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Chemical Kinetics - Order and molecularity of a reaction

Grade 12ICSEChemistry

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

🔑Concepts

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The Rate Law is an expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation: Rate=k[A]x[B]yRate = k[A]^x[B]^y.

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

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Order of a reaction can be 0,1,2,30, 1, 2, 3 and even a fraction. It is an experimental quantity and cannot be determined simply by looking at the balanced equation.

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Molecularity is the number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

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Molecularity is a theoretical concept and can only be a positive whole number (1,2,1, 2, or 33). It cannot be zero or a fraction.

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In a Complex Reaction (a reaction proceeding in several steps), the overall rate of the reaction is controlled by the slowest step, which is called the Rate Determining Step (RDS).

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For complex reactions, the order is given by the slowest step and the molecularity of the slowest step is usually the same as the order of the overall reaction.

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Difference: Order is applicable to elementary as well as complex reactions, whereas molecularity is applicable only to elementary reactions.

📐Formulae

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

n=x+y (where n is the overall order of reaction)n = x + y \text{ (where } n \text{ is the overall order of reaction)}

Units of Rate Constant (k)=(mol L−1)1−ns−1\text{Units of Rate Constant (k)} = (mol \, L^{-1})^{1-n} s^{-1}

For Zero Order (n=0): k units=mol L−1s−1\text{For Zero Order (n=0): } k \text{ units} = mol \, L^{-1} s^{-1}

For First Order (n=1): k units=s−1\text{For First Order (n=1): } k \text{ units} = s^{-1}

For Second Order (n=2): k units=mol−1L s−1\text{For Second Order (n=2): } k \text{ units} = mol^{-1} 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=12+32=42=2n = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2

Explanation:

The overall order is the sum of the exponents of the concentration terms in the rate law. Since the sum is 22, the reaction is of the second order.

Problem 2:

Identify the reaction order from the given rate constant: k=2.3×10−5 L mol−1s−1k = 2.3 \times 10^{-5} \, L \, mol^{-1} s^{-1}.

Solution:

Units=L mol−1s−1=(mol L−1)−1s−1Units = L \, mol^{-1} s^{-1} = (mol \, L^{-1})^{-1} s^{-1} Comparing with (mol L−1)1−ns−1(mol \, L^{-1})^{1-n} s^{-1}: 1−n=−1  ⟹  n=21-n = -1 \implies n = 2

Explanation:

By analyzing the units of the rate constant, we can determine the order. L mol−1s−1L \, mol^{-1} s^{-1} corresponds to a second-order reaction.

Problem 3:

For a reaction A+B→ProductA + B \rightarrow Product, the rate law is Rate=k[A]2[B]Rate = k[A]^2[B]. What is the change in rate if the concentration of AA is doubled and BB is halved?

Solution:

Rate1=k[A]2[B]Rate_1 = k[A]^2[B] Rate2=k[2A]2[12B]=k(4[A]2)(12[B])=2k[A]2[B]Rate_2 = k[2A]^2[\frac{1}{2}B] = k(4[A]^2)(\frac{1}{2}[B]) = 2k[A]^2[B] Rate2=2×Rate1Rate_2 = 2 \times Rate_1

Explanation:

Doubling [A][A] increases the rate by a factor of 22=42^2 = 4 (second order in AA). Halving [B][B] decreases the rate by a factor of 1/21/2 (first order in BB). The net effect is 4×0.5=24 \times 0.5 = 2 times the original rate.