Chemistry Equilibrium

Question

At a certain temperature, 0.4011 mol of N, and 1.681 mol of H, are placed in a 5.00 L container

N2(g) +3 H2 (g) = 2 NH, (g)

At equilibrium, 0.1401 mol of N2, is present. Calculate the equilibrium constant, Ke.

Answer 1

First, you really need a CORRECT equation, which would be N2(g) + 3H2(g) <==> 2NH3(g)

Set up an ICE table:

N2(g) + 3 H2 (g) ==> 2 NH3 (g)

0.4011....1.681............0........Initial

-x...........-3x...............+2x......Change

0.4011-x...1.681-3x....2x.......Equilibrium

At equilibrium N2 = 0.4011 -x = 0.1401 so we can find x:

x = 0.261

Final concentrations are:

[NH3] = 2 x 0.261 = 0.783

[N2] = 0.4011

[H2] = 1.681 - 3x = 1.681 - 0.783 = 0.898

Keq = [NH3]² / [N2][H2]³ = (0.783)² / (0.4011)(0.898)³

Keq = 2.11