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Chemistry - Kinetics and Energetics (Rates of Reaction and Enthalpy)

Grade 10IB

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

🔑Concepts

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

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Factors Affecting Rate: Concentration/Pressure increases collision frequency; Temperature increases both frequency and the proportion of particles with E>EaE > E_a; Surface Area increases the sites available for collision; Catalysts provide an alternative pathway with a lower EaE_a.

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Rate of Reaction: The change in concentration of a reactant or product per unit time, often expressed in mol⋅dm−3⋅s−1mol \cdot dm^{-3} \cdot s^{-1}.

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Exothermic Reactions: Reactions that release heat energy to the surroundings. The enthalpy of products is less than the enthalpy of reactants, resulting in a negative enthalpy change (ΔH<0\Delta H < 0).

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Endothermic Reactions: Reactions that absorb heat energy from the surroundings. The enthalpy of products is greater than the enthalpy of reactants, resulting in a positive enthalpy change (ΔH>0\Delta H > 0).

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Enthalpy Change (ΔH\Delta H): The heat energy change measured under constant pressure. It is calculated as ΔH=Hproducts−Hreactants\Delta H = H_{products} - H_{reactants}.

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Activation Energy (EaE_a): The minimum energy required for colliding particles to react by breaking existing chemical bonds.

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Bond Enthalpy: Breaking bonds is an endothermic process (requires energy, +ΔH+\Delta H), while forming bonds is an exothermic process (releases energy, −ΔH-\Delta H).

📐Formulae

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

ΔH=Hproducts−Hreactants\Delta H = H_{\text{products}} - H_{\text{reactants}}

q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T

ΔHreaction=∑Bond Enthalpybroken−∑Bond Enthalpyformed\Delta H_{\text{reaction}} = \sum \text{Bond Enthalpy}_{\text{broken}} - \sum \text{Bond Enthalpy}_{\text{formed}}

Rate=Volume of gas producedtime taken\text{Rate} = \frac{\text{Volume of gas produced}}{\text{time taken}}

💡Examples

Problem 1:

In a reaction between MgMg and HClHCl, 48 cm348 \text{ cm}^3 of H2H_2 gas was collected in 20 seconds20 \text{ seconds}. Calculate the average rate of reaction in cm3s−1\text{cm}^3\text{s}^{-1}.

Solution:

Rate=48 cm320 s=2.4 cm3s−1\text{Rate} = \frac{48 \text{ cm}^3}{20 \text{ s}} = 2.4 \text{ cm}^3\text{s}^{-1}

Explanation:

The rate is determined by dividing the total volume of gas produced by the time interval.

Problem 2:

Calculate the enthalpy change for a reaction where the total energy required to break bonds in reactants is 2450 kJ/mol2450 \text{ kJ/mol} and the total energy released when forming bonds in products is 3100 kJ/mol3100 \text{ kJ/mol}. Is the reaction exothermic or endothermic?

Solution:

ΔH=2450 kJ/mol−3100 kJ/mol=−650 kJ/mol\Delta H = 2450 \text{ kJ/mol} - 3100 \text{ kJ/mol} = -650 \text{ kJ/mol}

Explanation:

Since ΔH\Delta H is negative (ΔH<0\Delta H < 0), the reaction is exothermic. More energy is released during bond formation than is absorbed during bond breaking.

Problem 3:

Given that 50 g50 \text{ g} of water (c=4.18 J/g∘Cc = 4.18 \text{ J/g}^{\circ}\text{C}) increased in temperature by 10∘C10^{\circ}\text{C} during a reaction, calculate the heat energy (qq) absorbed by the water.

Solution:

q=50 g⋅4.18 J/g∘C⋅10∘C=2090 Jq = 50 \text{ g} \cdot 4.18 \text{ J/g}^{\circ}\text{C} \cdot 10^{\circ}\text{C} = 2090 \text{ J}

Explanation:

Using the specific heat capacity formula q=mcΔTq = mc\Delta T, we multiply the mass, the specific heat capacity of water, and the temperature change.