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Chemistry - Chemical reactions (Rates of reaction and Reversible reactions)

Grade 9IGCSE

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, volume, or mass of a reactant or product per unit time.

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Collision Theory states that for a reaction to occur, particles must collide with sufficient energy, known as the Activation Energy (EaE_a), and the correct orientation.

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Increasing the surface area of solid reactants increases the frequency of collisions, thereby increasing the reaction rate.

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Increasing the concentration of solutions or pressure of gases increases the number of particles per unit volume, leading to more frequent collisions.

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Increasing temperature increases the kinetic energy of particles, leading to more frequent collisions and, more importantly, a higher proportion of collisions having energy ≥Ea\ge E_a.

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A catalyst increases the rate of reaction by providing an alternative pathway with a lower Activation Energy (EaE_a), without being used up in the process.

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A reversible reaction is one where the products can react together to reform the original reactants, represented by the symbol ⇌\rightleftharpoons.

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Dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remains constant.

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Le Chatelier's Principle states that if a change (in temperature, pressure, or concentration) is applied to a system at equilibrium, the system will shift its position to counteract that change.

📐Formulae

Rate of reaction=Amount of reactant usedTime taken\text{Rate of reaction} = \frac{\text{Amount of reactant used}}{\text{Time taken}}

Rate of reaction=Amount of product formedTime taken\text{Rate of reaction} = \frac{\text{Amount of product formed}}{\text{Time taken}}

N2(g)+3H2(g)⇌2NH3(g)ΔH=−92 kJ/molN_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92 \text{ kJ/mol}

CuSO4⋅5H2O(s)⇌CuSO4(s)+5H2O(l)CuSO_4 \cdot 5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(l)

💡Examples

Problem 1:

In an experiment between magnesium and excess hydrochloric acid, 48 cm348\text{ cm}^3 of hydrogen gas (H2H_2) was collected in 22 minutes. Calculate the average rate of reaction in cm3/s\text{cm}^3/\text{s}.

Solution:

Rate=48 cm3120 s=0.4 cm3/s\text{Rate} = \frac{48\text{ cm}^3}{120\text{ s}} = 0.4\text{ cm}^3/\text{s}

Explanation:

First, convert the time from minutes to seconds (2×60=120 s2 \times 60 = 120\text{ s}). Then divide the total volume of gas produced by the time taken.

Problem 2:

Predict the effect of increasing the pressure on the equilibrium position of the reaction: 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g).

Solution:

The equilibrium will shift to the right (towards the products side).

Explanation:

There are 33 moles of gas on the left (22 of SO2SO_2 and 11 of O2O_2) and 22 moles of gas on the right (SO3SO_3). According to Le Chatelier's Principle, increasing pressure shifts the equilibrium to the side with fewer moles of gas to reduce the pressure.

Problem 3:

How does a catalyst affect the position of equilibrium in the Haber process: N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)?

Solution:

A catalyst has no effect on the position of equilibrium.

Explanation:

A catalyst increases the rate of both the forward and the reverse reactions equally. Therefore, while it helps reach equilibrium faster, it does not change the final yield or the position of the equilibrium.