Answer: The balanced chemical equation for the given reaction is:
2C(s) + N2(g) + 5H2(g) ⇌ 2CH3NH2(g)
The equilibrium constant expression for the reaction is:
Kc = [CH3NH2]^2/([N2][H2]^5)
At the beginning of the reaction, the concentration of N2 is 1.0 mol/2.00 L = 0.50 M and the concentration of H2 is 2.0 mol/2.00 L = 1.0 M. The concentration of C(s) is not given, but it is assumed to be large enough that its concentration does not change significantly during the reaction. Let the concentration of CH3NH2 at equilibrium be x mol/L. Then, according to the stoichiometry of the reaction, the equilibrium concentrations of N2 and H2 are (0.50 - x) mol/L and (1.0 - 5x) mol/L, respectively.
Substituting these values into the equilibrium constant expression and solving for x, we get:
Kc = [CH3NH2]^2/([N2][H2]^5)
1.8×10−6 = x^2/[(0.50 - x)(1.0 - 5x)^5]
1.8×10−6 (0.50 - x)(1.0 - 5x)^5 = x^2
1.8×10−6 (0.50 - x)(1 - 5x)^5 = x^2
This is a cubic equation that can be solved numerically to find the value of x, which represents the equilibrium concentration of CH3NH2. Using a numerical solver, we find that x = 5.42×10^-4 M. Therefore, the equilibrium concentrations of N2 and CH3NH2 are 0.50 - x = 0.499 M and x = 5.42×10^-4 M, respectively.
If the concentration of H2 is doubled, its concentration at equilibrium becomes 2.0 M - 5x. Substituting this new value into the equilibrium constant expression, we get:
K'c = [CH3NH2]^2/([N2][H2]^5)
K'c = (x^2)/[(0.50 - x)(2.0 - 5x)^5]
The value of K'c is different from Kc because it depends on the new concentration of H2. To find the ratio of K'c to Kc, we can divide the two expressions:
K'c/Kc = [(x^2)/[(0.50 - x)(2.0 - 5x)^5]] / [(x^2)/[(0.50 - x)(1.0 - 5x)^5]]
K'c/Kc = [(2.0 - 5x)^5]/[(1.0 - 5x)^5]
Substituting x = 5.42×10^-4, we get:
K'c/Kc = [(2.0 - 5(5.42×10^-4))^5]/[(1.0 - 5(5.42×10^-4))^5]
K'c/Kc = 1.31
Therefore, if the concentration of H2 is doubled, the equilibrium constant Kc increases by a factor of 1.31.