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Q.

For a reaction $Δ G^{\circ}$ is $- 10 J / m o l$ then $K_{e q}$ for the reaction is

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a

$e^{- 10 R T}$

b

$e^{10}$

c

$e^{10 / R T}$

d

$e^{- 10}$

answer is C.

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Detailed Solution

We know,

$∆ G ° = - R T \ln K_{e q}$

$- 10 = - R T \ln K_{e q}$

$\ln K_{e q} = \frac{10}{R T}$

$K_{e q} = e^{\frac{10}{R T}}$

Therefore, the correct option is (C).

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