The equilibrium constant #K_c# for the reaction #N_2(g)# + #3H_2(g)# #rightleftharpoons# #2NH_3(g)# at #450^oC# is 0.159. What is the equilibrium composition when 1 mol #N_2# is mixed with 3 mol #H_2# in a 2.00-L vessel? Thank you so much!

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Answer 1 · 2018-03-02T16:40:31

2.07 mol of NH3

Kc = #[NH3]^2/([N2]^1[H2]^3)#

[NH3] = #sqrt(Kc*[N2]^1 * [H2]^3#

Plugging the value of N2 and H2 as 1 and 3 mol respectively, we get the equilibrium amount of NH3 as 2.07 mol.