If 3.365 g of ethanol ($l$) burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The $|\Delta \mathrm{H}_{\mathrm{f}}^{\circ}|$ of ethanol at 298.15 K is _____ $\times 10^{2}\mathrm{~kJ\,mol}^{-1}$. (Nearest integer) Given: Standard enthalpy of combustion of graphite $=-393.5\mathrm{~kJ\,mol}^{-1}$ Standard enthalpy of formation of water ($l$) $=-285.8\mathrm{~kJ\,mol}^{-1}$ Molar mass in $\mathrm{gmol}^{-1}$ of C, H, O are 12, 1 and 16 respectively.

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Chemical Thermodynamics

If 3.365 g of ethanol (

l

) burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The

∣Δ H_{f}^{\circ} ∣

of ethanol at 298.15 K is _____

\times 1 0^{2} kJ mol^{- 1}

. (Nearest integer)
Given:
Standard enthalpy of combustion of graphite

= - 393.5 kJ mol^{- 1}

Standard enthalpy of formation of water (

l

)

= - 285.8 kJ mol^{- 1}

Molar mass in

gmol^{- 1}

of C, H, O are 12, 1 and 16 respectively.

Answer: 282

Moles of ethanol

= \frac{3.365}{46} = 0.07315

. The bomb-calorimeter (constant-volume) heat gives

Δ U_{c} = - \frac{99.472}{0.07315} = - 1359.8 kJ mo l^{- 1}

. Converting to enthalpy with

Δ n_{g} = 2 - 3 = - 1

Δ H_{c} = Δ U_{c} + Δ n_{g} R T = - 1359.8 - 2.48 = - 1362.3 kJ mo l^{- 1}

. For

C_{2} H_{5} OH + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O

Δ H_{c} = 2 (- 393.5) + 3 (- 285.8) - Δ H_{f}^{\circ}

, so

Δ H_{f}^{\circ} = - 1644.4 - (- 1362.3) = - 282.1 kJ mo l^{- 1}

. Hence

∣Δ H_{f}^{\circ} ∣ \approx 282 kJ mo l^{- 1}

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer