The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be 3263.9 kJ mol^–1 at 25°C. Calculate the heat of combustion of benzene at constant pressure. 

The reaction is:  C₆H₆(l) + 7 1/2O₂(g) → 6CO₂(g) + 3H₂O(l) 

In this reaction, O₂ is the only gaseous reactant and CO₂ is the only gaseous product.

∴   Δn_g = n_p − n_r = 6 − 7 1/2 = −1 1/2 =−3/2

Also, we are given  ΔU  (or  q_v) = 3263.9  kJ mol⁻¹

T = 25°C = 298 K

R = 8.314 J K⁻¹ mol⁻¹ = 8.314/1000 kJ K⁻¹mol⁻¹ 

ΔH (or q_p) = ΔU + Δn_gRT = −3263.9 kJ mol⁻¹ +(−3/2 mol) (8.314/1000 kJ K⁻¹mol⁻¹) (298 K) 

= −3263.9 − 3.7 kJ mol⁻¹ = −3267.6 kJ mol⁻¹