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Chemical Energetics - Exothermic and endothermic reactions

Grade 12A LevelChemistry

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

🔑Concepts

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Enthalpy change (ΔH\Delta H) is the heat energy change measured under constant pressure conditions. It is expressed in units of kJ⋅mol−1kJ \cdot mol^{-1}.

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Exothermic reactions release thermal energy to the surroundings. Consequently, the temperature of the surroundings increases, and the enthalpy of the products is lower than the reactants (ΔH<0\Delta H < 0).

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Endothermic reactions absorb thermal energy from the surroundings. The temperature of the surroundings decreases, and the enthalpy of the products is higher than the reactants (ΔH>0\Delta H > 0).

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Activation Energy (EaE_a) is the minimum kinetic energy that colliding particles must possess in order for a chemical reaction to occur.

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Bond breaking is an endothermic process (requires energy), whereas bond forming is an exothermic process (releases energy).

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In an exothermic reaction, the energy released when new bonds are formed is greater than the energy required to break existing bonds.

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In an endothermic reaction, the energy required to break existing bonds is greater than the energy released when new bonds are formed.

📐Formulae

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

ΔH=∑Bond Energies (Reactants)−∑Bond Energies (Products)\Delta H = \sum \text{Bond Energies (Reactants)} - \sum \text{Bond Energies (Products)}

q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T

ΔH=−qn\Delta H = \frac{-q}{n}

💡Examples

Problem 1:

Calculate the enthalpy change (ΔH\Delta H) for the combustion of methane: CH4(g)+2O2(g)→CO2(g)+2H2O(g)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g). Use the following average bond energies: C−H=413 kJ/molC-H = 413 \, kJ/mol, O=O=498 kJ/molO=O = 498 \, kJ/mol, C=O=805 kJ/molC=O = 805 \, kJ/mol, O−H=464 kJ/molO-H = 464 \, kJ/mol.

Solution:

  1. Energy to break bonds (Reactants): 4×(C−H)+2×(O=O)=4(413)+2(498)=1652+996=2648 kJ/mol4 \times (C-H) + 2 \times (O=O) = 4(413) + 2(498) = 1652 + 996 = 2648 \, kJ/mol.
  2. Energy released forming bonds (Products): 2×(C=O)+4×(O−H)=2(805)+4(464)=1610+1856=3466 kJ/mol2 \times (C=O) + 4 \times (O-H) = 2(805) + 4(464) = 1610 + 1856 = 3466 \, kJ/mol.
  3. ΔH=2648−3466=−818 kJ/mol\Delta H = 2648 - 3466 = -818 \, kJ/mol.

Explanation:

Since ΔH\Delta H is negative (−818 kJ/mol-818 \, kJ/mol), the combustion of methane is an exothermic reaction. More energy is released during the formation of C=OC=O and O−HO-H bonds in the products than is required to break the C−HC-H and O=OO=O bonds in the reactants.

Problem 2:

A student heats 100 g100 \, g of water using the combustion of an alcohol. The temperature of the water increases from 20∘C20^{\circ}C to 45∘C45^{\circ}C. Calculate the heat energy qq absorbed by the water. (Specific heat capacity of water c=4.18 J⋅g−1⋅K−1c = 4.18 \, J \cdot g^{-1} \cdot K^{-1})

Solution:

q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T q=100 g×4.18 J⋅g−1⋅K−1×(45−20) Kq = 100 \, g \times 4.18 \, J \cdot g^{-1} \cdot K^{-1} \times (45 - 20) \, K q=100×4.18×25=10,450 J=10.45 kJq = 100 \times 4.18 \times 25 = 10,450 \, J = 10.45 \, kJ

Explanation:

The formula q=mcΔTq = mc\Delta T calculates the total heat transferred to the water. Since the temperature increased, the reaction providing the heat must be exothermic.