MHT CET · Chemistry · Thermodynamics (C)
For \(\mathrm{NaCl}_{(\mathrm{s})}\) enthalpy of solution is \(4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and lattice enthalpy is \(790 \mathrm{~kJ} \mathrm{~mol}^{-1}\). What is hydration enthalpy of \(\mathrm{NaCl}\) ?
- A \(786 \mathrm{~kJ}\)
- B \(794 \mathrm{~kJ}\)
- C \(-786 \mathrm{~kJ}\)
- D \(-794 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(C) \(-786 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \Delta_{\text {soln }} \mathrm{H} & =\Delta_{\mathrm{L}} \mathrm{H}+\Delta_{\text {hyd }} \mathrm{H} \\ \therefore \quad \Delta_{\text {hyd }} \mathrm{H} & =\Delta_{\text {soln }} \mathrm{H}-\Delta_{\mathrm{L}} \mathrm{H} \\ & =4 \mathrm{~kJ} \mathrm{~mol}^{-1}-790 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & =-786 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\)
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