Question

question 42
given the following tabulated data, calculate the enthalpy of reaction (δhᵣₓₙ) for the combustion of ethane:
2 c₂h₆(g) + 7 o₂(g) → 4 co₂(g) + 6 h₂o(l)
δhᵣₓₙ = ?

compound	δhᵣ° (kj mol⁻¹)
c₂h₆(g)	-85.0
c₃h₈(g)	-103.9
co(g)	-110.5
co₂(g)	-393.5
h₂o(l)	-285.8

options:
-1359 kj
-3119 kj
-764.3 kj
-3459 kj

Explanation:

Step1: Recall Hess's Law

Hess's Law states that \(\Delta H_{\text{rxn}}=\sum \Delta H_f^{\circ}(\text{products})-\sum \Delta H_f^{\circ}(\text{reactants})\). For the reaction \(2\text{C}_2\text{H}_6(\text{g}) + 7\text{O}_2(\text{g})
ightarrow 4\text{CO}_2(\text{g})+6\text{H}_2\text{O}(\text{l})\), \(\text{O}_2(\text{g})\) has \(\Delta H_f^{\circ} = 0\) (standard enthalpy of formation of elements in their standard state is 0).

Step2: Identify \(\Delta H_f^{\circ}\) values

• \(\Delta H_f^{\circ}(\text{C}_2\text{H}_6(\text{g}))=-85.0\ \text{kJ/mol}\)
• \(\Delta H_f^{\circ}(\text{CO}_2(\text{g}))=-393.5\ \text{kJ/mol}\)
• \(\Delta H_f^{\circ}(\text{H}_2\text{O}(\text{l}))=-285.8\ \text{kJ/mol}\)
• \(\Delta H_f^{\circ}(\text{O}_2(\text{g})) = 0\ \text{kJ/mol}\)

Step3: Calculate \(\sum \Delta H_f^{\circ}(\text{products})\)

Products: \(4\ \text{mol}\ \text{CO}_2\) and \(6\ \text{mol}\ \text{H}_2\text{O}\).
\(\sum \Delta H_f^{\circ}(\text{products})=4\times\Delta H_f^{\circ}(\text{CO}_2)+6\times\Delta H_f^{\circ}(\text{H}_2\text{O})\)
\(= 4\times(-393.5)+6\times(-285.8)\)
\(=-1574 - 1714.8=-3288.8\ \text{kJ}\)

Step4: Calculate \(\sum \Delta H_f^{\circ}(\text{reactants})\)

Reactants: \(2\ \text{mol}\ \text{C}_2\text{H}_6\) and \(7\ \text{mol}\ \text{O}_2\).
\(\sum \Delta H_f^{\circ}(\text{reactants})=2\times\Delta H_f^{\circ}(\text{C}_2\text{H}_6)+7\times\Delta H_f^{\circ}(\text{O}_2)\)
\(=2\times(-85.0)+7\times0=-170\ \text{kJ}\)

Step5: Calculate \(\Delta H_{\text{rxn}}\)

\(\Delta H_{\text{rxn}}=\sum \Delta H_f^{\circ}(\text{products})-\sum \Delta H_f^{\circ}(\text{reactants})\)
\(=-3288.8-(-170)=-3288.8 + 170=-3118.8\approx - 3119\ \text{kJ}\) (close to the option -3119 kJ)

Answer:

-3119 kJ (corresponding to the option " -3119 kJ")