Calculate standard internal energy change for $\mathrm{OF}_{2(\mathrm{~g})}+\mathrm{H}_2…

Calculate standard internal energy change for $\mathrm{OF}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \longrightarrow 2 \mathrm{HF}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})}$ at 300 K, if $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ of $\mathrm{OF}_{2(\mathrm{~g})}, \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ and $\mathrm{HF}_{(\mathrm{g})}$ are 20, -250 and $\quad-270 \quad \mathrm{~kJ} \quad \mathrm{~mol}^{-1} \quad$ respectively. $\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
  1. $\quad-307.50 \mathrm{~kJ}$
  2. -342.48 kJ
  3. -412.00 kJ
  4. -214.48 kJ

Solution

$\begin{aligned} \Delta_{\mathrm{f}} \mathrm{H}^{\circ} & =\sum \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text { of product }-\sum \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text { of Reactants } \\ & =[(2 \times-270+0)-(20-250)] \mathrm{kJ} \mathrm{mol}^{-1} \\ & =-310 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \\ \Delta \mathrm{H}^{\circ} & =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ \Delta \mathrm{U}^{\circ} & =\Delta \mathrm{H}^{\circ}-\Delta \mathrm{ng}_{\mathrm{g}} \mathrm{RT} \\ & =-310 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}-(3-2) \times \\ & 8.314 \mathrm{~J} \mathrm{~K} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \\ \Delta \mathrm{U}^{\circ} & =-307.50 \mathrm{~kJ} \quad\end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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