The standard enthalpy of combustion of liquid benzene, C6H6(l) + 15/2 O2(g) -> 6CO2(g) + 3H2O(l), is -3267 kJ mol^-1 at 298 K. Find Delta U for the reaction in kJ mol^-1 (nearest integer). Use R = 8.314 J mol^-1K^-1.

JEE Main levelNumerical answerDifficulty 2 of 3td081

The standard enthalpy of combustion of liquid benzene, $C X_{6} H X_{6} (l) + \frac{15}{2} O X_{2} (g) 6 CO X_{2} (g) + 3 H X_{2} O (l)$ , is $- 3267 kJ mo l^{- 1}$ at $298 K$ . Find $Δ U$ for the reaction in $kJ mo l^{- 1}$ (nearest integer). Use $R = 8.314 J mo l^{- 1} K^{- 1}$ .

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Δ n_{g} = 6 - 7.5 = - 1.5

Δ U = Δ H - Δ n_{g} RT = - 3267 - (- 1.5) (8.314) (298) /1000 = - 3267 + 3.72 = - 3263.28 kJ mo l^{- 1}