Consider the following date Heat of combustion of $\mathrm{H}_{2}(\mathrm{~g})=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Heat of combustion of $\mathrm{C}(\mathrm{s})=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Heat of combustion of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})=-1234.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The heat of formation of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ is (-) ____ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer)

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Chemical Thermodynamics

Consider the following date

Heat of combustion of

H_{2} (g) = - 241.8 kJ mol^{- 1}

Heat of combustion of

C (s) = - 393.5 kJ mol^{- 1}

Heat of combustion of

C_{2} H_{5} OH (l) = - 1234.7 kJ mol^{- 1}

.

The heat of formation of

C_{2} H_{5} OH (l)

is (-) ____

kJ mol^{- 1}

(Nearest integer)

Answer: 278

The enthalpy of formation corresponds to the reaction

2 C (s) + 3 H_{2} (g) + \frac{1}{2} O_{2} (g) \to C_{2} H_{5} OH (l)

. Forming ethanol from its elements is equivalent to burning the elements and then reversing the combustion of ethanol, so

Δ H_{f} = 2 Δ H_{c} (C) + 3 Δ H_{c} (H_{2}) - Δ H_{c} (C_{2} H_{5} OH)

. Substituting the data,

Δ H_{f} = 2 (- 393.5) + 3 (- 241.8) - (- 1234.7) = - 787.0 - 725.4 + 1234.7 = - 277.7 kJ mo l^{- 1}

. The heat of formation of ethanol is therefore

- 278 kJ mo l^{- 1}

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer