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

Grade 11IGCSEChemistry

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

🔑Concepts

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 dm3s1\text{mol dm}^{-3}\text{s}^{-1} or g/s\text{g/s}.

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.

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.

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.

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.

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.

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}}

RateFrequency 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 dm32.0\text{ mol dm}^{-3} HClHCl. Calculate the average rate of reaction in cm3s1\text{cm}^3\text{s}^{-1}.

Solution:

Rate=50 cm320 s=2.5 cm3s1\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.

Rate of reaction (Concentration, Temperature, Surface area, Catalysts) Revision - Grade 11…