Question

For combustion of 1 mole of liquid benzene at 298 K, the heat of reaction at constant volume is –3264.2 kJ. What is the heat of combustion at constant pressure?
(\(R = 8.314\ \text{J K}^{-1}\text{ mol}^{-1}\))

• A. –2439.2 kJ mol\(^{-1}\)
• B. –816.9 kJ mol\(^{-1}\)
• C. –3267.9 kJ mol\(^{-1}\)
• D. –1633.9 kJ mol\(^{-1}\)

Hint:

Use \(\Delta H = \Delta U + \Delta nRT\) carefully by counting only gaseous moles.

Answer 1

The Correct Option is C

Step 1: Write the relation between \(\Delta H\) and \(\Delta U\).
\[ \Delta H = \Delta U + \Delta nRT \]
Step 2: Write balanced combustion equation of benzene.
\[ \mathrm{C_6H_6(l) + \tfrac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)} \]
Step 3: Calculate change in moles of gas.
Moles of gaseous products = 6
Moles of gaseous reactants = 7.5
\[ \Delta n = 6 - 7.5 = -1.5 \]
Step 4: Substitute values.
\[ \Delta H = -3264.2 + (-1.5)(8.314 \times 298 \times 10^{-3}) \]
\[ \Delta H = -3264.2 - 3.7 \]
\[ \Delta H = -3267.9\ \text{kJ mol}^{-1} \]
Step 5: Conclusion.
The heat of combustion at constant pressure is –3267.9 kJ mol\(^{-1}\).