A mixture of 0.119 mol of NO, 0.192 mol of H2, and 0.186 mol of H2O is placed in a 1.0 L reaction vessel at 300 K. The following equilibrium is established:

Question

A mixture of 0.119 mol of NO, 0.192 mol of H₂, and 0.186 mol of H₂O is placed in a 1.0 L reaction vessel at 300 K. The following equilibrium is established:

2NO(g) + 2H₂(g) ↔ N₂(g) + 2H₂O(g)

At equilibrium the molarity of NO is found to be 0.069 M. Determine K_c for this reaction.

Answer 1

2NO(g) + 2H₂(g) <==>  N₂(g) + 2H₂O(g)

0.119.........0.192...............0...............0.186.........Initial

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

0.119-2x....0.192-2x..........x.............0.186+2x.....Equilibrium

Since [NO] = 0.069, we know that 0.119-2x = 0.069

Solving for x, we have...

2x = 0.119 - 0.069

x = 0.025

Therefore @ equilibrium..

[NO] = 0.069 M

[H₂] = 0.142 M

[N₂] = 0.025 M

[H₂O] = 0.236 M

Kc = [N₂][H2O]² / [NO]²[H₂]²

Kc = (0.025)(0.236)² / (0.069)²(0.142)² = 0.001392 / 0.00009600

Kc = 14.5