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Question
For a certain reaction ΔH0 is −224 kJ and ΔS0 is −153 J K−1. At what temperature the change over from spontaneous to non-spontaneous will occur?
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Solution
Given:
ΔH0 = −224 kJ,
ΔS0 = −153 kJ K−1 = −0.153 kJ K−1
To find: T = ?
Formula:
∴ `ΔS^0 = (ΔH^0)/T`
∴ T = `-(ΔH^0)/(ΔS^0)`
= `- (-224 kJ)/(-0.153 kJ k^-1)`
∴ T = + 1464 K
Since ΔH0 and ΔS0 are both negative, the reaction is spontaneous at low temperature. A change over will occur at 1464 K. The reaction is spontaneous below 1464 K.
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