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Question

For the reaction in equilibrium
(I) CO(g)+12O2(g)CO2(g)
(II) 2CO(g)+O2(g)2CO2(g)
Then,

Equilibrium constant Free energy change
K1 ΔG1
K2 ΔG2


A
K2=K21, ΔG2=2ΔG1
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B
K2=2K1, ΔG2=2ΔG1
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C
K1=2K2, ΔG1=2ΔG1
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D
K2=K21, ΔG2=2ΔG21
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Solution

The correct option is A K2=K21, ΔG2=2ΔG1
For the reaction,
CO(g)+12O2(g)CO2(g) ...(1)
K1=[CO2][CO][O2]12
Again for the reaction,
2CO(g)+O2(g)2CO2(g)
K2=[CO2]2[CO]2[O2]=K21

We know,
ΔG1=2.303RTlogK1
ΔG2=2.303RTlogK2=2.303RTlogK21
ΔG2=2×(2.303RTlogK1)=2ΔG1


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