For the reaction H_2 F_2( g) -> H_2( g)+ F_2( g) Delta U=-59.6 kJ mol^-1 at 27^° C. The enthalpy change for the above reaction is (-) __________ kJ mol^-1 [nearest integer] Given: R=8.314 J K^-1 mol^-1.

H_2 F_2( g) -> H_2( g)+ F_2( g) Delta U=-59.6 kJ mol^-1 at 27^° C Delta H =Delta U+Delta n_g RT \ =-59.6+(1 x 8.314 x 300 / 1000) \ =-57.10 kJ mol^-1

For the reaction H2 F2g arrow H2gF2g Delta U596 kJ mol1... | null (Chemistry) | DivineJEE

Exam

Chemistry

For the reaction

$\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$

$\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$ .

The enthalpy change for the above reaction is ( $-$ ) __________ $\mathrm{kJ} \,\mathrm{mol}^{-1}$ [nearest integer]

Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ .

Correct Answer:

\mathrm{H}_{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})

\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}

27^{\circ} \mathrm{C}

\begin{aligned} \Delta \mathrm{H} &=\Delta \mathrm{U}+\Delta \mathrm{n}_{g} \mathrm{RT} \\ &=-59.6+\frac{1 \times 8.314 \times 300}{1000} \\ &=-57.10 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

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