Question

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia

Answer 1

Answer:

0.327 M

Explanation:

Step 1: Write the balanced equation

2 NH₃(g) ⇄ N₂(g) + 3H₂(g)

Step 2: Make an ICE chart

        2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

I              x             0            0

C          -2y            +y         +3y

E         x-2y             y           3y

Step 3: Find the value of y

The concentration of N₂ at equilibrium is 0.0161 M. Then,

y = 0.0161

Step 4: Find the value of x

The concentration of NH₃ at equilibrium is 0.295 M. Then,

x-2y = 0.295

x-2(0.0161) = 0.295

x = 0.327