At a particular temparature, the equilibrium constant, Kc for the reaction, 2 NO(g) + 2 H2(g) <=> N2(g) + 2 H2O(g) is 0.7689. Calculate the value of Kc for the reaction 2 N2(g) + 4 H2O(g) <=> 4 NO(g) + 4 H2(g)

Kc= 1.691

Explanation:

The equilibrium given  in this question  along its Kc is:

2 NO(g) + 2 H₂(g) <=> N₂(g) + 2 H₂O(g)     Kc = 0.7689

Now, they are asking us to find the value of the equilibrium constant Kc for:

2 N₂(g) + 4 H₂O(g) <=> 4NO(g) + 4 H₂(g)    Kc=?

This second equation is the reverse of the first multiplied by two. Therefore its equilibrium constant will the inverse square of the first one.

Kc = (  1 / 0.7689 )² = 1.691

Just remeber that the equilibrium constant for the reverse reaction is the inverse of the constant for the forward reaction.

Also if we multiply the coefficients of the balanced equation by a factor, we must raise thje equilibrium constant to the power of that factor.

the answer is d. 74.02