Question

A mixture consisting of 0.250 M N2(g) and 0.500 M H2(g) reaches equilibrium according to the equation N2(g) + 3 H2(g) → 2 NH3(g). At equilibrium, the concentration of ammonia is 0.150 M. Calculate the concentration of N2(g) at equilibrium.A. 0.0750 MB. 0.350 MC .0.425 MD. 0.275 ME. 0.150 M

Answer 1

Answer:

The concentration of N2 at the equilibrium is 0.175 M (it's not between the options)

Explanation:

Step 1: Data given

Mixture contains 0.250 M N2 and 0.500 M H2

At the equilibrium [NH3] = 0.150 M

Step 2: The balanced equation

N2(g) + 3 H2(g) → 2 NH3(g)

Step 3: The initial concentrations:

[N2] = 0.250M

[H2] = 0.500 M

[NH3] = 0 M

Since [NH3] gained 0.15 M in going to equilibrium, then H2 lost (0.150)*(3/2) = 0.225  

[H2] at the equilibrium = 0.500 -0.225 = 0.275 M

N2 lost (0.150/2)M = 0.075 M

[N2]at the equilibrium = 0.250 - 0.075 = 0.175 M

The concentration of N2 at the equilibrium is 0.175 M (it's not between the options)