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A 100mL reaction vessel initially contains 2.60x10^-2 moles of NO and 1.30x10^-2 moles of H2. At equilibrium the concentration of NO in the vessel is 0.161M. At equilibrium the vessel contains N2, H2O and H2. What is the value of the equilibrium constant Kc for the following reactions?2H2(g) + 2NO(g) <---> 2 H2O(g) + N2(g)

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sebassandin

Answer:

[tex]Kc=17.7[/tex]

Explanation:

Hello,

In this case, for the given reaction:

[tex]2H_2(g) + 2NO(g)\rightleftharpoons 2 H_2O(g) + N_2(g)[/tex]

In such a way, the initial concentrations are:

[tex][H_2]_0=\frac{1.30x10^{-2}mol}{0.1L}=0.130M[/tex]

[tex][NO]_0=\frac{2.60x10^{-2}mol}{0.1L}=0.260M[/tex]

Thus, at equilibrium, the change [tex]x[/tex], due to the chemical reaction extent, turns out:

[tex]x=\frac{[NO]_{0}-[NO]_{eq}}{2} =\frac{0.260M-0.161M}{2}=0.049M[/tex]

Thus, the rest of the concentrations at equilibrium are:

[tex][H_2]_{eq}=0.130M-2(0.049M)=0.032M[/tex]

[tex][H_2O]_{eq}=2(0.049M)=0.098M[/tex]

[tex][N_2]_{eq}=0.049M[/tex]

In such a way, the equilibrium constant for the reaction, result as follows, even when on the statement the NO is excluded, because it participates in the equilibrium:

[tex]Kc=\frac{[H_2O]_{eq}^2[N_2]_{eq}}{[H_2]_{eq}^2[NO]_{eq}^2}=\frac{(0.098M)^2(0.049M)}{(0.032M)^2(0.161M)^2} \\\\Kc=17.7[/tex]

Best regards.

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