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2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) the initial concentrations are 5.25 M...

2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g)

the initial concentrations are 5.25 M H2, 9.50 M NO, and no H2O or N2. At equilibrium, [N2] = 1.00 M.

a. Calculate the concentrations of H2, NO, and H2O at equilibrium.
b. Calculate the value of K under the reaction conditions at equilibrium.

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Answer #1

a)

Given at equilibrium,

[N2] = 1.00

SO,

x = 1.00 M

[H2] = 5.25 - 2x

= 5.25 - 2*1.00

= 3.25 M

[NO] = 9.50 - 2x

= 9.50 - 2*1.00

= 7.50 M

[H2O] = 2x

= 2*1.00

= 2.00 M

Answer:

[H2] = 3.25 M

[NO] = 7.50 M

[H2O] = 2.00 M

b)

Kc = [N2][H2O]^2 / [NO]^2 [H2]^2

= 1.00 * 2.00^2 / (7.50^2 * 3.25^2)

= 4.00 / 594.14

= 6.73*10^-3

Answer: 6.73*10^-3

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