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Chemical Kinetics - Rate of a reaction

Grade 12ICSEChemistry

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. It is always a positive quantity.

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The Average Rate of reaction is the change in concentration of a species divided by the time interval Δt\Delta t over which the change occurs: ravg=−Δ[R]Δt=Δ[P]Δtr_{avg} = -\frac{\Delta [R]}{\Delta t} = \frac{\Delta [P]}{\Delta t}.

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The Instantaneous Rate is the rate of reaction at a particular moment in time, determined by the slope of the tangent to the concentration-time curve at that instant: rinst=−d[R]dt=d[P]dtr_{inst} = -\frac{d[R]}{dt} = \frac{d[P]}{dt}.

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Stoichiometry plays a crucial role; for a reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, the rate is expressed as: −1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt-\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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Units of reaction rate are typically expressed as 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, units can be atm⋅s−1atm \cdot s^{-1}.

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Factors affecting reaction rates include the concentration of reactants, temperature, presence of a catalyst, and the surface area of solid reactants.

📐Formulae

r=−Δ[R]Δt=Δ[P]Δtr = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t}

rinst=lim⁡Δt→0(±Δ[C]Δt)=d[C]dtr_{inst} = \lim_{\Delta t \to 0} \left( \pm \frac{\Delta[C]}{\Delta t} \right) = \frac{d[C]}{dt}

r=−1n1d[A]dt=1n2d[B]dtr = -\frac{1}{n_1}\frac{d[A]}{dt} = \frac{1}{n_2}\frac{d[B]}{dt}

Rate Unit=[Concentration]×[Time]−1\text{Rate Unit} = [\text{Concentration}] \times [\text{Time}]^{-1}

💡Examples

Problem 1:

For 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 2.8×10−3 mol⋅L−1⋅s−12.8 \times 10^{-3} \, mol \cdot L^{-1} \cdot s^{-1}. Calculate the rate of disappearance of N2O5N_2O_5.

Solution:

From the stoichiometry of the reaction, the rate is given by: r=−12d[N2O5]dt=14d[NO2]dtr = -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt}. Given d[NO2]dt=2.8×10−3 mol⋅L−1⋅s−1\frac{d[NO_2]}{dt} = 2.8 \times 10^{-3} \, mol \cdot L^{-1} \cdot s^{-1}. Therefore, −d[N2O5]dt=24×(2.8×10−3)=1.4×10−3 mol⋅L−1⋅s−1-\frac{d[N_2O_5]}{dt} = \frac{2}{4} \times (2.8 \times 10^{-3}) = 1.4 \times 10^{-3} \, mol \cdot L^{-1} \cdot s^{-1}.

Explanation:

The rate of disappearance is related to the rate of formation by their respective stoichiometric coefficients. We multiply the rate of formation by the ratio of the coefficient of the reactant to the coefficient of the product.

Problem 2:

In the reaction H2+I2→2HIH_2 + I_2 \rightarrow 2HI, the concentration of H2H_2 decreases from 0.50 mol⋅L−10.50 \, mol \cdot L^{-1} to 0.30 mol⋅L−10.30 \, mol \cdot L^{-1} in 1010 minutes. Calculate the average rate of the reaction in mol⋅L−1⋅s−1mol \cdot L^{-1} \cdot s^{-1}.

Solution:

Δ[H2]=0.30−0.50=−0.20 mol⋅L−1\Delta [H_2] = 0.30 - 0.50 = -0.20 \, mol \cdot L^{-1}. Time interval Δt=10×60=600 s\Delta t = 10 \times 60 = 600 \, s. ravg=−Δ[H2]Δt=−−0.20600=3.33×10−4 mol⋅L−1⋅s−1r_{avg} = -\frac{\Delta [H_2]}{\Delta t} = -\frac{-0.20}{600} = 3.33 \times 10^{-4} \, mol \cdot L^{-1} \cdot s^{-1}.

Explanation:

Average rate is calculated by dividing the change in concentration by the total time taken in seconds to maintain standard SI units.