Question

Heat of combustion of \(\text{C(s)}\), \(\text{H}_2(g)\) and \(\text{C}_2\text{H}_6(g)\) are \(-x_1\), \(-x_2\) and \(-x_3\) respectively. Hence, the heat of formation of \(\text{C}_2\text{H}_6(g)\) is

• A. \(-x_1 - x_2 + x_3\)
• B. \(-2x_1 - 3x_2 + x_3\)
• C. \(x_1 + x_2 - x_3\)
• D. \(-x_3 + 2x_1 + 3x_2\)

Hint:

Hess’s law allows calculation of enthalpy change using alternate reaction pathways.

Answer 1

The Correct Option is B

Step 1: Write the formation reaction of ethane.
\[ 2\text{C(s)} + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \] Step 2: Apply Hess’s law using combustion reactions.
Combustion of reactants and products gives: \[ 2\text{C} + 2\text{O}_2 \rightarrow 2\text{CO}_2 \quad (\Delta H = -2x_1) \] \[ 3\text{H}_2 + \frac{3}{2}\text{O}_2 \rightarrow 3\text{H}_2\text{O} \quad (\Delta H = -3x_2) \] \[ \text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \quad (\Delta H = -x_3) \] Step 3: Calculate heat of formation.
\[ \Delta H_f = (-2x_1 - 3x_2) - (-x_3) \] \[ \Delta H_f = -2x_1 - 3x_2 + x_3 \] Step 4: Conclusion.
Thus, the heat of formation of ethane is \(-2x_1 - 3x_2 + x_3\).