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Derivation of Kp and Kc

Equilibrium c...

Question

Equilibrium constant $K_{c}$ for the reaction will be:

0.66 M

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0.56 M

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0.46 M

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0.36 M

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Solution

The correct option is A $0.66 M$
$K_{p} = K_{e} (R T)^{l n g}$
$K_{P} = K_{c} (R T)^{1}$
$K_{c} = \frac{38}{57.47} = 0.6621 M$ (Ans)

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