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Chemical Kinetics - Order of a Reaction

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Order of a Reaction is defined as the sum of the powers of the concentration of the reactants in the rate law expression. For a general reaction where Rate=k[A]x[B]yRate = k[A]^x[B]^y, the overall order n=x+yn = x + y.

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Order is an experimental quantity. It cannot be predicted solely from the stoichiometric coefficients of a balanced chemical equation.

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The value of the order can be zero, an integer, or even a fraction. A zero-order reaction means the rate of reaction is independent of the concentration of reactants.

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For elementary reactions, the order is usually equal to the molecularity. However, for complex reactions (multi-step), the overall order is determined by the slowest step, called the Rate Determining Step.

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The units of the rate constant kk depend on the order of the reaction. For an nthn^{th} order reaction, the units are given by (mol L−1)1−ns−1(\text{mol L}^{-1})^{1-n} \text{s}^{-1}.

📐Formulae

Rate=k[A]x[B]yRate = k[A]^x[B]^y

n=x+yn = x + y

Unit of k=(molL)1−ns−1\text{Unit of } k = \left( \frac{\text{mol}}{\text{L}} \right)^{1-n} \text{s}^{-1}

Zero Order Unit: mol L−1s−1\text{Zero Order Unit: } \text{mol L}^{-1} \text{s}^{-1}

First Order Unit: s−1\text{First Order Unit: } \text{s}^{-1}

Second Order Unit: mol−1L s−1\text{Second Order Unit: } \text{mol}^{-1} \text{L s}^{-1}

💡Examples

Problem 1:

Calculate the overall order of a reaction which has the rate expression: Rate=k[A]1/2[B]3/2Rate = k[A]^{1/2}[B]^{3/2}.

Solution:

n=2n = 2

Explanation:

The overall order nn is calculated by summing the exponents of the concentrations in the rate law: n=12+32=42=2n = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2. Therefore, the reaction is of the second order.

Problem 2:

Identify the reaction order for a reaction where the rate constant is k=2.3×10−5 L mol−1 s−1k = 2.3 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}.

Solution:

n=2n = 2

Explanation:

We use the general unit formula (mol L−1)1−ns−1(\text{mol L}^{-1})^{1-n} \text{s}^{-1}. Comparing this to the given unit L mol−1s−1\text{L mol}^{-1} \text{s}^{-1} (which is (mol L−1)−1s−1(\text{mol L}^{-1})^{-1} \text{s}^{-1}), we set 1−n=−11-n = -1. Solving for nn, we get n=2n = 2. Thus, it is a second-order reaction.

Problem 3:

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled?

Solution:

Ratenew=4×RateoldRate_{new} = 4 \times Rate_{old}

Explanation:

The rate law is Rate=k[R]2Rate = k[R]^2. If the new concentration [R′]=2[R][R'] = 2[R], then Rate′=k(2[R])2=4k[R]2Rate' = k(2[R])^2 = 4k[R]^2. Therefore, the rate increases by 44 times.