MHT CET · Chemistry · Thermodynamics (C)
Calculate heat of formation of \(\mathrm{HCl}\) gas from following reaction.
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})} ; \Delta \mathrm{H}=-194 \mathrm{~kJ} \)
- A \(-143 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(-92 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(-97 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(-97 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{HCl}_{(\mathrm{g})}, \Delta \mathrm{H}=-194 \mathrm{~kJ} \\ & \Delta \mathrm{H}_{\text {reaction }}=2 \Delta \mathrm{H}_{\mathrm{f}(\mathrm{HCl})}-\Delta \mathrm{H}_{\mathrm{f}\left(\mathrm{H}_2\right)}-\Delta \mathrm{H}_{\mathrm{f}\left(\mathrm{Cl}_2\right)} \\ & -194=2 \Delta \mathrm{H}_{\mathrm{f}(\mathrm{HCl})}-\mathrm{O}-\mathrm{O} \\ & \Delta \mathrm{H}_{\mathrm{f}(\mathrm{HCl})}=-194 / 2 \\ & =-97 \mathrm{~kJ} / \mathrm{mol}\end{aligned}\)
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