The difference between heats of reaction at constant pressure and constant volume of the following reaction would be;

2C₆H₆ (l)+ 15O₂ ---> 12CO₂ (g ) + 6H₂O(l) at 25° C in kJ is

To find the difference between the heat of reaction at constant pressure (ΔH) and at constant volume (ΔU) we use the thermodynamic relation:

$\Delta H\; -\; \Delta U\; =\; \Delta n\_\{gas\}·R·T$

where:

• Δn_gas = (moles of gaseous products) – (moles of gaseous reactants)
• R = 8.314 J·mol⁻¹·K⁻¹ (the ideal‑gas constant)
• T = 298 K (25 °C)

First, count the gaseous molecules in the reaction:

Reactants: 15 mol O₂(g) (the liquid benzene does not count as gas).

Products: 12 mol CO₂(g) (the liquid water is not gas).

Therefore

$\Delta n\_\{gas\}\; =\; 12\; -\; 15\; =\; -3$

Now substitute the values into the equation:

$\Delta H\; -\; \Delta U\; =\; (-3)\; \times \; (8.314 J·mol⁻¹·K⁻¹)\; \times \; (298 K)$

Calculate the product:

$(8.314\; \times \; 298)\; \approx \; 2478 J·mol⁻¹$

$\Delta H\; -\; \Delta U\; =\; -3\; \times \; 2478 J\; \approx \; -7434 J$

Convert joules to kilojoules:

$-7434 J\; =\; -7.434 kJ\; \approx \; -7.43 kJ$

Thus the heat of reaction at constant pressure is lower than the heat of reaction at constant volume by about 7.43 kJ. The required difference is:

$\Delta H\; -\; \Delta U\; =\; -7.43 kJ$