Final answer:
Kc for the reaction NO(g) + H2(g) ⇌ ½ N2(g) + H2O(g) is calculated by taking the square root of the original Kc, resulting in 26.3. Kc for the reaction 2N2(g) + 4H2O(g) ⇌ 4NO(g) + 4H2(g) is found by taking the reciprocal and squaring it, leading to 2.1 × 10^-5.
Step-by-step explanation:
To calculate the equilibrium constants (Kc) for the given reactions, we need to use the relationship between the different stoichiometries of the reactions and the original Kc that was provided.
For the given reaction 2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g), the Kc is given as 6.9 × 10^2.
(a) For the reaction NO(g) + H2(g) ⇌ ½ N2(g) + H2O(g), each of the concentrations will be raised to the power of their stoichiometric coefficients in the balanced equation, and since the reaction is halved, we take the square root of the original Kc:
Kc = √(6.9 × 10^2) = 26.3.
(b) For the reaction 2N2(g) + 4H2O(g) ⇌ 4NO(g) + 4H2(g), we are essentially reversing the original reaction and doubling it. Since the reverse reaction's Kc is the reciprocal of the forward reaction's Kc, and then squared because it's doubled, we get:
Kc = 1 / (6.9 × 10^2)^2 = 2.1 × 10^-5.