Question

For the combustion of 1 mole of liquid benzene at 298K, the heat of reaction at constant pressure is -3268 kJ mol\(^{-1}\), what is the heat of combustion at constant volume?

• A. - 3264.2 kJ mol\(^{-1}\)
• B. - 1632 kJ mol\(^{-1}\)
• C. - 6728 kJ mol\(^{-1}\)
• D. - 672.8 kJ mol\(^{-1}\)

Hint:

When converting between heat at constant pressure and constant volume, remember to include the effect of volume change.

Answer 1

The Correct Option is A

Step 1: Understanding the relation.
The heat of combustion at constant volume is related to the heat at constant pressure using the formula: \[ \Delta H = \Delta U + P \Delta V \] where \( \Delta H \) is the heat of reaction at constant pressure and \( \Delta U \) is the heat at constant volume.

Step 2: Applying the formula.
Given that \( R = 8.314 \times 10^{-3} \, \text{kJ mol}^{-1} \text{K}^{-1} \), we can calculate the difference between the heat of combustion at constant pressure and constant volume. Using the given values, we get: \[ \Delta U = \Delta H - P \Delta V = -3268 + 4.8 = -3264.2 \, \text{kJ mol}^{-1} \]
Step 3: Conclusion.
The correct answer is (A) - 3264.2 kJ mol\(^{-1}\), as it is the heat of combustion at constant volume.