At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction.

2 NH3(g) N2(g) + 3 H2(g)
At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?

4 answers

(NH3)@ start = 4.0/2L = 2.0M
(NH3)@ end = 2.0mol/2L = 1M
...........2NH3 ==> N2 + 3H2
initial....2.0......0.....0
change.....-2x......x.....3x
equil.....2-2x=1.....x.....3x
For NH3 at end, 2-2x = 1 and x = 1/2; then N2 = 1/2 and H2 = 1.5
Substitute and solve for K.

1.6875

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1.69 sucka

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