Br2}]^{1/2}

\text{Rate}=k\,[\ce{H2}][\ce{Br2}]^{1/2}

Worked example: rate law from tripling and doubling experiments

The reaction \ce{A + B -> products} was studied at constant temperature. Use the initial rates in the table to find the order in \ce{A} and in \ce{B}, write the rate law, find the value and unit of k, and predict the initial rate when [\ce{A}]=0.20\,M and [\ce{B}]=0.50\,M.

Experiment	Initial [\ce{A}] (\mathrm{mol\,dm^{-3}})	Initial [\ce{B}] (\mathrm{mol\,dm^{-3}})	Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})
1	0.10	0.10	2.0\times10^{-3}
2	0.30	0.10	1.8\times10^{-2}
3	0.10	0.20	4.0\times10^{-3}

Order in \ce{A}: compare experiments 1 and 2. [\ce{B}] is constant and [\ce{A}] becomes three times larger, so the rate ratio tells the power of 3. Do not assume the rate also triples: it rises nine times.
\dfrac{1.8\times10^{-2}}{2.0\times10^{-3}}=9=3^{x}\;\Rightarrow\;x=2
Order in \ce{B}: compare experiments 1 and 3. [\ce{A}] is constant and [\ce{B}] doubles, and the rate doubles too.
\dfrac{4.0\times10^{-3}}{2.0\times10^{-3}}=2=2^{y}\;\Rightarrow\;y=1
Combine the orders into the rate law. The overall order is 2+1=3.
\text{Rate}=k\,[\ce{A}]^{2}[\ce{B}]
Put the data of experiment 1 into the rate law and solve for k.
k=\dfrac{\text{Rate}}{[\ce{A}]^{2}[\ce{B}]}=\dfrac{2.0\times10^{-3}}{(0.10)^{2}(0.10)}=\dfrac{2.0\times10^{-3}}{1.0\times10^{-3}}=2.0
Find the unit of k by dividing the unit of rate by the unit of concentration cubed.
\dfrac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{(\mathrm{mol\,dm^{-3}})^{3}}=\mathrm{dm^{6}\,mol^{-2}\,s^{-1}}
Predict the new rate with the same k (the temperature is unchanged).
\text{Rate}=2.0\times(0.20)^{2}\times(0.50)=2.0\times0.040\times0.50=4.0\times10^{-2}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

\text{Rate}=k[\ce{A}]^{2}[\ce{B}],\quad k=2.0\,\mathrm{dm^{6}\,mol^{-2}\,s^{-1}},\quad \text{rate}=4.0\times10^{-2}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

Concept assessment exercise 7.1, question 1: Phosgene

Phosgene is a toxic gas, prepared by the reaction of carbon monoxide with chlorine: \ce{CO}(g)+\ce{Cl2}(g)\to\ce{COCl2}(g). The following data were obtained for a kinetic study of this reaction. Write the rate law for this reaction.

Experiment	Initial [\ce{CO}]	Initial [\ce{Cl2}]	Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})
1	1.000	0.100	1.29\times10^{-29}
2	0.100	0.100	1.30\times10^{-30}
3	0.100	1.000	1.30\times10^{-30}

Compare experiments 1 and 2. [\ce{Cl2}] is constant at 0.100 while [\ce{CO}] changes from 0.100 to 1.000 (ten times larger). The rate changes from 1.30\times10^{-30} to 1.29\times10^{-29}.
\dfrac{\text{rate}_1}{\text{rate}_2}=\dfrac{1.29\times10^{-29}}{1.30\times10^{-30}}\approx10=10^{x}\;\Rightarrow\;x=1
Compare experiments 2 and 3. [\ce{CO}] is constant at 0.100 while [\ce{Cl2}] is increased ten times, from 0.100 to 1.000. The rate does not change at all (1.30\times10^{-30} in both).
\dfrac{\text{rate}_3}{\text{rate}_2}=1=10^{y}\;\Rightarrow\;y=0
So the reaction is first order in \ce{CO} and zero order in \ce{Cl2} (overall order 1).
\text{Rate}=k\,[\ce{CO}]

\text{Rate}=k\,[\ce{CO}]

Concept assessment exercise 7.1, question 2: NO2 and CO

The reaction \ce{NO2 + CO -> NO + CO2} is second order in \ce{NO2} and is independent of the concentration of \ce{CO}. Write the rate law for the reaction. What is the overall order of the reaction?

Second order in \ce{NO2} means [\ce{NO2}] is raised to the power 2.
\text{Rate}\propto[\ce{NO2}]^{2}
Independent of \ce{CO} means zero order in \ce{CO}, so [\ce{CO}]^0=1 and the term does not appear.
\text{Rate}=k\,[\ce{NO2}]^{2}[\ce{CO}]^{0}=k\,[\ce{NO2}]^{2}
The overall order is the sum of the exponents.
2+0=2

\text{Rate}=k\,[\ce{NO2}]^{2};\quad\text{overall order}=2

7.2.3Half-life method for determining rate constant

Half-life (t_{1/2}): The time it takes for the concentration of a reactant to be reduced to half. The relationship between the half-life and the rate constant depends on the order of the reaction.

For a reaction of order n, with [\ce{A}]_0 the initial concentration of the reactant:

Half-lives for zero, first and second order reactions
Order n	Half-life	Depends on initial concentration?
0	t_{1/2}=\dfrac{[\ce{A}]_0}{2k}	Yes, directly proportional to [\ce{A}]_0
1	t_{1/2}=\dfrac{0.693}{k}	No
2	t_{1/2}=\dfrac{1}{k\,[\ce{A}]_0}	Yes, inversely proportional to [\ce{A}]_0

Rate constant of a first order reaction from its half-life

The reaction \ce{A + 2B -> C} follows first order kinetics and its half-life is 30\,\mathrm{s}. Find the rate constant.

Given

t_{1/2} = 30\,\mathrm{s} half-life
n = 1 first order

To find

For a first order reaction the half-life is t_{1/2}=0.693/k.
30=\dfrac{0.693}{k}
Rearrange for k.
k=\dfrac{0.693}{30}=0.0231\,\mathrm{s^{-1}}

k=0.0231\,\mathrm{s^{-1}}

Worked example: concentration left after several half-lives

A first order reaction \ce{A -> products} has k=0.0693\,\mathrm{min^{-1}}. The initial concentration of \ce{A} is 0.64\,M. Find (a) the half-life, (b) the concentration left after 30\,\mathrm{min}, (c) the time for the concentration to fall to 0.020\,M, and sketch the concentration-time graph.

Given

k = 0.0693\,\mathrm{min^{-1}} first order
[\ce{A}]_0 = 0.64\,M initial concentration

To find

t_{1/2}
[\ce{A}]\text{ at }30\,\mathrm{min}
\text{time to reach }0.020\,M

(a) For a first order reaction the half-life depends only on k.
t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{0.0693}=10.0\,\mathrm{min}
(b) Count how many half-lives fit into 30\,\mathrm{min}, then halve the concentration that many times.
\dfrac{30}{10}=3\text{ half-lives:}\quad0.64\to0.32\to0.16\to0.08\,M
(c) Find how many times 0.64\,M must be halved to reach 0.020\,M. The ratio is a power of two.
\dfrac{0.64}{0.020}=32=2^{5}\;\Rightarrow\;5\text{ half-lives}
Each half-life is 10\,\mathrm{min} because t_{1/2} of a first order reaction does not depend on the concentration.
t=5\times10\,\mathrm{min}=50\,\mathrm{min}
(d) Draw the curve through 0.64, 0.32, 0.16 and 0.08\,M at 0, 10, 20 and 30\,\mathrm{min}. The curve is steepest at the start and flattens, because the rate is proportional to the concentration that is left.

Concentration of A against time for a first order reaction with t_{1/2}=10\,\mathrm{min}. Every 10\,\mathrm{min} the concentration halves, whatever its starting value.

t_{1/2}=10.0\,\mathrm{min},\quad[\ce{A}]_{30}=0.08\,M,\quad t=50\,\mathrm{min}

7.3Effect of temperature on the rate of reactionsOften examined

The reaction rate generally increases with increasing temperature. According to collision theory, the rate of a reaction is proportional to the number of collisions between the reacting molecules. An increase in temperature increases the average kinetic energy of the molecules, and so the average speed of the reacting molecules. This raises the collision frequency, that is the number of effective collisions, and the reaction speeds up.

But only effective collisions produce a reaction. For a collision to be effective the molecules must have the activation energy and must be correctly oriented. At normal temperatures very few molecules have this activation energy. Not all reactant molecules have the same energy at a given temperature: most of them have medium energy, and some have a higher than average kinetic energy. The number of molecules with kinetic energy at least equal to E_a at temperature T is proportional to the shaded area under the Maxwell-Boltzmann kinetic energy curve.

The Maxwell-Boltzmann distribution curve shows how the available energy is distributed among the molecules of a gas at a constant temperature (Figure 7.2a).

Figure 7.2a: Maxwell-Boltzmann curve of kinetic energy. Only the molecules in the shaded tail, with energy at least E_a, are able to react.

The distribution curve shows that most molecules do not have the required activation energy at T_1, a lower temperature. Only some of the molecules have the required energy and are able to react. When the temperature is raised from T_1 to T_2 the energy of the molecules also increases. The proportion of molecules with the required activation energy therefore increases (Figure 7.2b), and the reaction rate increases.

Figure 7.2b: Maxwell-Boltzmann curve of kinetic energy at two temperatures. At the higher temperature T_2 the peak is lower and shifted to the right, and a greater fraction of molecules has enough energy to react (shaded).

It was found that, in general, the reaction rate increases two to three times for every 10\,\mathrm{K} increase in temperature.

Arrhenius equation

Arrhenius (1889) studied the effect of temperature on reaction rates. He found that the effect of temperature on the rate of reaction is given by the following equation, known as the Arrhenius equation:

k=A\,e^{-E_a/RT}

k: the rate constant
E_a: the energy of activation
R: the gas constant, R=8.3143\,\mathrm{J\,K^{-1}\,mol^{-1}}
T: the absolute temperature, in kelvin
A: a constant known as the Arrhenius constant, related to the frequency of collisions and the orientation of the reacting molecules

The rate constant k varies with temperature. It increases with temperature, which in turn increases the rate of the reaction.

Worked example: effect of a 10 K rise on the rate constant

For a reaction with E_a=50\,\mathrm{kJ\,mol^{-1}}, use the Arrhenius equation k=Ae^{-E_a/RT} with R=8.3143\,\mathrm{J\,K^{-1}\,mol^{-1}} to find by what factor k increases when the temperature is raised from 300\,\mathrm{K} to 310\,\mathrm{K}. Take A as the same at both temperatures.

Given

E_a = 50\,\mathrm{kJ\,mol^{-1}}=50\,000\,\mathrm{J\,mol^{-1}} convert to joules to match R
T_1 = 300\,\mathrm{K}
T_2 = 310\,\mathrm{K}

To find

k_2/k_1

Write the equation at each temperature.
k_1=A\,e^{-E_a/RT_1},\qquad k_2=A\,e^{-E_a/RT_2}
Divide. The constant A cancels, so only the exponentials are left.
\dfrac{k_2}{k_1}=e^{-E_a/RT_2}\div e^{-E_a/RT_1}=\exp\!\left[\dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\right]
Work out each part of the exponent.
\dfrac{E_a}{R}=\dfrac{50\,000}{8.3143}=6014\,\mathrm{K},\qquad \dfrac{1}{300}-\dfrac{1}{310}=1.075\times10^{-4}\,\mathrm{K^{-1}}
Multiply to get the exponent, then evaluate.
6014\times1.075\times10^{-4}=0.647\;\Rightarrow\;\dfrac{k_2}{k_1}=e^{0.647}=1.9
Check directly: e^{-E_a/RT} is 1.97\times10^{-9} at 300\,\mathrm{K} and 3.76\times10^{-9} at 310\,\mathrm{K}, and 3.76/1.97=1.9.
Meaning: a rise of only 10\,\mathrm{K} almost doubles k, which agrees with the book's rule that the rate increases two to three times for every 10\,\mathrm{K}. The exponent is small, but T sits in a negative exponent, so a small rise in T gives a large gain in k.

\dfrac{k_2}{k_1}\approx1.9

7.4The mechanism of a chemical reactionOften examined

The path that reactants take to form products in a chemical reaction is called the mechanism. A reaction rate equation is very useful because it gives information about the reaction mechanism. A reaction can take place in one step or in several steps. If a reaction proceeds in two or more steps, one of the steps is the slowest, and the rate of the slowest step determines the overall reaction rate. This is because it sets a limit to the rate at which the overall reaction can occur: no reaction can proceed more slowly than its rate-determining step. All the other steps in the mechanism are usually fast.

Rate-determining step: The slowest step in the reaction mechanism, which determines the overall rate of the reaction.
Reaction intermediate: A species that is produced in one step of a chemical reaction and consumed in another step. It does not appear in the overall reaction.
Reaction mechanism: The sequence of elementary steps that describe the reaction.

Example 7.4: Information from the rate law of NO2 + CO

For the reaction \ce{NO2}(g)+\ce{CO}(g)\to\ce{NO}(g)+\ce{CO2}(g) the rate is \text{Rate}=k[\ce{NO2}]^{2}. What information do you get from this about the rate-determining step?

The reaction is second order with respect to \ce{NO2} and zero order with respect to \ce{CO}, so it is independent of the concentration of \ce{CO}.
Two molecules of \ce{NO2} are involved in the rate-determining step, and no \ce{CO} is.
The overall equation shows one \ce{NO2} only, so the reaction must proceed in more than one step.
The proposed mechanism has a slow first step with two \ce{NO2} molecules, and a fast second step in which \ce{CO} reacts.
\begin{aligned}\text{Step I:}\quad&\ce{NO2}(g)+\ce{NO2}(g)\xrightarrow{\text{slow}}\ce{NO3}(g)+\ce{NO}(g)\\\text{Step II:}\quad&\ce{NO3}(g)+\ce{CO}(g)\xrightarrow{\text{fast}}\ce{NO2}(g)+\ce{CO2}(g)\end{aligned}
Add the steps and cancel \ce{NO3} and one \ce{NO2} from both sides. The sum is the overall reaction, so the mechanism is acceptable.
\ce{NO2}+\ce{CO}\longrightarrow\ce{NO}+\ce{CO2}
Step I is the rate-determining step. \ce{NO3} does not appear in the overall reaction, so it is a reaction intermediate.

\text{Step I (slow): }\ce{2NO2 -> NO3 + NO};\quad\text{intermediate: }\ce{NO3}

Example 7.5: Hypochlorite ion decomposition

Hypochlorite ion \ce{ClO-} in aqueous solution decomposes to chlorate ion \ce{ClO3-} and chloride ion: 3\ce{ClO-}(aq)\to\ce{ClO3-}(aq)+2\ce{Cl-}(aq). The rate of the reaction is second order in \ce{ClO-}, \text{Rate}=k[\ce{ClO-}]^{2}. The following two-step mechanism is consistent with the rate law. Select the rate-determining step.

\begin{aligned}\text{Step I:}\quad&\ce{ClO-}(aq)+\ce{ClO-}(aq)\longrightarrow\ce{ClO2-}(aq)+\ce{Cl-}(aq)\\\text{Step II:}\quad&\ce{ClO2-}(aq)+\ce{ClO-}(aq)\longrightarrow\ce{ClO3-}(aq)+\ce{Cl-}(aq)\\\hline\text{Overall:}\quad&3\ce{ClO-}(aq)\longrightarrow\ce{ClO3-}(aq)+2\ce{Cl-}(aq)\end{aligned}

The rate law is \text{Rate}=k[\ce{ClO-}]^2, so two \ce{ClO-} ions must take part in the rate-determining step.
Step I has two \ce{ClO-} ions; Step II has only one.
So Step I is the rate-determining step. The species \ce{ClO2-} is formed in Step I and used up in Step II, so it is a reaction intermediate.

\text{Step I is the rate-determining step}

Example 7.6: Rate law and the rate-determining step

The rate law \text{Rate}=k[\ce{NO2}][\ce{O3}] for the reaction 2\ce{NO2}(g)+\ce{O3}\to\ce{N2O5}(g)+\ce{O2}(g) indicates that one molecule of \ce{NO2} and one molecule of \ce{O3} participate in the rate-determining step. What general rule does this show?

Read the rate law: each reactant appears to the first power, so the rate-determining step contains one \ce{NO2} and one \ce{O3}.
\text{Rate}=k[\ce{NO2}]^{1}[\ce{O3}]^{1}
Compare with the overall equation: it has two \ce{NO2}, but only one is in the slow step. The second \ce{NO2} reacts in a fast step later.
So the rate law includes the concentration of each reactant raised to the power that equals the coefficient of that reactant in the equation for the rate-determining step, not in the overall equation.

\text{Powers in the rate law}=\text{coefficients in the rate-determining step}

Concept assessment exercise 7.2: Which mechanism fits the rate law?

For the reaction 2\ce{NO2}(g)+\ce{F2}(g)\to2\ce{NO2F}(g) the rate law is \text{Rate}=k[\ce{NO2}][\ce{F2}]. Which of the following mechanisms is consistent with the rate law? (a) 2\ce{NO2}+\ce{F2}\to2\ce{NO2F}. (b) \ce{NO2}+\ce{F2}\to\ce{NO2F}+\ce{F} (fast step); \ce{NO2}+\ce{F}\to\ce{NO2F} (slow step). (c) \ce{NO2}+\ce{F2}\to\ce{NO2F}+\ce{F} (slow step); \ce{NO2}+\ce{F}\to\ce{NO2F} (fast step). (d) \ce{F2}\to2\ce{F} (slow step); 2\ce{NO2}+2\ce{F}\to2\ce{NO2F} (fast step).

The rate law needs one \ce{NO2} and one \ce{F2} in the slow (rate-determining) step. Test each option by writing the rate law of its slow step.
(a) is a single step with two \ce{NO2} and one \ce{F2}. Its rate would be k[\ce{NO2}]^2[\ce{F2}]. This does not match.
\text{Rate}=k[\ce{NO2}]^{2}[\ce{F2}]
(b) The slow step is \ce{NO2}+\ce{F}. It contains \ce{F}, an intermediate, not \ce{F2}. The rate is not k[\ce{NO2}][\ce{F2}]. This does not match.
\text{Rate}=k[\ce{NO2}][\ce{F}]
(c) The slow step is \ce{NO2}+\ce{F2}, one molecule of each. Its rate is the observed rate law. The fast step uses up the intermediate \ce{F}. Adding the two steps gives 2\ce{NO2}+\ce{F2}\to2\ce{NO2F}. This matches.
\text{Rate}=k[\ce{NO2}][\ce{F2}]
(d) The slow step is \ce{F2}\to2\ce{F}, so the rate would be k[\ce{F2}], with no \ce{NO2}. This does not match.
\text{Rate}=k[\ce{F2}]

\text{Mechanism (c) is consistent with the rate law}

Example 7.7: Experimental rate law from a two-step mechanism

NO reacts with \ce{H2} according to 2\ce{NO}(g)+2\ce{H2}(g)\to\ce{N2}(g)+2\ce{H2O}(g). The mechanism involves two steps: 2\ce{NO}+\ce{H2}\to\ce{N2}+\ce{H2O2} (slow) and \ce{H2O2}+\ce{H2}\to2\ce{H2O} (fast). Write the experimental rate law for this reaction.

The overall rate is set by the slow step, the first one: 2\ce{NO}+\ce{H2}\to\ce{N2}+\ce{H2O2}.
For the rate-determining step the powers equal the coefficients in that step: two molecules of \ce{NO} and one of \ce{H2}. The fast step does not appear in the rate law.
\text{Rate}=k\,[\ce{NO}]^{2}[\ce{H2}]
Check the mechanism: adding the two steps gives 2\ce{NO}+2\ce{H2}\to\ce{N2}+2\ce{H2O}, and \ce{H2O2} cancels, so it is the reaction intermediate.

\text{Rate}=k\,[\ce{NO}]^{2}[\ce{H2}]

Worked example: choosing between two mechanisms

The reaction \ce{A2 + 2B -> 2AB} has the experimental rate law \text{Rate}=k[\ce{A2}]. Two mechanisms are proposed. Mechanism I: \ce{A2 -> 2A} (slow), then \ce{2A + 2B -> 2AB} (fast). Mechanism II: \ce{A2 + B -> AB + A} (slow), then \ce{A + B -> AB} (fast). Which mechanism fits? Name the intermediate and say what happens to the rate if [\ce{B}] is tripled.

Check that each mechanism adds up to the overall equation. Add the steps and cancel the species that appear on both sides.
\begin{aligned}&\text{I: }\ce{A2 + 2A + 2B -> 2A + 2AB}\;\Rightarrow\;\ce{A2 + 2B -> 2AB}\\ &\text{II: }\ce{A2 + 2B + A -> 2AB + A}\;\Rightarrow\;\ce{A2 + 2B -> 2AB}\end{aligned}
Both give the correct overall equation, so the rate law must decide. The rate law follows from the slow (rate-determining) step only.
\text{I: Rate}=k[\ce{A2}]\qquad\text{II: Rate}=k[\ce{A2}][\ce{B}]
Compare with the experiment, \text{Rate}=k[\ce{A2}]. Mechanism II predicts a first order dependence on \ce{B}, which is not observed. Mechanism I matches.
The intermediate is the species made in one step and used up in a later step, and it is absent from the overall equation. In mechanism I it is the atom \ce{A}.
The rate law has no [\ce{B}] term, so \ce{B} is zero order. Tripling [\ce{B}] leaves the rate unchanged, because \ce{B} reacts only in the fast step after the slow step.
\text{Rate}=k[\ce{A2}]\;\Rightarrow\;\text{no change}

\text{Mechanism I; intermediate }\ce{A};\ \text{rate unchanged on tripling }[\ce{B}]

7.4.1Potential energy diagram and reaction mechanismOften examined

A potential energy diagram (reaction path diagram) shows how the energy of the reacting system changes along the path from reactants to products. A reaction that takes place in several steps shows one peak for each elementary step, and the highest peak belongs to the slowest step, the rate-determining step.

Example 7.8: Proposing a mechanism from an energy diagram and a rate law

The potential energy diagram for the reaction between \ce{NO2} and \ce{F2} is shown in Figure 7.3. The experimental rate law for the reaction is \text{Rate}=k[\ce{NO2}][\ce{F2}] for 2\ce{NO2}(g)+\ce{F2}(g)\to2\ce{NO2F}(g). Propose a reaction mechanism.

Figure 7.3: Potential energy diagram for the reaction between \ce{NO2} and \ce{F2}. Step 1, with the higher peak, is the slow rate-determining step.

Step 1: number of elementary steps. The diagram shows two peaks, so the mechanism must involve two elementary steps.
Step 2: the rate-determining step. The activation energy of step 1 is greater than that of step 2, so step 1 is the slow, rate-determining step.
Step 3: molecules in the slow step. The rate law shows that one \ce{NO2} and one \ce{F2} molecule are involved in this step.
Step 4: propose the two steps. The overall reaction needs two \ce{NO2} and one \ce{F2} to give two \ce{NO2F}. In the rate-determining step only one \ce{NO2} and one \ce{F2} need to react, to give one \ce{NO2F} and a reaction intermediate (here the \ce{F} atom). In the second step the intermediate reacts with the other \ce{NO2} molecule to give the second \ce{NO2F}.
\begin{aligned}&\ce{NO2}+\ce{F2}\longrightarrow\ce{NO2F}+\ce{F}\quad\text{(slow)}\\&\ce{NO2}+\ce{F}\longrightarrow\ce{NO2F}\quad\text{(fast)}\end{aligned}
Step 5: add the steps to get the overall reaction, which must be the same as the reaction under consideration. The intermediate \ce{F} appears on both sides and cancels.
\ce{2NO2 + F2 -> 2NO2F}

\ce{NO2 + F2 -> NO2F + F}\ (\text{slow}),\quad \ce{NO2 + F -> NO2F}\ (\text{fast})

Concept assessment exercise 7.3, question 1: H2 and NO at 700 degrees C

The following data were collected for the reaction between \ce{H2} and \ce{NO} at 700^\circ\mathrm{C}: 2\ce{H2}+2\ce{NO}\to2\ce{H2O}+\ce{N2}. Suggest a plausible mechanism that is consistent with the rate law. (Hint: assume the oxygen atom is a reaction intermediate.) The potential energy diagram for this reaction is given in Figure 7.4.

Experiment	[\ce{H2}]	[\ce{NO}]	Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})
1	0.010	0.025	2.4\times10^{-6}
2	0.0050	0.025	1.2\times10^{-6}
3	0.010	0.0125	0.6\times10^{-6}

Experiment	1	2	3
[\ce{H2}]	0.010	0.0050	0.010
[\ce{NO}]	0.025	0.025	0.0125
Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})	2.4\times10^{-6}	1.2\times10^{-6}	0.6\times10^{-6}

Figure 7.4: Potential energy diagram for the reaction between \ce{H2} and \ce{NO}. Two peaks, so two elementary steps, the first (higher) being the slow step.

Find the order in \ce{H2} from experiments 1 and 2. [\ce{NO}] is constant and [\ce{H2}] is halved, from 0.010 to 0.0050. The rate halves, from 2.4\times10^{-6} to 1.2\times10^{-6}. So the reaction is first order in \ce{H2}.
\dfrac{1.2\times10^{-6}}{2.4\times10^{-6}}=\dfrac12=\left(\dfrac12\right)^{x}\;\Rightarrow\;x=1
Find the order in \ce{NO} from experiments 1 and 3. [\ce{H2}] is constant and [\ce{NO}] is halved, from 0.025 to 0.0125. The rate falls to one quarter, from 2.4\times10^{-6} to 0.6\times10^{-6}. So the reaction is second order in \ce{NO}.
\dfrac{0.6\times10^{-6}}{2.4\times10^{-6}}=\dfrac14=\left(\dfrac12\right)^{y}\;\Rightarrow\;y=2
The rate law is therefore third order overall.
\text{Rate}=k\,[\ce{H2}][\ce{NO}]^{2}
The potential energy diagram has two peaks, so there are two elementary steps, and the first peak is the higher one, so the first step is slow. The slow step must contain one \ce{H2} and two \ce{NO} molecules. With an O atom as intermediate, one way to write it is that the first step makes \ce{N2}, one \ce{H2O} and an O atom.
2\ce{NO}+\ce{H2}\longrightarrow\ce{N2}+\ce{H2O}+\ce{O}\quad\text{(slow)}
The O atom (intermediate) is used up in the second, fast step by the remaining \ce{H2}.
\ce{O}+\ce{H2}\longrightarrow\ce{H2O}\quad\text{(fast)}
Check: adding the steps gives 2\ce{NO}+2\ce{H2}\to\ce{N2}+2\ce{H2O}, which is the overall reaction, and the O atom cancels.
2\ce{NO}+2\ce{H2}\longrightarrow\ce{N2}+2\ce{H2O}

\begin{aligned}&\text{Rate}=k[\ce{H2}][\ce{NO}]^{2}\\ &\ce{2NO + H2 -> N2 + H2O + O}\ (\text{slow})\\ &\ce{O + H2 -> H2O}\ (\text{fast})\end{aligned}

Concept assessment exercise 7.3, question 2: Catalyst, intermediate and slow step

The following mechanism has been proposed for a reaction: \ce{H2O2}(aq)+\ce{I-}(aq)\to\ce{H2O}(l)+\ce{OI-}(aq) and \ce{OI-}(aq)+\ce{H2O2}(aq)\to\ce{H2O}(l)+\ce{O2}(g)+\ce{I-}(aq). Choose the catalyst, the reaction intermediate and the rate-determining step.

Add the two steps and cancel what appears on both sides. The overall reaction is 2\ce{H2O2}\to2\ce{H2O}+\ce{O2}.
2\ce{H2O2}\longrightarrow2\ce{H2O}+\ce{O2}
The catalyst is used in an early step and regenerated in a later step, and is absent from the overall equation. \ce{I-} is used in step 1 and re-formed in step 2.
\text{catalyst: }\ce{I-}
The intermediate is formed in one step and used up in a later step. \ce{OI-} is formed in step 1 and consumed in step 2.
\text{intermediate: }\ce{OI-}
The rate-determining step is step 1, \ce{H2O2}+\ce{I-}\to\ce{H2O}+\ce{OI-}. The book does not say which step is slow. Step 1 is taken as the slow step because it is the one that uses the catalyst \ce{I-} together with the reactant \ce{H2O2} (rate =k[\ce{H2O2}][\ce{I-}]), and the ion \ce{OI-} is then used up quickly.

\text{catalyst }\ce{I-};\ \text{intermediate }\ce{OI-};\ \text{slow step: step 1}

Concept assessment exercise 7.3, question 3: Is the mechanism consistent?

The rate law for 2\ce{H2}+2\ce{NO}\to\ce{N2}+2\ce{H2O} is \text{Rate}=k[\ce{H2}][\ce{NO}]^{2}. Is the following mechanism consistent with the rate law? Argue. \ce{H2}+\ce{NO}\to\ce{H2O}+\ce{N} (slow); \ce{N}+\ce{NO}\to\ce{N2}+\ce{O} (fast); \ce{O}+\ce{H2}\to\ce{H2O} (fast).

Add the three steps to see if they give the overall reaction. \ce{N} and \ce{O} cancel.
\ce{H2}+\ce{NO}+\ce{N}+\ce{NO}+\ce{O}+\ce{H2}\to\ce{H2O}+\ce{N}+\ce{N2}+\ce{O}+\ce{H2O}\;\Rightarrow\;2\ce{H2}+2\ce{NO}\to\ce{N2}+2\ce{H2O}
The steps do add up to the overall equation, so the mechanism passes this test.
Now use the slow step. It contains one \ce{H2} and one \ce{NO}, so the rate law it predicts is
\text{Rate}=k[\ce{H2}][\ce{NO}]
The observed rate law has [\ce{NO}]^2, second order in \ce{NO}. The predicted rate law is second order overall, not third. So the mechanism does not agree with the experiment.

\text{Not consistent: the slow step predicts }\text{Rate}=k[\ce{H2}][\ce{NO}]

Concept assessment exercise 7.3, question 4: N2O decomposition

The rate law for 2\ce{N2O}\to2\ce{N2}+\ce{O2} is \text{Rate}=k[\ce{N2O}]. The reaction occurs in two elementary steps. Assume an O atom is a reaction intermediate. Write a mechanism for the reaction.

The rate law is first order in \ce{N2O} only, so the slow step contains one \ce{N2O} molecule. The O atom is formed in this step.
\ce{N2O}\longrightarrow\ce{N2}+\ce{O}\quad\text{(slow)}
The O atom is used up in the second, fast step, by a second \ce{N2O} molecule. This step does not affect the rate law.
\ce{N2O}+\ce{O}\longrightarrow\ce{N2}+\ce{O2}\quad\text{(fast)}
Check: adding the two steps, the O atom cancels and the result is the overall reaction.
2\ce{N2O}\longrightarrow2\ce{N2}+\ce{O2}

\ce{N2O -> N2 + O}\ (\text{slow}),\quad\ce{N2O + O -> N2 + O2}\ (\text{fast})

7.5Activation energyOften examined

Chemical reactions involve the breaking and making of chemical bonds, and these changes are accompanied by changes in energy. Collision theory has been proposed to explain the observed kinetics of reactions. For a chemical reaction to occur, the combining atoms or molecules must collide with one another. These collisions may be effective or ineffective, depending on the energy and orientation of the colliding particles.

An effective collision can take place only if the energy of the colliding particles is high enough to overcome the repulsion between the electrons around the reacting particles. Proper orientation means that at the time of collision the atoms which are required to make new bonds should collide with each other.

Activation energy (E_a): The minimum amount of energy, in addition to the average kinetic energy, which the particles must possess for effective collisions.

No reaction occurs if the energy of the reacting particles is lower than the activation energy. The speed of a reaction therefore depends on its activation energy. The higher the activation energy, the lower the reaction rate, because only a small fraction of the molecules have enough energy to react. If the activation energy is small, a large number of molecules can cause effective collisions, so the reaction rate is higher.

Consider the reaction between \ce{A2} and \ce{B2} molecules to form a new molecule \ce{AB}. If the energy of these molecules is equal to or greater than the activation energy, the collision breaks their bonds and new bonds are formed. In an effective collision the molecules form an unstable species called the activated complex. Being a high-energy species it is short-lived and quickly breaks down into products. The activated complex is also called the transition state (Figure 7.5).

Figure 7.5: Effective collision of molecules. The activated complex (dashed bonds) is the unstable high-energy species that breaks down into the products.

In an efficient collision the colliding molecules move closer to each other and slow down just before the collision. Their kinetic energy decreases, and this leads to a corresponding increase in their potential energy. Activation energy appears in the form of hills between reactants and products. Molecules must first climb over this energy barrier before they can roll down the hill to form products. Only molecular collisions with the right activation energy and orientation can do this. If they lack the right energy they will not reach the top of the hill and will fall back chemically unchanged.

A potential energy diagram can also be used to understand enthalpy changes in chemical reactions. The reaction enthalpy is equal to the difference between the energies of the reactants and the products. In an exothermic reaction the products are at a lower energy level than the reactants. In an endothermic process the products have a higher energy level than the reactants. In both reactions the activation energy (E_a) is an energy barrier that must be overcome before products can be formed. If the activation energy is not available to the reacting particles, the reaction will not start. Endothermic reactions require a constant source of energy to complete the reaction. Figure 7.6 shows the energy profile of exothermic and endothermic reactions.

Figure 7.6: (a) Exothermic reaction, (b) endothermic reaction. E_{a,f} is the forward and E_{a,r} the reverse activation energy.

7.6CatalysisOften examined

Many industrial reactions are carried out at high temperature to maximise the amount of product that can be made in a given time. High temperature reactions bring safety concerns, and many chemical species are not stable at high temperature. A different method of increasing the rates of chemical reactions would therefore be useful. Another way to increase the reaction rate is to change its mechanism in a way that lowers the activation energy of the rate-determining step. This can be done by introducing a catalyst.

Catalyst: A substance which accelerates a chemical reaction but remains chemically unchanged at the end of the reaction.
Catalysis: The phenomenon of a catalyst accelerating a reaction.

A catalyst provides a new mechanism for the reaction with a lower energy of activation (Figure 7.7). The catalyst therefore increases the rate of a reaction by decreasing its energy of activation. A catalyst has no effect on the total thermodynamics, or the enthalpy, of the reaction. For this reason a catalyst cannot be used to bring about a chemical reaction which is not favoured thermodynamically.

Figure 7.7: Effect of a catalyst on the energy of activation. The catalysed path has a lower peak, but the reactants and products (and so \Delta H) are the same.

Example: the ozone layer

In the stratosphere the conversion of an ozone molecule by an oxygen atom into two \ce{O2} molecules occurs. This reaction has a high activation energy:

\ce{O3 + O -> 2O2}\qquad E_a=17.1\,\mathrm{kJ\,mol^{-1}}

Chlorofluorocarbons diffuse up into the stratosphere. These compounds absorb short wavelength ultraviolet light from the sun, which breaks carbon-chlorine bonds and produces chlorine atoms. Cl atoms catalyse the reaction by a mechanism that needs less activation energy:

\begin{aligned}\ce{O3 + Cl &-> O2 + ClO}&&E_a=2.1\,\mathrm{kJ\,mol^{-1}}\\\ce{O + ClO &-> O2 + Cl}&&E_a=0.4\,\mathrm{kJ\,mol^{-1}}\\\hline\text{Net reaction: }\ce{O3 + O &-> 2O2}&&E_a=2.5\,\mathrm{kJ\,mol^{-1}}\end{aligned}

Path	Activation energy (\mathrm{kJ\,mol^{-1}})
Direct: \ce{O3 + O -> 2O2}	17.1
Catalysed by Cl atoms (net)	2.5 (steps: 2.1 and 0.4)

The direct reaction between \ce{O3} and \ce{O} has a much higher activation energy than the chlorine-catalysed reaction.

Worked example: drawing the effect of a catalyst

A reaction has E_a=120\,\mathrm{kJ\,mol^{-1}} and \Delta H=-45\,\mathrm{kJ\,mol^{-1}}. A catalyst gives a new path with E_a=70\,\mathrm{kJ\,mol^{-1}}. (a) Draw both energy profiles. (b) State \Delta H for the catalysed reaction. (c) By what factor does the catalyst increase k at 300\,\mathrm{K}? Take A as unchanged.

(a) Draw the energy axis and the reaction path axis. Put the reactants at one level. The products are lower by 45\,\mathrm{kJ} because the reaction is exothermic. Draw one hill whose top is 120\,\mathrm{kJ} above the reactants, then a lower hill of 70\,\mathrm{kJ} between the same start and end levels (Figure below).

The catalysed path (accent) has a lower peak. Reactants and products are at the same levels in both paths, so \Delta H is the same.

(b) The start and end levels do not change, and \Delta H is the difference between them. A catalyst changes only the path, not the energy of reactants and products.
\Delta H=-45\,\mathrm{kJ\,mol^{-1}}\ \text{(unchanged)}
(c) Write the Arrhenius equation for each path and divide. A cancels. Use E_a in joules.
\dfrac{k_{\text{cat}}}{k_{\text{uncat}}}=\dfrac{e^{-70\,000/RT}}{e^{-120\,000/RT}}=e^{50\,000/RT}
Evaluate at T=300\,\mathrm{K} with R=8.3143\,\mathrm{J\,K^{-1}\,mol^{-1}}.
\dfrac{50\,000}{8.3143\times300}=20.05\;\Rightarrow\;e^{20.05}\approx5\times10^{8}
A 50\,\mathrm{kJ\,mol^{-1}} lower barrier makes the reaction about five hundred million times faster at this temperature, without any change in the energy released.

\Delta H=-45\,\mathrm{kJ\,mol^{-1}},\quad k_{\text{cat}}/k_{\text{uncat}}\approx5\times10^{8}

7.7Gibbs free energy change and feasibility of a reaction

The free energy change, \Delta G^\circ, is a measure used to predict whether a reaction is feasible. A feasible reaction is one which, once started, will go to completion without any energy being supplied to it.

If \Delta G^\circ is negative or equal to zero, the reaction is feasible.
If \Delta G^\circ is positive, the reaction is not feasible.

\Delta G^\circ	Feasibility
negative (or zero)	feasible
positive	not feasible

Formula sheet

Average rate of reaction

\text{Rate}=\dfrac{\text{change in concentration}}{\text{time taken}}

Change in concentration of a reactant (decrease) or of a product (increase) over a time interval.

Unit: \mathrm{mol\,dm^{-3}\,s^{-1}}

Rate in terms of reactant and product

\text{Rate}=\dfrac{dx}{dt}=-\dfrac{d[\ce{A}]}{dt}=+\dfrac{d[\ce{B}]}{dt}

For \ce{A -> B}: d[\ce{A}] and d[\ce{B}] are the small changes in concentration in the small time dt. The minus sign is for the reactant that is used up.

Unit: \mathrm{mol\,dm^{-3}\,s^{-1}}

Rate law (one reactant)

\text{Rate}=k\,[\ce{A}]^{x}

k is the rate constant and x the order with respect to A. When [\ce{A}]=1\,\mathrm{M}, the rate equals k.

Unit: depends on the order of the reaction

Rate law (general reaction)

\text{Rate}=k\,[\ce{A}]^{x}[\ce{B}]^{y},\qquad\text{overall order}=x+y

x and y are the orders with respect to A and B, found by experiment (not from the balanced equation).

Unit: depends on the overall order

Half-life, zero order

t_{1/2}=\dfrac{[\ce{A}]_0}{2k}

[\ce{A}]_0 is the initial concentration of the reactant.

Unit: time (s)

Half-life, first order

t_{1/2}=\dfrac{0.693}{k}

Independent of the initial concentration.

Unit: time (s)

Half-life, second order

t_{1/2}=\dfrac{1}{k\,[\ce{A}]_0}

[\ce{A}]_0 is the initial concentration of the reactant.

Unit: time (s)

Arrhenius equation

k=A\,e^{-E_a/RT}

k rate constant, A Arrhenius constant, E_a activation energy, R=8.3143\,\mathrm{J\,K^{-1}\,mol^{-1}}, T absolute temperature.

k has the unit of the rate constant; E_a in \mathrm{J\,mol^{-1}}; T in \mathrm{K}

Key terms

Chemical kinetics: The study of the rates, mechanisms and factors that affect the rate of chemical reactions.
Mechanism: The path (the sequence of all the chemical steps) by which the reactants form the products in a chemical reaction.
Rate of reaction: The change in concentration of reactants or products per unit time. It is the decrease in concentration of reactants per unit time or the increase in concentration of products per unit time.
Instantaneous rate: The rate of a reaction over an infinitesimally small period of time, dx/dt, at a given time.
Rate law (rate equation): An expression which shows how the reaction rate is related to the concentration of the reactants, each raised to a power found by experiment.
Rate constant: The proportionality constant k in the rate law. It is the rate of the reaction when the molar concentration of each reactant is unity.
Order of reaction: The sum of all the exponents to which the molar concentration terms in the rate equation are raised. The book also defines it as the number of molecules of reactants participating in the rate-determining step.
Zero order reaction: A reaction whose rate is independent of the concentration of the reactant molecules.
First order reaction: A reaction whose rate is directly proportional to the first power of the concentration of a single reactant.
Pseudo first order reaction: A bimolecular reaction in which the solvent is in excess and its concentration remains constant, so it does not take part in the rate-determining step.
Method of initial rates: Finding the order by changing the concentration of one reactant at a time, keeping everything else constant, and comparing the initial reaction rates.
Half-life (t_{1/2}): The time taken for the concentration of a reactant to be reduced to half.
Collision theory: The rate of a reaction is proportional to the number of collisions between the reacting molecules, and only effective collisions (with activation energy and correct orientation) lead to a reaction.
Effective collision: A collision in which the molecules have at least the activation energy and the correct orientation, so that a reaction takes place.
Maxwell-Boltzmann distribution curve: A curve showing how the available kinetic energy is distributed among the molecules of a gas at a constant temperature.
Arrhenius equation: k=Ae^{-E_a/RT}, the equation that gives the effect of temperature on the rate constant.
Arrhenius constant (A): A constant in the Arrhenius equation, related to the frequency of collisions and the orientation of the reacting molecules.
Activation energy (E_a): The minimum amount of energy, in addition to the average kinetic energy, that the particles must possess for effective collisions.
Activated complex (transition state): The unstable, short-lived, high-energy species formed in an effective collision, which quickly breaks down into the products.
Rate-determining step: The slowest step in the reaction mechanism, which determines the overall rate of the reaction.
Reaction intermediate: A species produced in one step of a reaction and consumed in another step; it does not appear in the overall reaction.
Catalyst: A substance that accelerates a chemical reaction but remains chemically unchanged at the end of it, by providing a new mechanism with lower activation energy.
Homogeneous catalysis: Catalysis in which the catalyst and the reactants are in the same phase.
Heterogeneous catalysis: Catalysis in which the catalyst and the reactants are in different phases.
Enzymes: Catalysts in living organisms.
Feasible reaction: A reaction which, once started, goes to completion without any energy being supplied to it. It has \Delta G^\circ negative or zero.

Multiple choice questionsp. 161–163

The rate of a reaction \_\_\_\_\_\_ as the reaction proceeds.

AIncreases
BDecreasesAnswer
CRemains the same
DMay increase or decrease

Why

Reactant molecules are used up, so their concentrations fall. The rate depends on these concentrations, so the rate falls as well. The book says the rate decreases continuously during a reaction (Figure 7.1).

(c) is wrong because the rate is never constant during a reaction, and (d) is wrong because the rate falls steadily rather than changing unpredictably.

The unit of the rate constant is the same as that of the rate of reaction in \_\_\_\_\_\_ order reaction.

AFirst
BSecond
CThird
DZeroAnswer

Why

For a zero order reaction the rate law is rate =k, so k and the rate have the same units, \mathrm{mol\,dm^{-3}\,s^{-1}}.

For the other orders the units differ: first order k is \mathrm{s^{-1}}, second order \mathrm{dm^3\,mol^{-1}\,s^{-1}} and third order \mathrm{dm^6\,mol^{-2}\,s^{-1}}.

iii2026 Annual I, Set D, MCQ 3

For the reaction \ce{2A + B ->[slow] C}, the rate law for the reaction is

Arate =k[\mathrm{A}]^2[\mathrm{B}]Answer
Brate =k[\mathrm{A}][\mathrm{B}]
Crate =k[\mathrm{C}]
DNone of these

Why

The single slow step is the rate determining step, and for a single step the rate law follows its molecularity: each reactant appears with its stoichiometric coefficient. \ce{2A} gives [\mathrm{A}]^2 and \ce{B} gives [\mathrm{B}]^1.

(b) uses the wrong power of A, and (c) wrongly uses the product C, which does not appear in the rate law of a reactant-driven step.

For the reaction \ce{2A + B -> C + D}, the rate law is rate =k[\mathrm{A}]^2. The order of the reaction in B is

AFirst
BSecond
CThird
DZeroAnswer

Why

B does not appear in the rate law, so the rate does not depend on [\mathrm{B}]. A reactant that does not appear in the rate law has order 0 with respect to it. The coefficient of B in the equation does not give its order.

The overall order is 2+0=2. The book prints the options as "(b) (c) Third"; the third option is read as (c) Third.

The activation energy for a reaction can be

AIncreased by increasing temperature
BIncreased by decreasing temperature
CDecreased by increasing concentration of reactants
DNone of theseAnswer

Why

The activation energy E_a is a fixed property of a reaction pathway. Temperature and concentration change how many molecules have energy at least equal to E_a, but they do not change E_a itself. So (a), (b) and (c) are all wrong.

Only a catalyst changes E_a, by providing a new mechanism with a lower activation energy.

Rate law for the reaction \ce{R-X + H2O -> R-OH + HX} is rate =k[\mathrm{R{-}X}]. The rate of reaction will be doubled when

AConcentration of \ce{H2O} is doubled
BConcentration of R-X is reduced to half
CConcentration of R-X is doubledAnswer
DNone of these

Why

The rate is first order in R-X, so rate \propto [\mathrm{R{-}X}]. Doubling [\mathrm{R{-}X}] doubles the rate, which makes (c) correct.

Water does not appear in the rate law, so doubling its concentration leaves the rate unchanged (a). Halving R-X halves the rate (b).

vii

The rate of an enzyme catalysed reaction is independent of

AConcentration of substrateAnswer
BConcentration of product
CEnzyme
DTemperature

Why

The book states that reactions catalysed by enzymes also follow zero order kinetics. At high substrate concentration the enzyme is saturated, so the rate is fixed by the amount of enzyme and does not depend on the substrate concentration.

The rate does depend on the enzyme (c) and on temperature (d), which affects the rate constant.

Note. The book prints no key for this item. (a) is chosen from the book's zero order statement for enzyme reactions.

viii

If a reaction proceeds in such a way that the order of reaction is independent of the reactants concentration, the overall order would be

AFirst
BSecond
CThird
DZeroAnswer

Why

A reaction whose order does not depend on reactant concentration is a zero order reaction (book section 7.2.1). The overall order is the sum of the exponents in the rate law, which here is 0.

Reactions with high activation energy are usually

AFast
BSlowAnswer
CExothermic
DReversible

Why

At a given temperature only a small fraction of molecules have energy at least equal to a high E_a, so there are few effective collisions and the reaction is slow. Being exothermic or reversible is a property of \Delta H or of the equilibrium, not of E_a.

In a reversible reaction a catalyst lowers the activation energy of the

AForward reaction
BReverse reaction
CForward as well as reverse reactionAnswer
DForward reaction but increases for the reverse reaction

Why

A catalyst provides a new mechanism with a lower activation energy for the same pathway, and this applies in both directions. Because it does not change \Delta H, it lowers E_a of the forward and the reverse reaction equally, so (a), (b) and (d) are wrong.

Short questionsp. 163–164

23 marks

What is chemical kinetics? How do you differentiate chemical kinetics from chemical equilibrium?

Solution

Chemical kinetics is the study of the rates of chemical reactions, the factors that affect them and the mechanisms by which reactions occur. Rate information is what is used to infer the mechanism of a reaction.

Chemical equilibrium describes the state of a reversible reaction in which the forward and reverse rates are equal, so the concentrations no longer change. The two topics answer different questions:

Kinetics asks how fast a reaction goes and by which steps it goes. It uses rate, time and mechanism.
Equilibrium asks how far a reaction goes and what the final composition is. It uses the concentrations at equilibrium and says nothing about time.

33 marks

Explain effects of concentration, temperature and surface area on reaction rates.

Solution

Concentration: a higher concentration of a reactant increases the rate, because the rate law is rate =k[\mathrm{A}]^x. More molecules per unit volume means more collisions per second, so the rate rises with [\mathrm{A}].

Temperature: a rise in temperature increases the average kinetic energy, so molecules move faster and collide more often. A larger fraction of molecules also has energy at least equal to E_a (the Maxwell-Boltzmann curve). The book notes that the rate roughly doubles or triples for every 10\,\mathrm{K} rise. The Arrhenius equation k=Ae^{-E_a/RT} shows the same effect on the rate constant.

Surface area: a solid reactant reacts at its surface. Powdering it increases the exposed surface, so more collisions take place per second with the other reactant and the rate increases (see question 11).

43 marks

Evaluate that increase in collision energy by increasing the temperature can improve the collision frequency.

Solution

The statement is partly correct. Raising the temperature increases the average kinetic energy of the molecules, so they move faster and collide more often. The collision frequency therefore does increase.

The main effect, however, is on the energy of the molecules rather than on the collisions themselves. Only collisions with energy at least equal to E_a are effective. A rise in temperature shifts the Maxwell-Boltzmann curve, so a much larger fraction of molecules have E\geq E_a, and this is the larger cause of the faster reaction.

Conclusion: increasing the temperature improves the collision frequency and the proportion of effective collisions, so the rate increases. The statement is accepted in part, but the energy of the molecules, not just the collision frequency, is the decisive factor.

52025 Annual I, Set D, Q.2 (xii) OR3 marks

Draw energy diagrams that represent the activation energy and show the effect of a catalyst.

Solution

The energy profile shows the energy of the reactants, the activation energy barrier and the energy of the products. The height of the peak above the reactants is the activation energy E_a. An exothermic reaction is drawn with products below the reactants, as in Figure 7.6 (a).

Figure 7.7: Energy profile with and without a catalyst. The catalysed curve (accent) has a lower peak, so its E_a is smaller. The reactants and products are at the same energy in both curves.

The catalyst does not change the energies of the reactants or the products, so \Delta H is the same with and without it. Only the height of the barrier changes.

62025 Annual I, Set D, Q.2 (xii) OR3 marks

What is the effect of a catalyst on the following: the rate of reaction; the energy of activation?

Solution

Rate of reaction: a catalyst increases the rate. It provides a new mechanism with a lower activation energy, so a larger fraction of molecules has enough energy to react and the rate constant increases. The catalyst itself is chemically unchanged at the end of the reaction.

Energy of activation: a catalyst lowers the energy of activation of the rate determining step (Figure 7.7). It does not change the enthalpy of the reaction, so it cannot make an unfavourable reaction feasible.

72 marks

The reaction of an alkyl halide, R-X with water is as follows: \ce{R-X + H2O -> R-OH + HX}. If the reaction were a single step process, what would you predict the rate law to be?

Solution

In a single (elementary) step the order with respect to each reactant equals its stoichiometric coefficient, and both R-X and water take part. The predicted rate law is:

\text{rate}=k[\ce{R-X}][\ce{H2O}]

Water is normally present in large excess, so its concentration hardly changes. The book (section 7.2.1 f) calls such a bimolecular reaction pseudo first order, with rate =k[\ce{R-X}]. The prediction above applies to a single step.

82 marks

The reaction of a compound A and B to give C and D was found to be second order in A and second order overall. Write rate expression for the reaction.

Solution

The order in A is 2. The overall order is 2, so the order in B is 2-2=0. B therefore does not appear in the rate expression.

\text{rate}=k[\mathrm{A}]^2[\mathrm{B}]^0=k[\mathrm{A}]^2

93 marks

Defend the following statements.

Solution

The three statements are defended in parts (a) to (c).

(a) A very small amount of catalyst may prove sufficient to carry out a reaction.

A catalyst is not used up. It takes part in the mechanism and is regenerated at the end of each cycle, so a small amount can convert a large quantity of reactants. Each catalyst site can carry out the reaction many times, which is why a tiny amount is enough.

(b) The reaction rate decreases every moment.

The reaction rate depends on the concentrations of the reactants. As the reaction proceeds these concentrations fall continuously, so the rate falls continuously (Figure 7.1). The book notes that the rate is never constant during a reaction. The statement is true for reactions whose rate depends on reactant concentration, but a zero order reaction has a constant rate until its reactant is used up.

For a second order reaction, rate =k[\mathrm{A}]^2. Rearranging gives k=\dfrac{\text{rate}}{[\mathrm{A}]^2}.

k=\dfrac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{\mathrm{mol^2\,dm^{-6}}}=\mathrm{mol^{-1}\,dm^{3}\,s^{-1}}=\mathrm{dm^3\,mol^{-1}\,s^{-1}}

The statement is correct.

113 marks

Explain why powdered Zn reacts faster with an acid than a piece of Zn.

Solution

Powdered zinc has a much larger surface area exposed to the acid, so the acid molecules have many more places to collide with zinc atoms at the same time. The rate of reaction depends on the frequency of effective collisions, so the collision frequency per second is higher and the reaction is faster.

Both samples have the same amount of zinc and acid and the same temperature. Only the surface area differs, so the difference in rate comes from the surface area alone.

123 marks

Explore the impact of catalysts on the energetics of chemical reactions.

Solution

A catalyst lowers the activation energy of the reaction by providing an alternative mechanism. A lower energy barrier means more molecules have enough energy to react at a given temperature, so the reaction is faster and can be carried out at a lower temperature. Energy is saved, which is why many industrial reactions use catalysts.

A catalyst does not change the energies of the reactants and products. The enthalpy change \Delta H, and so the free energy change \Delta G^\circ, is unaffected. A reaction with positive \Delta G^\circ cannot be made feasible by a catalyst, because the catalyst does not change the thermodynamics. The book gives the example of the ozone reaction, where chlorine atoms catalyse a route with a lower activation energy.

133 marks

How does the concept of activation energy relate to speed of a chemical reaction?

Solution

Activation energy is the minimum energy that colliding particles must have, in addition to their average kinetic energy, for an effective collision. Only a small fraction of molecules have this energy at a given temperature.

A high activation energy means only a small fraction of collisions are effective, so the reaction is slow. A low activation energy means a large fraction of molecules can react, so the reaction is fast. The rate constant depends on E_a through k=Ae^{-E_a/RT}.

Numerical problemsp. 164–165

104 marks

For the reaction \ce{A + B -> AB}, the following data were obtained (the table columns read as [\mathrm{A}], [\mathrm{B}] and initial rate). Illustrate the rate equation for the reaction.

Experiment	[A] (mol dm^{-3})	[B] (mol dm^{-3})	Initial rate (mol dm^{-3} s^{-1})
1	0.10	0.01	1.00\times10^{-5}
2	0.10	0.02	2.00\times10^{-5}
3	0.20	0.01	2.00\times10^{-5}
4	0.30	0.02	6.00\times10^{-5}

Solution

Given

r_1 = 1.00\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}} at [\mathrm{A}]=0.10 and [\mathrm{B}]=0.01\,\mathrm{mol\,dm^{-3}}

To find

the rate equation and the rate constant k

Order with respect to B. Keep [\mathrm{A}] constant (experiments 1 and 2). Doubling [\mathrm{B}] from 0.01 to 0.02 doubles the rate from 1.00\times10^{-5} to 2.00\times10^{-5}, so the rate is directly proportional to [\mathrm{B}].
\frac{r_2}{r_1}=\frac{2.00\times10^{-5}}{1.00\times10^{-5}}=2=\left(\frac{0.02}{0.01}\right)^1 \;\Rightarrow\; \text{order in B}=1
Order with respect to A. Keep [\mathrm{B}] constant (experiments 1 and 3). Doubling [\mathrm{A}] from 0.10 to 0.20 doubles the rate, so the order in A is also 1.
\frac{r_3}{r_1}=\frac{2.00\times10^{-5}}{1.00\times10^{-5}}=2=\left(\frac{0.20}{0.10}\right)^1 \;\Rightarrow\; \text{order in A}=1
Rate equation from the orders.
\text{rate}=k[\mathrm{A}][\mathrm{B}]
Rate constant from experiment 1.
k=\frac{r_1}{[\mathrm{A}][\mathrm{B}]}=\frac{1.00\times10^{-5}}{0.10\times0.01}=1.0\times10^{-2}\,\mathrm{mol^{-1}\,dm^{3}\,s^{-1}}
Check with experiment 4.
k[\mathrm{A}][\mathrm{B}]=1.0\times10^{-2}\times0.30\times0.02=6.0\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

\text{rate}=k[\mathrm{A}][\mathrm{B}],\qquad k=1.0\times10^{-2}\,\mathrm{mol^{-1}\,dm^{3}\,s^{-1}}

Book answer. Rate \propto [A][B] (the book gives no value of k).

Calculate the free energy change for the following reaction at 298\,\mathrm{K}: \ce{MgCO3(s) -> MgO(s) + CO2(g)}, with \Delta H^\circ=+117\,\mathrm{kJ\,mol^{-1}} and \Delta S^\circ=+0.175\,\mathrm{kJ\,K^{-1}\,mol^{-1}}.

Solution

The free energy change is found from \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ and tells us whether the reaction is feasible.

a. Is this reaction feasible at this temperature?

Given

\Delta H^\circ = +117\,\mathrm{kJ\,mol^{-1}}
\Delta S^\circ = +0.175\,\mathrm{kJ\,K^{-1}\,mol^{-1}}
T = 298\,\mathrm{K}

To find

\Delta G^\circ and feasibility

Use the relation between the free energy change, enthalpy change and entropy change (section 6.7 of the book).
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
Substitute the values. T\Delta S^\circ is in \mathrm{kJ\,mol^{-1}}, the same unit as \Delta H^\circ.
\Delta G^\circ=117-298\times0.175=117-52.15=64.85\,\mathrm{kJ\,mol^{-1}}
A positive \Delta G^\circ means the reaction is not feasible.
\Delta G^\circ\approx+64.9\,\mathrm{kJ\,mol^{-1}}>0

\Delta G^\circ\approx+64.9\,\mathrm{kJ\,mol^{-1}}: \text{not feasible at } 298\,\mathrm{K}

b. Define Gibbs free energy.

Gibbs free energy, G, is a thermodynamic quantity defined as G=H-TS. Its change \Delta G^\circ at constant temperature and pressure tells whether a reaction is feasible: a negative or zero value means feasible, and a positive value means not feasible.

c. Define enthalpy.

Enthalpy, H, is the heat content of a system at constant pressure. The enthalpy change \Delta H is the heat absorbed or evolved in a reaction at constant pressure. A positive \Delta H means an endothermic reaction (here MgCO_3 absorbs heat to decompose).

Note. The book prints no answer. The entropy change is printed as 0.175\,\mathrm{kJ\,K^{-1}\,mol^{-1}} (the printed text is partly covered by bold overlay). The reaction would become feasible only above T=\Delta H^\circ/\Delta S^\circ=117/0.175\approx669\,\mathrm{K}, where \Delta G^\circ becomes zero.

Consider the following reaction: \ce{2H2(g) + 2NO(g) -> 2H2O(g) + N2(g)}. This reaction is first order with respect to \ce{H2} and second order with respect to NO.

Solution

a. Write the rate expression for this reaction.

The order is 1 in \ce{H2} and 2 in NO, so the rate law is:

\text{rate}=k[\ce{H2}][\mathrm{NO}]^2

b. Predict the overall order of this reaction.

The overall order is the sum of the exponents: 1+2=3. The reaction is third order overall.

c. How would you expect the rate of this reaction to change if the concentration of NO is doubled?

The rate depends on [\mathrm{NO}]^2. Doubling [\mathrm{NO}] multiplies the rate by 2^2=4, so the rate becomes four times larger.

d. At a certain temperature k was found to be 220\,\mathrm{mol^{-2}\,dm^{6}\,s^{-1}} when [\ce{H2}]=[\mathrm{NO}]=1.54\times10^{-3}\,\mathrm{mol\,dm^{-3}}. Calculate the rate of reaction at this temperature.

Given

k = 220\,\mathrm{mol^{-2}\,dm^{6}\,s^{-1}}
[\ce{H2}] = 1.54\times10^{-3}\,\mathrm{mol\,dm^{-3}}
[\mathrm{NO}] = 1.54\times10^{-3}\,\mathrm{mol\,dm^{-3}}

To find

the rate r

Substitute into the rate expression from part (a).
r=k[\ce{H2}][\mathrm{NO}]^2
Cube the concentration (one power of \ce{H2} and two powers of NO give three powers in all).
r=220\times(1.54\times10^{-3})^3=220\times3.652\times10^{-9}
Multiply to get the rate.
r=8.03\times10^{-7}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

r\approx8.03\times10^{-7}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 21 mark

The area under a Boltzmann distribution curve represents:

ATemperature
BPressure
CTotal number of moleculesAnswer
DActivation energy

Why

The Maxwell-Boltzmann (Boltzmann distribution) curve plots the number of molecules against their kinetic energy. Every molecule of the sample lies somewhere under the curve, so the total area under the curve is the total number of molecules. This area stays the same when the temperature changes; only the shape of the curve changes (Figure 7.2b of the book).

Temperature (A) is fixed for one curve and appears as the label of the curve, not as its area. Pressure (B) is not plotted. Activation energy (D) is a position on the energy axis, and only the small shaded area beyond E_a gives the molecules able to react, not the whole area.

2026 Annual I, Set B, MCQ 31 mark

Which factor does NOT affect rate of reaction?

ATemperature
BConcentration
CCatalyst
DColor of reactantsAnswer

Why

The rate of a reaction depends on the temperature (A: more molecules with energy at least E_a), on the concentration of the reactants (B: more collisions per second) and on a catalyst (C: it lowers E_a). The colour of the reactants is only a physical property and has nothing to do with collision frequency, energy or orientation, so it does not affect the rate.

The surface area of a solid reactant is another factor (powdered zinc reacts faster than a lump), but colour is not.

2025 Annual I, Set B, MCQ 161 mark

Identify correct relationship for determination of half-life of a first order reaction:

At_{1/2} = 0.693/kAnswer
Bt_{1/2} = [A^{\circ}]/2k
Ct_{1/2} = 1/k[A^{\circ}]
Dt_{1/2} = 0.023\,[A^{\circ}]/k

Why

For a first order reaction the half-life is independent of the initial concentration and is t_{1/2}=0.693/k (book section 7.2.3, \ln2=0.693).

Option B, t_{1/2}=[A^\circ]/2k, is the zero order half-life. Option C, t_{1/2}=1/k[A^\circ], is the second order half-life. Option D contains the initial concentration, but a first order half-life does not depend on it, and its numerical factor is not 0.693.

2025 Annual I, Set B, Q.2 (xiv) OR3 marks

For the reaction \ce{N2O4 -> 2NO2}, the concentration of \ce{N2O4} is reduced from 0.1 M to 0.05 M in 60 sec. If reaction is 1st order, calculate rate constant for this reaction.

Solution

Given

\ce{N2O4}\to2\ce{NO2} = first order
[\ce{N2O4}]_0 = 0.1\,\mathrm{M}
[\ce{N2O4}] = 0.05\,\mathrm{M} after 60\,\mathrm{s}

To find

rate constant k

The concentration falls from 0.1\,\mathrm{M} to 0.05\,\mathrm{M}, exactly half. So 60\,\mathrm{s} is one half-life.
t_{1/2}=60\,\mathrm{s}
For a first order reaction t_{1/2}=0.693/k. Rearrange for k.
k=\dfrac{0.693}{t_{1/2}}
Substitute the half-life.
k=\dfrac{0.693}{60\,\mathrm{s}}=0.01155\,\mathrm{s^{-1}}

The unit of a first order rate constant is \mathrm{s^{-1}}.

k=1.16\times10^{-2}\,\mathrm{s^{-1}}

2026 Annual I, Set B, Q.36 marks

Phosgene is a toxic gas. This gas is prepared by the reaction of carbon monoxide with chlorine. \ce{CO(g) + Cl2(g) -> COCl2(g)} The following data were obtained for kinetic study of this reaction (table below). (a) Calculate the numerical value of the rate constant (k) for this reaction using initial rates. (b) Using half-life method calculate t_{1/2} of reaction.

Experiment	Initial [\ce{CO}]	Initial [\ce{Cl2}]	Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})
1	1.000	0.100	1.29\times10^{-29}
2	0.100	0.100	1.30\times10^{-30}
3	0.100	1.000	1.30\times10^{-30}

Solution

(a) Rate constant k from the initial rates.

Find the order in \ce{CO} from experiments 1 and 2. [\ce{Cl2}] is constant at 0.100 while [\ce{CO}] is ten times larger (0.100\to1.000). The rate rises from 1.30\times10^{-30} to 1.29\times10^{-29}, about ten times.
\dfrac{\text{rate}_1}{\text{rate}_2}=\dfrac{1.29\times10^{-29}}{1.30\times10^{-30}}\approx10=10^{x}\;\Rightarrow\;x=1
Find the order in \ce{Cl2} from experiments 2 and 3. [\ce{CO}] is constant at 0.100 while [\ce{Cl2}] is increased ten times (0.100\to1.000). The rate does not change.
\dfrac{\text{rate}_3}{\text{rate}_2}=1=10^{y}\;\Rightarrow\;y=0
Write the rate law. The reaction is first order in \ce{CO} and zero order in \ce{Cl2}, so it is first order overall.
\text{Rate}=k\,[\ce{CO}]
Put the data of experiment 1 into the rate law and solve for k.
k=\dfrac{\text{Rate}}{[\ce{CO}]}=\dfrac{1.29\times10^{-29}\,\mathrm{mol\,dm^{-3}\,s^{-1}}}{1.000\,\mathrm{mol\,dm^{-3}}}=1.29\times10^{-29}\,\mathrm{s^{-1}}
Check with experiment 2. The same value (within the rounding of the printed data) is obtained.
k=\dfrac{1.30\times10^{-30}}{0.100}=1.30\times10^{-29}\,\mathrm{s^{-1}}

The unit of k for a first order reaction is \mathrm{s^{-1}}.

k\approx1.3\times10^{-29}\,\mathrm{s^{-1}}

(b) Half-life of the reaction.

The reaction is first order, so its half-life does not depend on the concentration and is given by t_{1/2}=0.693/k.

Write the half-life formula for a first order reaction.
t_{1/2}=\dfrac{0.693}{k}
Substitute k=1.29\times10^{-29}\,\mathrm{s^{-1}}.
t_{1/2}=\dfrac{0.693}{1.29\times10^{-29}\,\mathrm{s^{-1}}}=5.37\times10^{28}\,\mathrm{s}

This is an astronomically long time (about 1.7\times10^{21} years). The printed rates are extremely small, so under these conditions the reaction is practically not taking place.

t_{1/2}\approx5.4\times10^{28}\,\mathrm{s}

Note. The rates are printed as 10^{-29} and 10^{-30} in the paper (the book's own version of this data); the ratio 1.29\times10^{-29}/1.30\times10^{-30}=9.9 is rounded to 10.

2025 Annual I, Set B, Q.6 OR7 marks

Explain the occurrence of a chemical reaction with reference to molecular collisions, activation energy, formation of activated complex and enthalpy changes in the reaction.

Solution

Molecular collisions

For a chemical reaction to occur, the reacting atoms or molecules must first collide. Not every collision gives products. A collision is effective only if (i) the colliding particles have enough energy to overcome the repulsion between their electron clouds and (ii) they have the proper orientation, so that the atoms which must form the new bonds meet each other. Most collisions are ineffective, and the particles simply bounce apart unchanged.

Activation energy

The activation energy E_a is the minimum amount of energy, in addition to the average kinetic energy, which the particles must possess for an effective collision. It acts as an energy barrier (a hill) between reactants and products. As the molecules approach, they slow down, their kinetic energy falls and their potential energy rises. Only molecules with energy at least equal to E_a can climb the hill; the others fall back unchanged. The higher the activation energy, the smaller the fraction of effective collisions and the slower the reaction.

Formation of the activated complex

In an effective collision of, for example, \ce{A2} and \ce{B2}, the old bonds begin to break while the new bonds begin to form. An unstable, high-energy species called the activated complex (transition state) is formed at the top of the energy hill. It is short-lived and at once breaks down into the products \ce{2AB}.

Effective collision of molecules: the activated complex (dashed bonds) is the unstable high-energy species that breaks down into the products.

Enthalpy changes in the reaction

A potential energy diagram shows the energy of the system along the reaction path. The reaction enthalpy is the difference between the energy of the products and that of the reactants, \Delta H=H_\text{products}-H_\text{reactants}.

Exothermic reaction: the products are at a lower energy level than the reactants, so energy is released and \Delta H is negative.
Endothermic reaction: the products are at a higher energy level than the reactants, so energy is absorbed and \Delta H is positive.

Energy profile of (a) an exothermic and (b) an endothermic reaction. E_{a,f} is the forward and E_{a,r} the reverse activation energy.

In both cases the activation energy is a barrier that must be overcome before the products can form. If the reacting particles do not have it, the reaction does not start. The height of the barrier decides how fast the reaction goes, while \Delta H tells how much heat is given out or taken in.

2026 Annual I, Set D, MCQ 21 mark

Increase in temperature causes molecules to:

ALose energy
BStay same
CGain kinetic energyAnswer
DStop moving

Why

Temperature is a measure of the average kinetic energy of the molecules. Raising the temperature increases the average kinetic energy, so the molecules gain kinetic energy and move faster. This is why more molecules reach the activation energy and the reaction rate rises (Maxwell-Boltzmann curve, section 7.3).

They do not lose energy (A) or stop moving (D), and they certainly do not stay the same (B), since a hotter sample has molecules with a higher average speed.

2026 Annual I, Set D, MCQ 31 mark

For the reaction 2A + B \xrightarrow{slow} C, the rate law for the reaction is:

A\text{rate}=k[A]^2[B]Answer
B\text{rate}=k[A][B]
C\text{rate}=k[C]
D\text{rate}=k[A][B]^2

Why

The slow step is the rate-determining step. Here the equation is given as one slow step, \ce{2A + B -> C}, so the powers in the rate law equal the coefficients of this step: [\ce{A}]^2 for \ce{2A} and [\ce{B}]^1 for \ce{B}. Hence \text{rate}=k[\ce{A}]^2[\ce{B}].

Option B uses the wrong power of A. Option D swaps the powers of A and B. Option C uses the product \ce{C}, which is not a reactant and does not appear in the rate law of the slow step.

2025 Annual I, Set D, MCQ 61 mark

Correct rate equation for given reaction is: \ce{H2 + Cl2 ->[sunlight] 2HCl}

ARate = k[\ce{H2}][\ce{Cl2}]
BRate = k[\ce{H2}]
CRate = k[\ce{Cl2}]
DRate = kAnswer

Why

The book states that the combination of \ce{H2} and \ce{Cl2} in the presence of sunlight is a zero order reaction. Its rate is independent of the concentrations of both reactants, because the rate is controlled by the light absorbed, so \text{rate}=k[\ce{H2}]^0[\ce{Cl2}]^0=k.

Option A, k[\ce{H2}][\ce{Cl2}], is the tempting wrong answer: it copies the coefficients of the balanced equation, but the order can only be found by experiment and is not read from the equation.

2025 Annual I, Set D, Q.2 (xi) OR3 marks

For the reaction \ce{2NO2 + F2 -> 2NO2F}, Rate = k[\ce{NO2}][\ce{F2}], propose the mechanism of reaction.

Solution

The rate law \text{Rate}=k[\ce{NO2}][\ce{F2}] is first order in each reactant, so the rate-determining (slow) step must contain one \ce{NO2} molecule and one \ce{F2} molecule. The overall equation has two \ce{NO2}, so the second \ce{NO2} must react in a fast step later.

Proposed mechanism

Slow step (rate-determining): one \ce{NO2} and one \ce{F2} react to give one \ce{NO2F} and an intermediate, the \ce{F} atom.
\ce{NO2 + F2 -> NO2F + F}\quad(\text{slow})
Fast step: the intermediate \ce{F} reacts with the second \ce{NO2}.
\ce{NO2 + F -> NO2F}\quad(\text{fast})
Add the two steps and cancel the intermediate \ce{F} from both sides. The sum is the overall reaction, so the mechanism is acceptable.
\ce{2NO2 + F2 -> 2NO2F}
Check the rate law from the slow step: one molecule of each reactant gives the experimental rate law.
\text{Rate}=k[\ce{NO2}][\ce{F2}]

Potential energy diagram for 2\ce{NO2}+\ce{F2}\to2\ce{NO2F}: two peaks, so two steps; the higher peak (step 1) is the slow, rate-determining step.

The atom \ce{F} is formed in step 1 and used up in step 2, so it is a reaction intermediate and does not appear in the overall equation or in the rate law.

\ce{NO2 + F2 -> NO2F + F}\ (\text{slow}),\quad \ce{NO2 + F -> NO2F}\ (\text{fast})

2025 Annual I, Set D, Q.2 (xii) OR3 marks

How does a catalyst affect the rate of a chemical reaction? Explain by drawing a reaction pathway diagram.

Solution

Effect of a catalyst

A catalyst increases the rate of a reaction. It provides a new mechanism (path) for the reaction with a lower activation energy. At the same temperature a larger fraction of the molecules then has energy at least equal to the new, smaller E_a, so more collisions are effective and the rate constant increases. The catalyst is chemically unchanged at the end of the reaction.

Reaction pathway diagram

Reaction pathway diagram: the catalysed path (accent) has a lower peak, so a smaller E_a, but the reactants, the products and \Delta H are the same.

The path without a catalyst has the higher peak (activation energy E_a).
The path with a catalyst has a lower peak (smaller E_a, accent curve).
The energy of the reactants and of the products is the same on both paths, so \Delta H is not changed by the catalyst.

Because a catalyst does not change the thermodynamics of the reaction, it cannot make a reaction that is not feasible (\Delta G^\circ positive) go.

2026 Annual I, Set D, Q.36 marks

Experiment	Initial [\ce{CO}]	Initial [\ce{Cl2}]	Initial rate (\mathrm{mol\,dm^{-3}\,s^{-1}})
1	1.000	0.100	1.29\times10^{-29}
2	0.100	0.100	1.30\times10^{-30}
3	0.100	1.000	1.30\times10^{-30}

Solution

(a) Rate constant k from the initial rates.

Find the order in \ce{CO} from experiments 1 and 2. [\ce{Cl2}] is constant at 0.100 while [\ce{CO}] is ten times larger (0.100\to1.000). The rate rises from 1.30\times10^{-30} to 1.29\times10^{-29}, about ten times.
\dfrac{\text{rate}_1}{\text{rate}_2}=\dfrac{1.29\times10^{-29}}{1.30\times10^{-30}}\approx10=10^{x}\;\Rightarrow\;x=1
Find the order in \ce{Cl2} from experiments 2 and 3. [\ce{CO}] is constant at 0.100 while [\ce{Cl2}] is increased ten times (0.100\to1.000). The rate does not change.
\dfrac{\text{rate}_3}{\text{rate}_2}=1=10^{y}\;\Rightarrow\;y=0
Write the rate law. The reaction is first order in \ce{CO} and zero order in \ce{Cl2}, so it is first order overall.
\text{Rate}=k\,[\ce{CO}]
Put the data of experiment 1 into the rate law and solve for k.
k=\dfrac{\text{Rate}}{[\ce{CO}]}=\dfrac{1.29\times10^{-29}\,\mathrm{mol\,dm^{-3}\,s^{-1}}}{1.000\,\mathrm{mol\,dm^{-3}}}=1.29\times10^{-29}\,\mathrm{s^{-1}}
Check with experiment 2. The same value (within the rounding of the printed data) is obtained.
k=\dfrac{1.30\times10^{-30}}{0.100}=1.30\times10^{-29}\,\mathrm{s^{-1}}

The unit of k for a first order reaction is \mathrm{s^{-1}}.

k\approx1.3\times10^{-29}\,\mathrm{s^{-1}}

(b) Half-life of the reaction.

The reaction is first order, so its half-life does not depend on the concentration and is given by t_{1/2}=0.693/k.

Write the half-life formula for a first order reaction.
t_{1/2}=\dfrac{0.693}{k}
Substitute k=1.29\times10^{-29}\,\mathrm{s^{-1}}.
t_{1/2}=\dfrac{0.693}{1.29\times10^{-29}\,\mathrm{s^{-1}}}=5.37\times10^{28}\,\mathrm{s}

This is an astronomically long time (about 1.7\times10^{21} years). The printed rates are extremely small, so under these conditions the reaction is practically not taking place.

t_{1/2}\approx5.4\times10^{28}\,\mathrm{s}

Note. The rates are printed as 10^{-29} and 10^{-30} in the paper (the book's own version of this data); the ratio 1.29\times10^{-29}/1.30\times10^{-30}=9.9 is rounded to 10.

Exam focus

Use initial-rate data to find the order, the rate law and the rate constant k. (numerical)
A 6-mark Section C question in both 2026 sets (Q.3, the phosgene data). Compare book numerical question 10 and Example 7.2.
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Half-life method: t_{1/2}=0.693/k for a first order reaction, and finding k from concentration data. (numerical)
Asked in 2026 Q.3 (b) (both sets), in 2025 Set B MCQ 16 and in 2025 Set B Q.2 (xiv) OR (k from 0.1\to0.05\,\mathrm{M} in 60\,\mathrm{s}).
Go to answer
Suggest a reaction mechanism consistent with a rate law, and name the slow (rate-determining) step and the intermediate. (short)
Asked in 2025 Set D Q.2 (xi) OR (\ce{2NO2 + F2}), and in 2026 Set D MCQ 3 (rate law of a slow step). Book exercise MCQ iii and short question 7 use the same rule.
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Explain the effect of a catalyst on the rate and draw the reaction pathway diagram. (short)
Asked in 2025 Set D Q.2 (xii) OR; it is the book's short questions 5, 6 and 12 and MCQ x.
Go to answer
Explain a reaction through collisions, activation energy, the activated complex and enthalpy change, with energy profiles. (long)
A 7-mark Section C question in 2025 Set B (Q.6 OR). Learn Figures 7.5 and 7.6 of the book.
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Explain the effect of temperature on the rate using the Maxwell-Boltzmann curve and the Arrhenius equation. (long)
Two MCQs in 2026 (Set B MCQ 2 on the area under the curve, Set D MCQ 2 on kinetic energy and temperature). Book short questions 3 and 4 ask the same idea.
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Order of reaction and the unit of k, zero order reactions and how the rate changes with concentration. (short)
2025 Set D MCQ 6 (\ce{H2 + Cl2} in sunlight is zero order, rate =k) and 2026 Set B MCQ 3 (factors that affect the rate). Book exercise MCQs ii, iv, vi and viii.
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Decide whether a reaction is feasible from \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ. (numerical)
Not in the 2025 or 2026 papers. It is the unit's last SLO and book numerical question 14 (the decomposition of \ce{MgCO3}).
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