JEE MAIN - Chemistry (2019 - 10th January Morning Slot - No. 23)
A process has $$\Delta $$H = 200 J mol–1 and $$\Delta $$S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :
4 K
20 K
5 K
12 K
Explanation
$$\Delta $$G = $$\Delta $$H $$-$$ T$$\Delta $$S
T = $${{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K$$
T = $${{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K$$
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