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Chemical Kinetics - Rate expression and reaction mechanism (HL only)

Grade 12IBChemistry

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

🔑Concepts

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The rate expression or rate law is an experimentally determined equation: rate=k[A]m[B]nrate = k[A]^m[B]^n, where kk is the rate constant, and mm and nn are the orders of reaction with respect to reactants AA and BB.

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The overall order of reaction is the sum of the individual orders (m+nm + n). The units of the rate constant kk depend on the overall order: (mol⋅dm−3)1−order⋅s−1(mol \cdot dm^{-3})^{1-order} \cdot s^{-1}.

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Reaction mechanisms consist of a series of elementary steps. The molecularity of an elementary step refers to the number of reactant particles involved (unimolecular, bimolecular, or termolecular).

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The Rate-Determining Step (RDS) is the slowest step in a multi-step reaction. The rate expression for the overall reaction is determined by the stoichiometry of the reactants in the RDS.

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Reaction intermediates are species that are produced in one step and consumed in a subsequent step. They do not appear in the overall chemical equation.

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If the RDS involves an intermediate formed from a prior fast equilibrium step, the concentration of the intermediate in the rate law must be substituted using the equilibrium constant expression of the fast step.

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The Arrhenius equation k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}} demonstrates the temperature dependence of the rate constant. A plot of ln⁡k\ln k against 1T\frac{1}{T} yields a straight line with a gradient of −EaR-\frac{E_a}{R}.

📐Formulae

rate=k[A]m[B]nrate = k[A]^m[B]^n

k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}

ln⁡k=−EaRT+ln⁡A\ln k = -\frac{E_a}{RT} + \ln A

ln⁡k1k2=EaR(1T2−1T1)\ln \frac{k_1}{k_2} = \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) units of R=8.31 J⋅K−1⋅mol−1R = 8.31 \, J \cdot K^{-1} \cdot mol^{-1}

units of k=(mol⋅dm−3)1−ns−1units \, of \, k = (mol \cdot dm^{-3})^{1-n} s^{-1}

💡Examples

Problem 1:

A reaction follows the mechanism: Step 1: NO2(g)+NO2(g)→k1NO3(g)+NO(g)NO_2(g) + NO_2(g) \xrightarrow{k_1} NO_3(g) + NO(g) (Slow) Step 2: NO3(g)+CO(g)→k2NO2(g)+CO2(g)NO_3(g) + CO(g) \xrightarrow{k_2} NO_2(g) + CO_2(g) (Fast) Identify the overall reaction and the rate expression.

Solution:

Overall reaction: NO2(g)+CO(g)→NO(g)+CO2(g)NO_2(g) + CO(g) \rightarrow NO(g) + CO_2(g). Rate expression: rate=k[NO2]2rate = k[NO_2]^2.

Explanation:

The overall reaction is found by summing the steps and cancelling intermediates (NO3NO_3). Since Step 1 is the Rate-Determining Step (RDS), the rate expression is derived solely from its reactants: rate=k[NO2][NO2]=k[NO2]2rate = k[NO_2][NO_2] = k[NO_2]^2. Note that COCO does not appear in the rate law because it is involved in a step after the RDS.

Problem 2:

For a reaction where the rate constant kk is 2.5×10−2 dm3⋅mol−1⋅s−12.5 \times 10^{-2} \, dm^3 \cdot mol^{-1} \cdot s^{-1}, determine the overall order and calculate the rate when [A]=0.20 mol⋅dm−3[A] = 0.20 \, mol \cdot dm^{-3} and [B]=0.50 mol⋅dm−3[B] = 0.50 \, mol \cdot dm^{-3} if the reaction is first order with respect to both.

Solution:

Overall order = 1+1=21 + 1 = 2. rate=(2.5×10−2)(0.20)(0.50)=2.5×10−3 mol⋅dm−3⋅s−1rate = (2.5 \times 10^{-2})(0.20)(0.50) = 2.5 \times 10^{-3} \, mol \cdot dm^{-3} \cdot s^{-1}.

Explanation:

The units of kk (dm3⋅mol−1⋅s−1dm^3 \cdot mol^{-1} \cdot s^{-1}) confirm it is a second-order reaction. Using the rate law rate=k[A]1[B]1rate = k[A]^1[B]^1, we substitute the given concentrations and rate constant to find the reaction rate.

Rate expression and reaction mechanism (HL only) Revision - Grade 12 Chemistry IB