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 H2(g) at equilibrium.

Answer 1

The concentration of H2 at the equilibrium is 0.275 M

Step-by-step explanation:

Step 1: Data given

Molarity of N2 = 0.250 M

Molarity of H2 = 0.500 M

At equilibrium, the concentration of ammonia is 0.150 M

Step 2: The balanced equation

N2(g) + 3 H2(g) → 2 NH3(g)

Step 3: The initial concentrations

[N2] = 0.250 M

[H2] = 0.500 M

[NH3] = 0 M

Step 4: The concentrations at the equilibrium

[N2] = 0.250 - X M

[H2] = 0.500 - 3X M

[NH3] = 2X M = 0.150

X = 0.150 / 2 = 0.075

[N2] = 0.250 - 0.075 M = 0.175 M

[H2] = 0.500 - 3X M = 0.275 M

[NH3] = 2X M = 0.150

Step 5: Calculate Kc

Kc = [NH3]² / [N2][H2]³

Kc = (0.150²) / (0.175 * 0.275³)

Kc = 1.70

The concentration of H2 at the equilibrium is 0.275 M

Answer 2

Equilibrium concentration of
`H_(2)` is 0.0275 M

Step-by-step explanation:

Construct an ICE table to calculate changes in concentration to establish equilibrium.

`N_(2)+3H_(2)\rightleftharpoons 2NH_(3)`

I(M): 0.250 0.500 0

C(M): -x -3x +2x

E(M): (0.250-x) (0.500-3x) (2x)

At equilibrium, concentration of ammonia,
`[NH_(3)]=2x=0.150M`

So,
`x=(0.150)/(2)M=0.075M`

Hence, equilibrium concentration of
`H_(2)`,
`[H_(2)]=(0.500-3x)M=[0.500-(3* 0.075)]M=0.275M`

So, equilibrium concentration of
`H_(2)` is 0.0275 M