The value of log10 K for a reaction A⇌B is: Given−△rH∘298K=−54.07kJmol−1,△rS∘298K=10JK−1mol−1 and R−8.314JK−1mol−1
A
10
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B
100
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C
5
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D
95
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Solution
The correct option is A 10 Given: △rH∘298K=−54.07kJmol−1,△rS∘298K=10JK−1mol−1,R−=8.314JK−1mol−1,T=298K We know the relation △G∘=△H∘−T△S∘ at equilibrium △G∘=0 0=−54.07×1000−298×100=−57050Jmol−1⇒−57050=−5705log10Klog10K=10