13. What is the Ke equilibrium-constant expression for the following equilibrium

Question

13. What is the Ke equilibrium-constant expression for the following equilibrium? A) [H20] B) [Fe][H20] [Fe][H20] D) [H2] H20 [Fe] H 14, Given the reaction 2NH,(g) At equilibrium, it was found that the concentration of H2 was 0.0591 M, the concentration of N2 was 0.0197 M, and the concentration of NH3 was 0.441 M. What is Ke for this equilibrium? A) 3.97 x 10 B) 1.58 x 10 s D) 2.24 x 10 E) 2.65 x 10 5. Consider the following equilibrium A(g) B20g) The concentration of A at equilibrium may be increased by A) decreasing the temperature. B) adding B2 to the system. D) increasing the pressure. E) adding AB2 to the system.

Explanation / Answer

13.

While writing expression for Kc, pure substances should be eliminated.

In the present question, FeO(s) and Fe(s) are to be eliminated.

Thus, Kc = [H2O] / [H2]

Option A is correct.

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14.

For the given chemical equation, the expression for Kc is

Kc = [H2]^3[N2] / [NH3]^2

Given, [H2] = 0.0591 M

[N2] = 0.0197 M

[NH3] = 0.441 M

Plugging these values in the Kc expression,

Kc = (0.0591)^3(0.0197) / (0.441)^2

Kc = 2.10 x 10^-5

Option C is correct.

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15.

Option E is correct.

Increasing concentration of the products moves the equilibrium towards the reactants side (That is the concentration of reactants increases). As A is the reactant, adding AB2(Product) to the system increases the concentration of A.

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