MHT CET · Chemistry · Thermodynamics (C)
For a reaction \(\Delta \mathrm{H}=-30 \mathrm{~kJ}\) and \(\Delta \mathrm{S}=-45 \mathrm{~J} \mathrm{~K}^{-1}\), at what temperature reaction changes from spontaneous to non spontaneous?
- A \(777 \cdot 0 \mathrm{~K}\)
- B \(675 \cdot 0 \mathrm{~K}\)
- C \(666 \cdot 6 \mathrm{~K}\)
- D \(375 \cdot 0 \mathrm{~K}\)
Answer & Solution
Correct Answer
(C) \(666 \cdot 6 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{H}=-30 \mathrm{~kJ}, \Delta \mathrm{S}=-45 \mathrm{~J} \mathrm{~K}^{-1}\)\(=-0.045 \mathrm{~kJ} \mathrm{~K}^{-1} \)
\( \therefore \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \mathrm{~kJ}}{-0.045 \mathrm{~kJ} \mathrm{~K}^{-1}}=666.67 \mathrm{~K} \)
Since \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are both negative, the reaction is spontaneous at low temperature and nonspontaneous above \(666.67 \mathrm{~K}\).
\( \therefore \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \mathrm{~kJ}}{-0.045 \mathrm{~kJ} \mathrm{~K}^{-1}}=666.67 \mathrm{~K} \)
Since \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are both negative, the reaction is spontaneous at low temperature and nonspontaneous above \(666.67 \mathrm{~K}\).
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