MHT CET · Chemistry · Thermodynamics (C)
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. \(\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\)
- A \(-307.50 \mathrm{~kJ}\)
- B -342.48 kJ
- C -412.00 kJ
- D -214.48 kJ
Answer & Solution
Correct Answer
(A) \(-307.50 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\(\Delta_{\mathrm{f}} \mathrm{H}^{\circ} =\sum \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text { of product }-\) \(\sum \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text { of Reactants } \)
\( =[(2 \times-270+0)-(20-250)] \mathrm{kJ} \mathrm{mol}^{-1} \)
\( =-310 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \)
\( \Delta \mathrm{H}^{\circ} =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \)
\( \Delta \mathrm{U}^{\circ} =\Delta \mathrm{H}^{\circ}-\Delta \mathrm{ng}_{\mathrm{g}} \mathrm{RT} \)
\( =-310 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}-(3-2) \times \)
\( 8.314 \mathrm{~J} \mathrm{~K} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \)
\( \Delta \mathrm{U}^{\circ} =-307.50 \mathrm{~kJ}\)
\( =[(2 \times-270+0)-(20-250)] \mathrm{kJ} \mathrm{mol}^{-1} \)
\( =-310 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \)
\( \Delta \mathrm{H}^{\circ} =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \)
\( \Delta \mathrm{U}^{\circ} =\Delta \mathrm{H}^{\circ}-\Delta \mathrm{ng}_{\mathrm{g}} \mathrm{RT} \)
\( =-310 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}-(3-2) \times \)
\( 8.314 \mathrm{~J} \mathrm{~K} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \)
\( \Delta \mathrm{U}^{\circ} =-307.50 \mathrm{~kJ}\)
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