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Chemical Reactions - Rate of reaction

Grade 12IGCSEChemistry

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

🔑Concepts

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The rate of reaction is defined as the change in concentration of a reactant or product per unit time, expressed as mol⋅dm−3⋅s−1mol \cdot dm^{-3} \cdot s^{-1}.

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Collision Theory: For a reaction to occur, particles must collide with a kinetic energy equal to or greater than the activation energy (EaE_a) and in the correct spatial orientation.

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Activation Energy (EaE_a): The minimum energy required by colliding particles to break bonds and initiate a chemical reaction.

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Effect of Concentration/Pressure: Increasing the concentration of reactants in solution or the pressure of gaseous reactants increases the number of particles per unit volume, which increases the frequency of collisions and thus the rate of reaction.

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Effect of Temperature: Increasing temperature increases the average kinetic energy of particles. According to the Maxwell-Boltzmann distribution, a much larger fraction of particles will have energy E≥EaE \geq E_a, leading to more frequent successful collisions.

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Catalysts: A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy (EaE_a). The catalyst is not consumed in the process.

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Surface Area: For heterogeneous reactions involving solids, increasing the surface area (e.g., by using a powder) increases the number of particles exposed to collisions, increasing the reaction rate.

📐Formulae

Average Rate=Δ[Concentration]Δt\text{Average Rate} = \frac{\Delta [\text{Concentration}]}{\Delta t}

Rate=Volume of gasTime taken\text{Rate} = \frac{\text{Volume of gas}}{\text{Time taken}}

Rate=k[A]m[B]n\text{Rate} = k[A]^m[B]^n

k=Ae−EaRTk = A e^{-\frac{E_a}{RT}}

💡Examples

Problem 1:

In the decomposition of hydrogen peroxide, 2H2O2(aq)→2H2O(l)+O2(g)2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g), a student collects 40 cm340\text{ cm}^3 of O2O_2 gas in 10 seconds10\text{ seconds}. Calculate the average rate of reaction in cm3s−1\text{cm}^3\text{s}^{-1}.

Solution:

Rate=ΔVΔt=40 cm310 s=4.0 cm3s−1\text{Rate} = \frac{\Delta V}{\Delta t} = \frac{40\text{ cm}^3}{10\text{ s}} = 4.0\text{ cm}^3\text{s}^{-1}

Explanation:

The rate of reaction is calculated by dividing the change in the volume of the gaseous product by the total time taken for that change.

Problem 2:

If the rate law for a reaction A+B→CA + B \rightarrow C is given by Rate=k[A]2[B]\text{Rate} = k[A]^2[B], calculate the factor by which the rate increases if the concentration of [A][A] is doubled and [B][B] remains constant.

Solution:

New Rate=k(2[A])2[B]=4(k[A]2[B])=4×Original Rate\text{New Rate} = k(2[A])^2[B] = 4(k[A]^2[B]) = 4 \times \text{Original Rate}

Explanation:

Since the reaction is second order with respect to AA, doubling the concentration of AA results in a 22=42^2 = 4 fold increase in the reaction rate.

Rate of reaction - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry