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A 100-mL reaction vessel initially contains 2. 60x 10^-2 moles of NO and 1. 30 x 10^-2 moles of H2. At equilibrium, the concentration of NO in the vessel is 0. 161 M. At equiibrium the vessel also contains N2, H2O, and H2. What is the value of the equilibrium constant Kc for the following reaction?

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Step-by-step explanation:

The balanced chemical equation for the reaction is:

2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)

At equilibrium, the concentration of NO is given as 0.161 M. We can assume that the concentration of H2 is also 0.161 M because the stoichiometric coefficient of H2 in the balanced equation is 2.

The initial moles of NO is 2.60 x 10^-2 moles, and the initial moles of H2 is 1.30 x 10^-2 moles. The total moles of NO and H2 at equilibrium is (2.60 x 10^-2 moles) - (2 mol of NO for every 2 mol of H2) = 2.60 x 10^-2 moles - 1.30 x 10^-2 moles = 1.30 x 10^-2 moles.

The total volume of the reaction vessel is 100 mL = 0.100 L. The concentration of N2 formed and the concentration of H2O formed can be assumed to be equal because the stoichiometric coefficient of N2 and H2O in the balanced equation is the same.

The concentration of N2 and H2O at equilibrium is (1.30 x 10^-2 moles) / (0.100 L) = 0.130 M.

The equilibrium expression for the reaction is:

Kc = [N2] * [H2O] / ([NO]^2 * [H2]^2)

Substituting the given values, we have:

Kc = (0.130 M) * (0.130 M) / (0.161 M)^2

Kc = 0.0169 M^2 / 0.0259 M^2

Kc = 0.653

Answer : Kc = 0.653

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