An equilibrium mixture at 225∘C contains 8.0×10−2M NH3 and 0.20M H2 for the reaction
3H2(g)+N2(g)←→2NH3(g)
If the Kc at this temperature is 1.7×102, what is the equilibrium concentration of N2?
Kc = [NH3]^2 / [N2] [H2]^3
so [N2] = [NH3]^2 / [H2]^3 Kc
To find the equilibrium concentration of N2, we need to use the given equilibrium concentrations of NH3 and H2, along with the equilibrium constant (Kc), to calculate the equilibrium concentration of N2.
The balanced chemical equation for the reaction is:
3H2(g) + N2(g) ⇌ 2NH3(g)
The equilibrium constant expression in terms of concentrations is:
Kc = [NH3]^2 / [H2]^3 * [N2]
Given that the equilibrium concentrations of NH3 and H2 are 8.0×10^(-2) M and 0.20 M, respectively, and Kc = 1.7×10^2, we can substitute these values into the equilibrium constant expression and solve for the unknown equilibrium concentration of N2.
Kc = (8.0×10^(-2))^2 / (0.20)^3 * [N2]
Rearranging the equation to solve for [N2]:
[N2] = Kc * (0.20)^3 / (8.0×10^(-2))^2
Now, we can plug in the given values and calculate the equilibrium concentration of N2.
[N2] = 1.7×10^2 * (0.20)^3 / (8.0×10^(-2))^2
Simplifying the equation:
[N2] = 1.7×10^2 * 0.008 / 0.0064
[N2] = 212.5 M
Therefore, the equilibrium concentration of N2 is 212.5 M.