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Difficulty: MediumStandard Enthalpy Changes and Hess's Law
Calculate the standard enthalpy of combustion of liquid carbon disulfide (CS2(l)\text{CS}_2(l)) in kJ mol−1\text{kJ mol}^{-1}, given the following standard enthalpies of formation:
ΔHf∘[CS2(l)]=+88 kJ mol−1\Delta H_f^\circ[\text{CS}_2(l)] = +88\text{ kJ mol}^{-1}
ΔHf∘[CO2(g)]=−394 kJ mol−1\Delta H_f^\circ[\text{CO}_2(g)] = -394\text{ kJ mol}^{-1}
ΔHf∘[SO2(g)]=−297 kJ mol−1\Delta H_f^\circ[\text{SO}_2(g)] = -297\text{ kJ mol}^{-1}
The balanced chemical equation for the combustion process is:
CS2(l)+3O2(g)→CO2(g)+2SO2(g)\text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g)

What is the standard enthalpy change of combustion in kJ mol−1\text{kJ mol}^{-1}?

Answer: -1076 kJ mol^-1

Answer

The standard enthalpy of combustion of liquid carbon disulfide is -1076 kJ mol^-1.
Applying Hess's law using standard enthalpies of formation gives ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}). For the combustion of carbon disulfide, this evaluates to [(−394)+2(−297)]−(+88)=−988−88=−1076 kJ mol−1[(-394) + 2(-297)] - (+88) = -988 - 88 = -1076\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Formulate the standard enthalpy of reaction equation using standard enthalpies of formation.
\Delta H^\circ = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})
According to Hess's Law, the net standard enthalpy change for a chemical process equals the sum of standard formation enthalpies of products minus reactants.
2
Substitute the provided standard enthalpy of formation values into the expression, taking into account stoichiometry.
\Delta H^\circ = [-394 + 2(-297)] - [88] = -1076\text{ kJ mol}^{-1}
Sulfur dioxide is formed with a mole ratio of 2, so its formation enthalpy must be doubled; oxygen gas has a formation enthalpy of zero.

Key Concept

Standard Enthalpy Changes and Hess's Law
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Topics

Standard Enthalpy Changes and Hess's Law