Question

The equilibrium constant, Kc, for the following reaction is 6.30 at 723K. 2NH3(g) N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 15.7 L container at 723K contains 0.284 mol of NH3(g) and 0.437 mol of N2, the equilibrium concentration of H2 is__________ M.

Answer 1

[NH3] = mol of NH3 / volume in L

= 0.284 mol / 15.7 L

= 0.01809 M

[N2] = mol of N2 / volume in L

= 0.437 mol / 15.7 L

= 0.02783 M

Now use:

Kc = [N2][H2]^3 / [NH3]^2

6.30 = (0.02783) * [H2]^3 / (0.01809)^2

[H2]^3 = 0.07408

[H2] = 0.420 M

Answer: 0.420 M