Question

Consider the following reaction 2NO(g) + 2 H2 = N2(g) + 2 H2O(g) Determine the value...

Consider the following reaction 2NO(g) + 2 H2 = N2(g) + 2 H2O(g)

Determine the value of the equilibrium constant, Kc, for the reaction. Initially, a mixture of 0.100 M NO, 0.050 M hydrogen gas, and 0.100 M of water vapor was allowed to reach equilibrium. At equilibrium the concentration of NO was found to be 0.0062 M.

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Answer #1

Reaction: 2NO(g) + 2 H2 = N2(g) + 2 H2O(g)

Initially:       0.1 ....    0.05 ... 0 ............0.1

Change:    -2x .......-2x ........+x ..........+2x

Final:       0.1-2x ....0.05-2x....x .........0.1+2x

Given at equilibrium,

[NO] = 0.0062 M

=> 0.1 - 2x = 0.062

=> x = 0.019 M

Therefore at equilibrium,

[H2] = 0.05 - 2x = 0.012 M

[N2] = x = 0.019 M

[H2O] = 0.1 + 2x = 0.138 M

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

=> Kc = (0.019) x (0.138)^2 / (0.062)^2 x (0.012)^2 = 653.7

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