MHT CET · Chemistry · Thermodynamics (C)
Calculate the value of \(\Delta G\) for following reaction at \(300 \mathrm{~K}\).
\(\mathrm{H}_2 \mathrm{O}_{(\mathrm{s})} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)}\)\((\Delta \mathrm{H}=7 \mathrm{~kJ}, \Delta \mathrm{S}=\) \(24.8 \mathrm{~J} \mathrm{~K}^{-1})\)
- A \(0.74 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(-0.82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(0.21 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(-0.44 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(-0.44 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{H} =7 \mathrm{~kJ} \)
\( \Delta \mathrm{S} =24.8 \mathrm{~J} \mathrm{~K}^{-1}=24.8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \)
\( \mathrm{~T} =300 \mathrm{~K} \)
\( \Delta \mathrm{G} =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \)
\( \therefore \Delta \mathrm{G} =7 \mathrm{~kJ}-\left(300 \mathrm{~K} \times 24.8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1}\right) \)
\( =7 \mathrm{~kJ}-7.44 \mathrm{~kJ}=-0.44 \mathrm{~kJ}\)
\( \Delta \mathrm{S} =24.8 \mathrm{~J} \mathrm{~K}^{-1}=24.8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \)
\( \mathrm{~T} =300 \mathrm{~K} \)
\( \Delta \mathrm{G} =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \)
\( \therefore \Delta \mathrm{G} =7 \mathrm{~kJ}-\left(300 \mathrm{~K} \times 24.8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1}\right) \)
\( =7 \mathrm{~kJ}-7.44 \mathrm{~kJ}=-0.44 \mathrm{~kJ}\)
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