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Question

The value of log10K for a reaction AB is:
Given: r H298K=54.07 kJ mol1,rSo298K=10 J K1 mol1 and R=8.314 J K1mol1; 2.303×8.314×298=5705).


A
5
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B
10
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C
95
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D
100
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Solution

The correct option is D 10
Given-
r H298K=54.07 kJ mol1 
rSo298K=10 J K1 mol1
R=8.314 J K1mol1
2.303×8.314×298=5705)
we know-
Go=HoTSo
=54.07×103(298)(10) J mol1
=57050 J mol1
We know, 
Go=2.303 RT log Ko
log10Ko=Go2.303RT=57050 J mol15705 J mol1=10
hence, log10Ko=10

Chemistry

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