Question

A 2 L reaction vessel contains NH3, N2 and H2 at equilibrium at a certain temperature....

A 2 L reaction vessel contains NH3, N2 and H2 at equilibrium at a certain temperature. The equilibrium concentrations are [NH3] = 0.25 M, [N2] = 0.11 M and [H2] = 1.91 M. Calculate the equilibrium constant Kc for the synthesis of ammonia as described in the following reaction. N2 (g) + 3H2 (g) 2 NH3 (g) If 0.12 moles of N2 is then added to the reaction flask. Calculate the new equilibrium concentrations.

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Answer #1

The balanced reaction

N2 + 3H2 = 2NH3

Equilibrium constant expression of the reaction

K = [NH3]2 / [N2] [H2]3

= (0.25*0.25) / (0.11)*(1.91)3

= 0.1557

New concentration of N2 added = moles/Volume

= 0.12/2 = 0.06

New concentration of N2 = 0.06 + 0.11 = 0.17 M

The balanced reaction with ICE TABLE

N2 + 3H2 = 2NH3

I 0.17 1.91 0.25

C - x - 3x +2x

E (0.17-x) (1.91-3x) (0.25+2x)

Equilibrium constant expression of the reaction

K = [NH3]2 / [N2] [H2]3

0.1557 = (0.25+2x)2 / (0.17-x)*(1.91-3x)3

0.1557(0.17-x)(1.91-3x)3 = (0.25+2x)2

(0.026469 - 0.1557x) (6.967871 - 27x3 - 32.8329x + 51.57x2) =

0.0625 + 4x2 + x

0.1844 - 0.715x3 - 0.816x + 1.365x2 - 1.085x + 4.2x4 + 5.11x2 + 8.03x3 - 0.0625 - 4x2 - x = 0

4.2x4 + 7.315x3 + 2.475x2 - 2.9x + 0.1219 = 0

x = 0.0439

At equilibrium

[NH3] = 0.25+2x = 0.25 + 2*0.0439 = 0.3378 M

[N2] = 0.17-x = 0.17 - 0.0439 = 0.1261 M

[H2] = 1.91-3x = 1.91 - 3*0.0439 = 1.7783

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