Question

What will be the standard internal energy change for the reaction at 298 K?

$O{F}_{2\left(g\right)}+H_2{O}_{\left(g\right)}\to O_{2\left(g\right)}+2H{F}_{\left(g\right)};△H^0$ = - 310 kJ.

• A. -312.5 kJ
• B. -125.03 kJ
• C. -310 kJ
• D. - 156 kJ

Answer 1

The correct option is C -310 kJ

$\Delta H=\Delta U+\Delta n_{g}RT$

$Here,\Delta n_g=1+2-1-1=1$

$\therefore \Delta U^{\circ }=H^{\circ }-\Delta n_{g}RT$

$\therefore \Delta U^{\circ }=(-310\times {10}^3)-(1\times 8.314\times 298)=-307522.4J$

$\therefore \Delta U\approx -310kJ$

Hence, the correct answer is option $\text{C}$.