# what would be the K(sub c) value for this equation 3N2(g)+6 H2O(g)⇌6 NO(g)+6 H2(g) if N2(g)+2 H2O(g)⇌2 NO(g)+2H2(g), Kc = 3.94 x 10-3?

what would be the K(sub c) value for this equation:

3N2(g)+6 H2O(g)⇌6 NO(g)+6 H2(g)

if N2(g)+2 H2O(g)⇌2 NO(g)+2H2(g), Kc = 3.94 x 10-3

### 1 Answer

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- RaisLv 65 years agoFavorite Answer
3N2(g)+6 H2O(g)⇌6 NO(g)+6 H2(g)

K'c = [NO(g)]^6 x [H2(g)]^6/{[N2(g)]^3 x [H2O(g)]^6 ------------- (1)

And for the following

N2(g)+2 H2O(g)⇌2 NO(g)+2H2(g)

Kc = [NO(g)]^2 x [H2(g)]^2/{[N2(g)] x [H2O(g)]^2 = 3.94 x 10^-3 --------- (2)

Now if we cube equation (2) we will get equation (1). Hence for the required reaction

K'c = Kc^3 = (3.94 x 10^-3)^3 = 6.12 x 10^-8

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