What is the value of the equilibrium constant at 500 °C for the formation of NH
3
according to the following
equation?
N
2
(
g
) + 3H
2
(
g
)
⇌ 2NH
3
(
g
)
An equilibrium mixture of NH
3
(
g
), H
2
(
g
), and N
2
(
g
) at 500 °C was found to contain 1.35
M
H
2
, 1.15
M
N
2
, and
4.12
×
10
−1
M
NH
3
.
To determine the value of the equilibrium constant (K) at 500 °C for the formation of NH3, you need to use the equilibrium concentrations of NH3, H2, and N2.
Given:
- [H2] = 1.35 M (concentration of H2)
- [N2] = 1.15 M (concentration of N2)
- [NH3] = 4.12 × 10^−1 M (concentration of NH3)
According to the balanced equation:
N2(g) + 3H2(g) ⇌ 2NH3(g)
The equilibrium constant expression (K) for this reaction is:
K = ([NH3]^2) / ([N2] * [H2]^3)
Substituting the given concentrations:
K = ([4.12 × 10^−1]^2) / ([1.15] * [1.35]^3)
Calculating the value of K will give you the answer.