Question

NH3 is normally encountered as a gas with a pungent odor. It is formed from hydrogen and nitrogen.
N2(g) + 3H2(g)  2NH3(g)

A chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0 litre reaction vessel in 1 : 3 ratio. The equilibrium constant for the reaction is 1.0  10–4, and the equilibrium value of [H2(g)] is 0.12 mol/L. Calculate the equilibrium value of [NH3.

Answer 1

Let initial [N2] = x and [H2] = 3x (1.0 L vessel so mol = mol·L−1). If y mol·L−1 of N2 reacts, at equilibrium:

[N2]eq = x − y
[H2]eq = 3x − 3y = 0.12 → x − y = 0.04 ⇒ [N2]eq = 0.04 M
[NH3]eq = 2y

Use Kc = [NH3]^2 / ([N2][H2]^3) = 1.0×10^−4. Substitute known [N2] and [H2]:

[NH3]^2 = Kc·[N2]·[H2]^3 = (1.0×10^−4)(0.04)(0.12^3)
= 6.912×10^−9

[NH3] = sqrt(6.912×10^−9) ≈ 8.3×10^−5 M

So [NH3]eq ≈ 8.3 × 10^−5 mol·L−1.