For the reaction $\mathrm{H}_{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$ $\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$. The enthalpy change for the above reaction is (-) \_\_\_\_ $\mathrm{kJ} \mathrm{mol}^{-1}$ [nearest integer] Given : $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Chemical Thermodynamics

For the reaction

H_{2} F_{2} (g) \to H_{2} (g) + F_{2} (g)

Δ U = - 59.6 kJ mol^{- 1}

2 7^{\circ} C

.
The enthalpy change for the above reaction is (-)
____

kJ mol^{- 1}

[nearest integer] Given :

R = 8.314 JK^{- 1} mol^{- 1}

Answer: 57

Using the relation

Δ H = Δ U + Δ n_{g} R T

, where

Δ n_{g}

is the change in number of moles of gas:

Δ n_{g} = (1 + 1) - 1 = 1

mol

Δ H = - 59.6 + (1) (8.314 \times 1 0^{- 3}) (300)

Δ H = - 59.6 + 2.49 = - 57.11 \approx - 57

kJ/mol

The answer is 57 (magnitude).

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Integer answer