7.2 Rate Law

The rate of a chemical reaction is defined as the instantaneous change in the concentration of a reactant or product at a given time. Experimental studies show that the rate of a reaction is proportional to the molar concentration of reactants, with each concentration term raised to a specific power. This relationship is known as the rate law.

For a general reaction: $\mathrm{A}\to \text{Product}$

The rate can be expressed as: $\text{Rate}\propto [A{]}^x$ $\text{Rate}=k[A{]}^x$

• $k$ is the rate constant, a proportionality constant specific to a particular reaction at a given temperature.
• The expression $\text{Rate}=k[A{]}^x$ is the rate law or rate equation.
• The exponent $x$ is the order of reaction with respect to reactant A, and it must be determined experimentally.

The rate constant (k) can be defined as the rate of reaction when the molar concentration of each reactant is unity (1 M). $\text{When }[A]=1\text{M},\text{Rate}=k$ The value of $k$ is independent of concentration and time but changes with temperature.

7.2.1 Order of Reaction and Rate Equation

The order of reaction is the sum of the exponents to which the molar concentration terms in the rate equation are raised. It reflects how the concentration of reactants affects the reaction rate and helps in determining the reaction mechanism.

The Mechanism of a Chemical Reaction→

Consider a general reaction: $aA+bB\to cC+dD$

The experimentally determined rate equation is: $\text{Rate}=k[A{]}^x[B{]}^y$

• $x$ is the order of reaction with respect to reactant A.
• $y$ is the order of reaction with respect to reactant B.
• The overall order of the reaction is the sum $x+y$.

Important Note: The exponents $x$ and $y$ are determined experimentally and may not be the same as the stoichiometric coefficients $a$ and $b$ from the balanced chemical equation.

For example, for the reaction: $2{\mathrm{NO}}_2+{\mathrm{O}}_3\to {\mathrm{N}}_2{\mathrm{O}}_5+{\mathrm{O}}_2$

The experimentally determined rate law is: $\text{Rate}=k[{\mathrm{NO}}_2][{\mathrm{O}}_3]$ In this case, the order with respect to ${\mathrm{NO}}_2$ is 1, even though its stoichiometric coefficient is 2. The overall order of the reaction is $1+1=2$.

Types of Order of Reactions

The order of a reaction can be a whole number, zero, or a fraction.

a) Zero-Order Reaction A reaction whose rate is independent of the concentration of the reactant(s).

• Rate Law: $\text{Rate}=k[\text{Reactant}{]}^0=k$
• Example: Decomposition of ammonia on a heated tungsten surface. $2{\mathrm{NH}}_{3(g)}\stackrel{\text{W}}{\to }{\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)}$ $\text{Rate}=k[{\mathrm{NH}}_3{]}^0=k$

b) First-Order Reaction A reaction whose rate is directly proportional to the first power of the concentration of a single reactant.

• Rate Law: $\text{Rate}=k[A{]}^1$
• Example 1: Thermal decomposition of dinitrogen pentoxide. $2{\mathrm{N}}_2{\mathrm{O}}_{5(g)}\to 2{\mathrm{N}}_2{\mathrm{O}}_{4(g)}+{\mathrm{O}}_{2(g)}$ $\text{Rate}=k[{\mathrm{N}}_2{\mathrm{O}}_5]$
• Example 2: Decomposition of ammonium nitrite. ${\mathrm{NH}}_4{\mathrm{NO}}_2\to {\mathrm{N}}_2+2{\mathrm{H}}_2\mathrm{O}$ $\text{Rate}=k[{\mathrm{NH}}_4{\mathrm{NO}}_2]$

c) Second-Order Reaction A reaction for which the sum of the exponents in the rate law is two.

• Rate Laws: $\text{Rate}=k[A{]}^2$ or $\text{Rate}=k[A][B]$
• Example 1: Decomposition of nitrogen dioxide. $2{\mathrm{NO}}_{2(g)}\to 2{\mathrm{NO}}_{(g)}+{\mathrm{O}}_{2(g)}$ $\text{Rate}=k[{\mathrm{NO}}_2{]}^2$
• Example 2: Reaction between nitrogen monoxide and ozone. ${\mathrm{NO}}_{(g)}+{\mathrm{O}}_{3(g)}\to {\mathrm{NO}}_{2(g)}+{\mathrm{O}}_{2(g)}$ $\text{Rate}=k[\mathrm{NO}][{\mathrm{O}}_3](\text{Overall order}=1+1=2)$

d) Third-Order Reaction A reaction for which the sum of the exponents in the rate law is three.

• Rate Laws: $\text{Rate}=k[A{]}^3$, $\text{Rate}=k[A{]}^2[B]$, etc.
• Example: Oxidation of nitrogen monoxide by oxygen. $2{\mathrm{NO}}_{(g)}+{\mathrm{O}}_{2(g)}\to 2{\mathrm{NO}}_{2(g)}$ $\text{Rate}=k[\mathrm{NO}{]}^2[{\mathrm{O}}_2](\text{Overall order}=2+1=3)$

e) Fractional-Order Reaction A reaction where the overall order is a fraction.

• Example: Formation of hydrogen bromide. ${\mathrm{H}}_{2(g)}+{\mathrm{Br}}_{2(g)}\to 2{\mathrm{HBr}}_{(g)}$ $\text{Rate}=k[{\mathrm{H}}_2][{\mathrm{Br}}_2{]}^{1/2}(\text{Overall order}=1+0.5=1.5)$

f) Pseudo First-Order Reaction A bimolecular reaction where one reactant (often the solvent) is present in such large excess that its concentration remains effectively constant. The reaction rate then appears to depend only on the concentration of the other reactant.

• Example: Hydrolysis of tert-butyl bromide in excess water. $({\mathrm{CH}}_3{)}_3{\mathrm{C}\text{-}\mathrm{Br}}_{(l)}+{\mathrm{H}}_2{\mathrm{O}}_{(\text{Excess})}\to ({\mathrm{CH}}_3{)}_3{\mathrm{C}\text{-}\mathrm{OH}}_{(l)}+{\mathrm{HBr}}_{(l)}$ $\text{Rate}=k^{\prime }[({\mathrm{CH}}_3{)}_3\mathrm{C}\text{-}\mathrm{Br}]$ Where $k^{\prime }=k[{\mathrm{H}}_2\mathrm{O}]$ is the pseudo first-order rate constant.

7.2.2 Determination of Reaction Order, Rate Law, and Rate Constant

The order of a reaction and the rate law are determined experimentally, not from the balanced equation. The most common method is the initial rates method.

Initial Rates Method: The initial rate of reaction is measured for several experiments in which the concentration of one reactant is varied while all others are kept constant.

Steps:

1. Measure the initial rate of reaction at different initial concentrations of each reactant.
2. Compare two experiments where only one reactant concentration changes.
3. Use the ratio of rates to find the order with respect to that reactant.
4. Repeat for each reactant to find the overall order.
5. Substitute known values of rate, concentrations, and orders into the rate law to calculate $k$.

Worked Example:

For the reaction $A+B\to \text{Products}$, the following data were collected:

Finding order w.r.t. A (compare Exp 1 and 2, $[B]$ constant): $\frac{{\text{Rate}}_2}{{\text{Rate}}_1}=\frac{k[0.20{]}^x[0.10{]}^y}{k[0.10{]}^x[0.10{]}^y}=2^x=\frac{4.0\times 10^{-3}}{2.0\times 10^{-3}}=2$ $\therefore x=1(\text{first order w.r.t. A})$

Finding order w.r.t. B (compare Exp 1 and 3, $[A]$ constant): $\frac{{\text{Rate}}_3}{{\text{Rate}}_1}=2^y=\frac{8.0\times 10^{-3}}{2.0\times 10^{-3}}=4$ $\therefore y=2(\text{second order w.r.t. B})$

Rate Law: $\text{Rate}=k[A][B{]}^2$ — Overall order = 3

Calculating k (using Experiment 1): $k=\frac{\text{Rate}}{[A][B{]}^2}=\frac{2.0\times 10^{-3}}{(0.10)(0.10{)}^2}=2.0{\text{dm}}^6{\text{mol}}^{-2}{\text{s}}^{-1}$