What is the equilibrium constant for the reaction N2H4 l H2

What is the equilibrium constant for the reaction,

N₂H₄ (l) + H₂ (g) 2 NH₃ (g)

given the reactions below?

N₂ (g) + 2 H₂ (g) N₂H₄ (l) K₁ = 2.4 x 10^–3

2 NH₃ (g) N₂ (g) + 3 H₂ (g) K₂ = 2.0 x 10^–3

Solution

Reactions given-

N2(g) + 2H2(g) <---> N2H4(l). K1= 2.4*10^-3

Inversing the reaction-

N2H4(l) <----> N2(g) + 2H2(g) . K1 = 1/2.4*10^-3 = 4.16*10^-4...1

Second reaction given-

2NH3(g) <---> N2(g) + 3H2(g) K2 = 2.0*10^-3

Inversing the reaction-

3H2(g) + N2(g) <--->2NH3(g) K2 = 1/2.0*10^-3 = 0.5*10^-3..(2)

Adding 1 and 2, we get-

N2H4(l)+H2(g) <---> 2NH3(g)

K = K1*K2 = 4.16*10^-4*0.5*10^-3 = 2.08*10^-7 Answer