# 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: $$ \mathrm{A} \rightarrow \text{Product} $$ 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). $$ \text{When } [A] = 1 M, \quad \text{Rate} = k $$ 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. Consider a general reaction: $$ aA + bB \rightarrow cC + dD $$ 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: $$ 2\mathrm{NO}_{2} + \mathrm{O}_{3} \rightarrow \mathrm{N}_{2}\mathrm{O}_{5} + \mathrm{O}_{2} $$ 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. $$ 2\mathrm{NH}_{3(g)} \xrightarrow{\text{W}} \mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} $$ $$ \text{Rate} = k[\mathrm{NH}_{3}]^0 = k $$ **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. $$ 2\mathrm{N}_{2}\mathrm{O}_{5(g)} \rightarrow 2\mathrm{N}_{2}\mathrm{O}_{4(g)} + \mathrm{O}_{2(g)} $$ $$ \text{Rate} = k[\mathrm{N}_{2}\mathrm{O}_{5}] $$ * **Example 2:** Decomposition of ammonium nitrite. $$ \mathrm{NH}_{4}\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} + 2\mathrm{H}_{2}\mathrm{O} $$ $$ \text{Rate} = k[\mathrm{NH}_{4}\mathrm{NO}_{2}] $$ **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. $$ 2\mathrm{NO}_{2(g)} \rightarrow 2\mathrm{NO}_{(g)} + \mathrm{O}_{2(g)} $$ $$ \text{Rate} = k[\mathrm{NO}_{2}]^2 $$ * **Example 2:** Reaction between nitrogen monoxide and ozone. $$ \mathrm{NO}_{(g)} + \mathrm{O}_{3(g)} \rightarrow \mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)} $$ $$ \text{Rate} = k[\mathrm{NO}][\mathrm{O}_{3}] \quad (\text{Overall order = 1+1=2}) $$ **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. $$ 2\mathrm{NO}_{(g)} + \mathrm{O}_{2(g)} \rightarrow 2\mathrm{NO}_{2(g)} $$ $$ \text{Rate} = k[\mathrm{NO}]^2[\mathrm{O}_{2}] \quad (\text{Overall order = 2+1=3}) $$ **e) Fractional-Order Reaction** A reaction where the overall order is a fraction. * **Example:** Formation of hydrogen bromide. $$ \mathrm{H}_{2(g)} + \mathrm{Br}_{2(g)} \rightarrow 2\mathrm{HBr}_{(g)} $$ $$ \text{Rate} = k[\mathrm{H}_{2}][\mathrm{Br}_{2}]^{1/2} \quad (\text{Overall order = } 1 + 0.5 = 1.5) $$ **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. $$ (\mathrm{CH}_{3})_{3}\mathrm{C-Br}_{(l)} + \mathrm{H}_{2}\mathrm{O}_{(\text{Excess})} \rightarrow (\mathrm{CH}_{3})_{3}\mathrm{C-OH}_{(l)} + \mathrm{HBr}_{(l)} $$ $$ \text{Rate} = k'[(\mathrm{CH}_{3})_{3}\mathrm{C-Br}] $$ Where $k' = k[\mathrm{H}_{2}\mathrm{O}]$ is the pseudo first-order rate constant. ### 7.2.2 Determination of Reaction Order, Rate Law, and Rate Constant The **method of initial rates** is used to determine the order of a reaction experimentally. In this method, the initial rate of reaction is measured while systematically varying the initial concentration of one reactant and keeping the others constant. ### Example 7.2: Determining Rate Law from Experimental Data The reaction between nitrogen (II) oxide and ozone is: $$ \mathrm{NO}_{(g)} + \mathrm{O}_{3(g)} \rightarrow \mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)} $$ The following data were obtained at 25°C. Determine the rate law for the reaction. | Experiment | Initial [NO] (M) | Initial [O₃] (M) | Initial Rate (M s⁻¹) | | :---: | :---: | :---: | :---: | | 1 | $1.00 \times 10^{-6}$ | $9.00 \times 10^{-6}$ | $1.98 \times 10^{-4}$ | | 2 | $2.00 \times 10^{-6}$ | $9.00 \times 10^{-6}$ | $3.96 \times 10^{-4}$ | | 3 | $1.00 \times 10^{-6}$ | $3.00 \times 10^{-6}$ | $6.60 \times 10^{-5}$ | **Solution:** 1. **Determine the order with respect to NO:** * Compare experiments 1 and 2, where $[O_3]$ is constant. * $[NO]$ doubles ($1.00 \times 10^{-6} \rightarrow 2.00 \times 10^{-6}$). * The initial rate also doubles ($1.98 \times 10^{-4} \rightarrow 3.96 \times 10^{-4}$). * Since the rate doubles when the concentration doubles, the reaction is **first order** with respect to NO. * $\text{Rate} \propto [\mathrm{NO}]^1$ 2. **Determine the order with respect to O₃:** * Compare experiments 1 and 3, where $[NO]$ is constant. * $[O_3]$ decreases by a factor of 3 ($9.00 \times 10^{-6} \rightarrow 3.00 \times 10^{-6}$). * The initial rate also decreases by a factor of 3 ($1.98 \times 10^{-4} \rightarrow 6.60 \times 10^{-5}$). * Since the rate changes proportionally to the concentration, the reaction is **first order** with respect to O₃. * $\text{Rate} \propto [\mathrm{O}_{3}]^1$ 3. **Write the overall rate law:** * Combining the dependencies, the rate law is: $$ \text{Rate} = k[\mathrm{NO}][\mathrm{O}_{3}] $$ * The overall reaction order is $1 + 1 = 2$ (second order). 4. **Calculate the rate constant (k):** * Use the data from any experiment (e.g., experiment 1) and substitute into the rate law. $$ k = \frac{\text{Rate}}{[\mathrm{NO}][\mathrm{O}_{3}]} $$ $$ k = \frac{1.98 \times 10^{-4} \, M \cdot s^{-1}}{(1.00 \times 10^{-6} \, M)(9.00 \times 10^{-6} \, M)} $$ $$ k = \frac{1.98 \times 10^{-4}}{9.00 \times 10^{-12}} \, M^{-1}s^{-1} = 2.2 \times 10^{7} \, M^{-1}s^{-1} $$ ### Example 7.3: Writing Rate Law from Given Orders The reaction $ \mathrm{H}_{2} + \mathrm{Br}_{2} \rightarrow 2\mathrm{HBr} $ is first order in $H_2$ and half order in $Br_2$. Write the rate law. **Solution:** 1. **Given Information:** * Rate is proportional to the first power of $[H_2]$: $\text{Rate} \propto [\mathrm{H}_{2}]^1$ * Rate is proportional to the half power of $[Br_2]$: $\text{Rate} \propto [\mathrm{Br}_{2}]^{1/2}$ 2. **Combine the proportionalities:** * The overall rate law combines these dependencies: $$ \text{Rate} = k[\mathrm{H}_{2}][\mathrm{Br}_{2}]^{1/2} $$ ### 7.2.3 Half-Life Method for Determining Rate Constant The **half-life ($t_{1/2}$)** of a reaction is the time required for the concentration of a reactant to decrease to half its initial value. The relationship between half-life and the rate constant $k$ depends on the reaction order. * **Zero-Order (n=0):** $t_{1/2} = \frac{[A_0]}{2k}$ * **First-Order (n=1):** $t_{1/2} = \frac{0.693}{k}$ * **Second-Order (n=2):** $t_{1/2} = \frac{1}{k[A_0]}$ *Where $[A_0]$ is the initial concentration of the reactant.* **Example:** A first-order reaction has a half-life of 30 s. Calculate its rate constant. 1. **Given values:** * $t_{1/2} = 30 \, s$ * Reaction is first order. 2. **Apply the formula:** * For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$. 3. **Show calculation:** $$ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30 \, s} = 0.0231 \, s^{-1} $$