A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2NO (g) + 2H2 (g) ⥫⥬==N2 (g) + 2H2O (g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentrations of H2, N2, and H2O.

Calculate the equilibrium concentration of N2

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

2NO (g) + 2H2 (g) ⇒ N2 (g) + 2H2O (g)

i mol 0.10 0.050 0.10

c mol - 0.038 - 0.038 + 0019 + 0.038

e mol 0.062 0.012 00.019 0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] = 0.062 M

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M