Consider the following data : Heat of formation of $\mathrm{CO}_{2}(\mathrm{~g})=-393.5\mathrm{~kJ}\mathrm{~mol}^{-1}$ Heat of formation of $\mathrm{H}_{2}\mathrm{O}(\mathrm{l})=-286.0\mathrm{~kJ}\mathrm{~mol}^{-1}$ Heat of combustion of benzene $=-3267.0\mathrm{~kJ}\mathrm{~mol}^{-1}$ The heat of formation of benzene is \_\_\_\_ $\mathrm{kJ}\mathrm{mol}^{-1}$. (Nearest integer)

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Chemical Thermodynamics

Consider the following data :

Heat of formation of

CO_{2} (g) = - 393.5 kJ mol^{- 1}

Heat of formation of

H_{2} O (l) = - 286.0 kJ mol^{- 1}

Heat of combustion of benzene

= - 3267.0 kJ mol^{- 1}

The heat of formation of benzene is ____

kJ mol^{- 1}

.
(Nearest integer)

Answer: 48

Benzene's combustion is

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)

, with

Δ H_{c} = - 3267.0 k J / m o l

. By Hess's law,

Δ H_{c} = [6Δ H_{f} (C O_{2}) + 3Δ H_{f} (H_{2} O)] - Δ H_{f} (C_{6} H_{6})

, since

O_{2}

has zero formation enthalpy. Substituting,

- 3267 = [6 (- 393.5) + 3 (- 286.0)] - Δ H_{f} (C_{6} H_{6}) = - 3219 - Δ H_{f} (C_{6} H_{6})

. Solving,

Δ H_{f} (C_{6} H_{6}) = - 3219 + 3267 = 48 k J / m o l

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Integer answer