The standard internal energy change of combustion of C6H6(l) at 298 K measured in a bomb calorimeter is -3263 kJ mol^-1 for C6H6(l) + 7.5 O2(g) -> 6CO2(g) + 3H2O(l). Taking gases as ideal and R = 8.314 J mol^-1K^-1, calculate Delta H for the combustion in kJ mol^-1 (answer to one decimal place, with sign).

JEE Main levelNumerical answerDifficulty 2 of 3td006

The standard internal energy change of combustion of $C X_{6} H X_{6} (l)$ at $298 K$ measured in a bomb calorimeter is $- 3263 kJ mo l^{- 1}$ for $C X_{6} H X_{6} (l) + 7.5 O X_{2} (g) 6 CO X_{2} (g) + 3 H X_{2} O (l)$ . Taking gases as ideal and $R = 8.314 J mo l^{- 1} K^{- 1}$ , calculate $Δ H$ for the combustion in $kJ mo l^{- 1}$ (answer to one decimal place, with sign).

Numerical answer type, answer in kJ/mol, tolerance 0.1.

Δ n_{g} = 6 - 7.5 = - 1.5

Δ H = Δ U + Δ n_{g} RT = - 3263 + (- 1.5) (8.314) (298) /1000 = - 3266.716 kJ mo l^{- 1}

, i.e.

- 3266.7

to one decimal.