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Chemical Reactions - Rate of reaction (Concentration, Temperature, Surface area, Catalysts)

Grade 11A LevelChemistry

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

🔑Concepts

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The rate of reaction is the change in concentration of a reactant or product per unit time. It is commonly expressed in units such as mol dm−3s−1\text{mol dm}^{-3}\text{s}^{-1} or g/s\text{g/s}.

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Collision Theory states that for a reaction to occur, reactant particles must collide with each other with a minimum amount of energy called the activation energy (EaE_a) and in the correct orientation.

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Increasing the concentration of a solution or the pressure of a gas increases the number of particles per unit volume. This leads to a higher frequency of effective collisions, thus increasing the reaction rate.

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Raising the temperature increases the average kinetic energy of the particles. This leads to a higher frequency of collisions and, crucially, a much larger fraction of particles possessing energy ≥Ea\ge E_a.

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For solid reactants, increasing the surface area (by grinding the solid into a powder) exposes more particles to the other reactant, increasing the frequency of collisions at the surface.

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A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy (EaE_a). The catalyst itself remains chemically unchanged at the end of the reaction.

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The rate of reaction can be measured by monitoring the volume of gas produced (VV), the decrease in mass of the reaction mixture (mm), or the time taken for a color change/precipitate to form.

📐Formulae

Average Rate=Change in ConcentrationTime Taken=Δ[C]Δt\text{Average Rate} = \frac{\text{Change in Concentration}}{\text{Time Taken}} = \frac{\Delta [C]}{\Delta t}

Rate (from gas volume)=Volume of gas producedTime taken\text{Rate (from gas volume)} = \frac{\text{Volume of gas produced}}{\text{Time taken}}

Rate∝Frequency of successful collisions\text{Rate} \propto \text{Frequency of successful collisions}

💡Examples

Problem 1:

In an experiment, 50 cm350\text{ cm}^3 of hydrogen gas (H2H_2) was collected in 20 s20\text{ s} when magnesium ribbon reacted with excess 2.0 mol dm−32.0\text{ mol dm}^{-3} HClHCl. Calculate the average rate of reaction in cm3s−1\text{cm}^3\text{s}^{-1}.

Solution:

Rate=50 cm320 s=2.5 cm3s−1\text{Rate} = \frac{50\text{ cm}^3}{20\text{ s}} = 2.5\text{ cm}^3\text{s}^{-1}

Explanation:

The rate is determined by dividing the total volume of gas produced by the total time taken for that production.

Problem 2:

Explain why the reaction between CaCO3(s)CaCO_3(s) and HCl(aq)HCl(aq) is faster when using 5 g5\text{ g} of marble chips (powder) compared to 5 g5\text{ g} of marble lumps.

Solution:

The powder has a much larger total surface area than the lumps for the same mass of CaCO3CaCO_3.

Explanation:

A larger surface area means more CaCO3CaCO_3 particles are exposed to the H+H^+ ions in the acid at any given time. This increases the frequency of collisions, which results in a higher rate of reaction.

Problem 3:

A reaction has an activation energy of EaE_a. If a catalyst is added, how does the energy profile change?

Solution:

The catalyst provides a path where the activation energy is Ea,catE_{a, \text{cat}}, such that Ea,cat<EaE_{a, \text{cat}} < E_a.

Explanation:

By lowering the energy barrier, a higher proportion of colliding particles have energy ≥Ea,cat\ge E_{a, \text{cat}}, leading to a greater number of successful collisions per second.