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Kc and Kp of Reverse Reaction

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The equilibrium constant Kc for the reaction shown below is 3.8 x 10-5 at 727°C.

I2(g) = 2 I(g)

Calculate Kc and Kp for the following equilibrium at the same temperature.

2 I(g)= I2(g)

Since they give us first rxn Kc do I first take the inverse (to get the Kc for the second rxn)and then figure the Kp? Is this right? Can you show me?

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This solution explains how to calculate Kc and Kp of a reverse reaction in equilibrium.

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