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Chemical Energetics - Exothermic and endothermic reactions

Grade 11IGCSEChemistry

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

🔑Concepts

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An exothermic reaction is one that releases heat energy to the surroundings, causing the temperature of the surroundings to increase. In these reactions, the enthalpy change ΔH\Delta H is negative (ΔH<0\Delta H < 0).

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An endothermic reaction is one that absorbs heat energy from the surroundings, causing the temperature of the surroundings to decrease. In these reactions, the enthalpy change ΔH\Delta H is positive (ΔH>0\Delta H > 0).

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Bond breaking is an endothermic process because energy must be supplied to overcome the forces of attraction between atoms.

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Bond making is an exothermic process because energy is released when new chemical bonds are formed.

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The overall enthalpy change of a reaction depends on the balance between the energy taken in to break bonds in the reactants and the energy released when new bonds are formed in the products.

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The Activation Energy (EaE_a) is the minimum energy that colliding particles must possess for a reaction to occur. It is represented as the 'hump' in a reaction profile diagram.

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In a reaction profile for an exothermic reaction, the reactants are at a higher energy level than the products. In an endothermic reaction, the reactants are at a lower energy level than the products.

📐Formulae

ΔH=Energy required to break bonds−Energy released making bonds\Delta H = \text{Energy required to break bonds} - \text{Energy released making bonds}

ΔH=∑(bond energies of reactants)−∑(bond energies of products)\Delta H = \sum (\text{bond energies of reactants}) - \sum (\text{bond energies of products})

Q=mcΔTQ = m c \Delta T

💡Examples

Problem 1:

Calculate the enthalpy change (ΔH\Delta H) for the combustion of methane: CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O. Given bond energies: C−H=413 kJ/molC-H = 413\text{ kJ/mol}, O=O=495 kJ/molO=O = 495\text{ kJ/mol}, C=O=799 kJ/molC=O = 799\text{ kJ/mol}, O−H=463 kJ/molO-H = 463\text{ kJ/mol}.

Solution:

  1. Energy in (breaking bonds): (4×413)+(2×495)=1652+990=2642 kJ/mol(4 \times 413) + (2 \times 495) = 1652 + 990 = 2642\text{ kJ/mol}. \n2. Energy out (forming bonds): (2×799)+(4×463)=1598+1852=3450 kJ/mol(2 \times 799) + (4 \times 463) = 1598 + 1852 = 3450\text{ kJ/mol}. \n3. ΔH=2642−3450=−808 kJ/mol\Delta H = 2642 - 3450 = -808\text{ kJ/mol}.

Explanation:

Since the energy released during bond making (3450 kJ/mol3450\text{ kJ/mol}) is greater than the energy required for bond breaking (2642 kJ/mol2642\text{ kJ/mol}), the reaction is exothermic, resulting in a negative enthalpy change.

Problem 2:

A reaction has a total energy of 432 kJ432\text{ kJ} absorbed to break reactant bonds and 340 kJ340\text{ kJ} released when product bonds are formed. State the enthalpy change and whether the reaction is exothermic or endothermic.

Solution:

ΔH=432 kJ−340 kJ=+92 kJ\Delta H = 432\text{ kJ} - 340\text{ kJ} = +92\text{ kJ}

Explanation:

Because the enthalpy change ΔH\Delta H is positive (+92 kJ+92\text{ kJ}), the reaction is endothermic. More energy was required to break the bonds than was released when new bonds formed.

Exothermic and endothermic reactions Revision - Grade 11 Chemistry IGCSE