For the reaction H_2 F_2( g ) → H_2( g )+ F_2( g ); Δ U=-59.6 kJ mol^-1 at 27^° C. The enthalpy change for the above reaction is (-) \ \ \ \ kJ…

For the reaction $$\begin{aligned} & \mathrm{H}_2 \mathrm{~F}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \\ & \Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { at } 27^{\circ} \mathrm{C} . \end{aligned} $$ The enthalpy change for the above reaction is (-) \_\_\_\_ $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$ [nearest integer] Given: $\displaystyle \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.