The standard Gibbs free energy change for the reaction: 2NOBr(g)⇌2NO(g)+Br2(g)
is +12.50 kJ/mole at 27∘C. What is the equilibrium constant for the reaction at 27∘C?
A
105
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B
10−5
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C
e5
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D
e−5
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Solution
The correct option is De−5 ΔG=−R×T×lnK 12.5×1000=−8.314×300×lnK lnK=5K=e−5