The reaction: N^2(g) + 3 H^2(g) 2 NH_3(g) At 400 degree C temperature K_c = 0.50

Question

The reaction: N^2(g) + 3 H^2(g) 2 NH_3(g) At 400 degree C temperature K_c = 0.500 Sufficient reactants and product are mixed in a 1.0 L container to produce these initial concentrations: [N_2] = 2.00 M, [H_2] = 3.00 M, and (NH_3) = 4.00 M. Calculate the value of Q. Predict if the mixture needs to shift towards more reactants or more product to reach equilibrium. Set up the ICE table, fill in all the terms but do not try to solve for "x" yet. What is the term for the H2 at equilibrium? (The term will have "x" in it.)

Explanation / Answer

N2(g) + 3H2(g) --------> 2NH3(g)

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

Qc = (4)2/2*(3)3

      = 16/54 = 0.29

KC   = 0.5

Kc>Qc The equilibrium shift to right side . The equilibrium shift ot more products side.

        N2(g) + 3H2(g) --------> 2NH3(g)

I       2             3                      4

C    +x           +3x     -2x

E 2+x    3+3x    4-2x

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

0.5   =    (4-2x)2/(2+x)(3+3x)3

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