8. An equilibrium mixture contains 1.50 mol of nitrogen, 4.45 mol of hydrogen an

Question

8. An equilibrium mixture contains 1.50 mol of nitrogen, 4.45 mol of hydrogen and 7.05 mol of ammonia in a 5.00 L reaction vessel. Calculate Kc for the Haber-Bosh process. N2(g) + 3 H2(g) 2 NH3(g) 8. An equilibrium mixture contains 1.50 mol of nitrogen, 4.45 mol of hydrogen and 7.05 mol of ammonia in a 5.00 L reaction vessel. Calculate Kc for the Haber-Bosh process. N2(g) + 3 H2(g) 2 NH3(g) 8. An equilibrium mixture contains 1.50 mol of nitrogen, 4.45 mol of hydrogen and 7.05 mol of ammonia in a 5.00 L reaction vessel. Calculate Kc for the Haber-Bosh process. N2(g) + 3 H2(g) 2 NH3(g)

Explanation / Answer

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

at equilibrium,

[N2] = 1.50 mol/5 L = 0.30 M

[H2] = 4.45 mol/5 L = 0.89 M

[NH3] = 7.05 mol/5 L = 1.41 M

So,

Kc = (1.41)^2/(0.30)(0.89)^3 = 9.40

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