krit.club logo

Chemical Energetics - Enthalpy changes

Grade 12IGCSEChemistry

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

🔑Concepts

•

Enthalpy change ΔH\Delta H is defined as the heat energy change measured under conditions of constant pressure. It is measured in kJ mol−1kJ\,mol^{-1}.

•

In an Exothermic reaction, energy is released to the surroundings, making ΔH\Delta H negative (ΔH<0\Delta H < 0). The temperature of the surroundings increases.

•

In an Endothermic reaction, energy is absorbed from the surroundings, making ΔH\Delta H positive (ΔH>0\Delta H > 0). The temperature of the surroundings decreases.

•

Standard Enthalpy of Formation ΔHf⊖\Delta H^\ominus_f is the enthalpy change when 1 mole1\text{ mole} of a compound is formed from its constituent elements in their standard states under standard conditions (298 K298\text{ K} and 100 kPa100\text{ kPa}).

•

Standard Enthalpy of Combustion ΔHc⊖\Delta H^\ominus_c is the enthalpy change when 1 mole1\text{ mole} of a substance is burned completely in excess oxygen under standard conditions.

•

Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final states are the same.

•

Average Bond Enthalpy is the energy required to break 1 mole1\text{ mole} of a given covalent bond in the gaseous state, averaged over a range of compounds.

•

Bond breaking is an endothermic process (requires energy), while bond making is an exothermic process (releases energy).

📐Formulae

q=m×c×ΔTq = m \times c \times \Delta T

ΔH=−qn\Delta H = \frac{-q}{n}

ΔHreaction=∑ΔHf⊖(products)−∑ΔHf⊖(reactants)\Delta H_{reaction} = \sum \Delta H^\ominus_f(\text{products}) - \sum \Delta H^\ominus_f(\text{reactants})

ΔHreaction=∑ΔHc⊖(reactants)−∑ΔHc⊖(products)\Delta H_{reaction} = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})

ΔH=∑Bond Enthalpiesbroken−∑Bond Enthalpiesformed\Delta H = \sum \text{Bond Enthalpies}_{broken} - \sum \text{Bond Enthalpies}_{formed}

💡Examples

Problem 1:

Calculate the enthalpy change for the reaction: C2H4(g)+H2(g)→C2H6(g)C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g). Given bond enthalpies: C=CC=C is 614 kJ/mol614\text{ kJ/mol}, C−CC-C is 348 kJ/mol348\text{ kJ/mol}, C−HC-H is 413 kJ/mol413\text{ kJ/mol}, and H−HH-H is 436 kJ/mol436\text{ kJ/mol}.

Solution:

Bonds broken: 1×(C=C)+1×(H−H)+4×(C−H)=614+436+(4×413)=2702 kJ/mol1 \times (C=C) + 1 \times (H-H) + 4 \times (C-H) = 614 + 436 + (4 \times 413) = 2702\text{ kJ/mol}. Bonds formed: 1×(C−C)+6×(C−H)=348+(6×413)=2826 kJ/mol1 \times (C-C) + 6 \times (C-H) = 348 + (6 \times 413) = 2826\text{ kJ/mol}. ΔH=2702−2826=−124 kJ/mol\Delta H = 2702 - 2826 = -124\text{ kJ/mol}.

Explanation:

The enthalpy change is calculated by subtracting the total energy released during bond formation from the total energy required for bond breaking. Since the result is negative, the reaction is exothermic.

Problem 2:

In a calorimetry experiment, 0.50 g0.50\text{ g} of hexane (C6H14C_6H_{14}, Mr=86.0M_r = 86.0) is burned to heat 200 g200\text{ g} of water. The temperature of the water rises from 20.0∘C20.0^\circ C to 45.0∘C45.0^\circ C. Calculate the enthalpy of combustion of hexane. (Specific heat capacity of water c=4.18 J g−1K−1c = 4.18\text{ J g}^{-1}K^{-1})

Solution:

q=mcΔT=200×4.18×(45.0−20.0)=20900 J=20.9 kJq = m c \Delta T = 200 \times 4.18 \times (45.0 - 20.0) = 20900\text{ J} = 20.9\text{ kJ}. Moles of hexane n=0.5086.0=0.00581 moln = \frac{0.50}{86.0} = 0.00581\text{ mol}. ΔHc=−20.90.00581=−3597 kJ/mol\Delta H_c = \frac{-20.9}{0.00581} = -3597\text{ kJ/mol}.

Explanation:

First, calculate the heat energy absorbed by the water using q=mcΔTq=mc\Delta T. Then, divide the heat (with a negative sign for an exothermic combustion) by the number of moles of the fuel burned to find the molar enthalpy of combustion.

Enthalpy changes - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry