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Chemical Kinetics - Rate of a Chemical Reaction

Grade 12CBSEChemistry

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

🔑Concepts

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The rate of a chemical reaction is defined as the change in the molar concentration of any one of the reactants or products per unit time.

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The units for the rate of reaction are typically mol⋅L−1⋅s−1mol \cdot L^{-1} \cdot s^{-1} or mol⋅dm−3⋅s−1mol \cdot dm^{-3} \cdot s^{-1}. For gaseous reactions, it can be expressed as atm⋅s−1atm \cdot s^{-1}.

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Average rate (ravr_{av}) is the rate measured over a specific time interval Δt\Delta t, expressed as rav=−Δ[R]Δt=Δ[P]Δtr_{av} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t}.

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Instantaneous rate (rinstr_{inst}) is the rate of reaction at a particular moment in time, calculated as the slope of the tangent to the concentration-time curve: rinst=−d[R]dt=d[P]dtr_{inst} = -\frac{d[R]}{dt} = \frac{d[P]}{dt}.

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For a general reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, the rates of disappearance of reactants and appearance of products are related by their stoichiometric coefficients: Rate=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}.

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The rate of a reaction is influenced by factors such as the concentration of reactants, temperature, presence of a catalyst, and the surface area of solid reactants.

📐Formulae

rav=Decrease in concentration of ReactantTime taken=−Δ[R]Δtr_{av} = \frac{\text{Decrease in concentration of Reactant}}{\text{Time taken}} = -\frac{\Delta [R]}{\Delta t}

rav=Increase in concentration of ProductTime taken=Δ[P]Δtr_{av} = \frac{\text{Increase in concentration of Product}}{\text{Time taken}} = \frac{\Delta [P]}{\Delta t}

rinst=−d[R]dt=d[P]dtr_{inst} = -\frac{d[R]}{dt} = \frac{d[P]}{dt}

Rate=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

💡Examples

Problem 1:

For the reaction R→PR \rightarrow P, the concentration of a reactant changes from 0.03M0.03 M to 0.02M0.02 M in 2525 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Solution:

Δ[R]=[R]2−[R]1=0.02M−0.03M=−0.01M\Delta [R] = [R]_2 - [R]_1 = 0.02 M - 0.03 M = -0.01 M. Δt=25min\Delta t = 25 min. rav=−Δ[R]Δt=−−0.01M25min=4×10−4M⋅min−1r_{av} = -\frac{\Delta[R]}{\Delta t} = -\frac{-0.01 M}{25 min} = 4 \times 10^{-4} M \cdot min^{-1}. In seconds: rav=4×10−4M60s=6.66×10−6M⋅s−1r_{av} = \frac{4 \times 10^{-4} M}{60 s} = 6.66 \times 10^{-6} M \cdot s^{-1}.

Explanation:

The average rate is the negative change in reactant concentration divided by the time interval. To convert to seconds, the rate in min−1min^{-1} is divided by 6060.

Problem 2:

In the reaction 2N2O5(g)→4NO2(g)+O2(g)2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g), the rate of formation of NO2NO_2 is 0.0072mol⋅L−1⋅s−10.0072 mol \cdot L^{-1} \cdot s^{-1}. Calculate the rate of disappearance of N2O5N_2O_5.

Solution:

From the stoichiometry: Rate=−12d[N2O5]dt=14d[NO2]dt\text{Rate} = -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt}. Given d[NO2]dt=0.0072mol⋅L−1⋅s−1\frac{d[NO_2]}{dt} = 0.0072 mol \cdot L^{-1} \cdot s^{-1}. Therefore, −d[N2O5]dt=24×0.0072=0.0036mol⋅L−1⋅s−1-\frac{d[N_2O_5]}{dt} = \frac{2}{4} \times 0.0072 = 0.0036 mol \cdot L^{-1} \cdot s^{-1}.

Explanation:

The rate of disappearance of N2O5N_2O_5 is related to the rate of appearance of NO2NO_2 by the ratio of their stoichiometric coefficients (24)(\frac{2}{4}).