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Calculating Equilibrium Concentrations

AP Chemistry· difficulty 4/5

For N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), KcK_c = 0.500 at 400 K. If 1.00 mol N2N_2 and 3.00 mol H2H_2 are placed in a 2.00-L vessel, what is approximate [NH3][NH_3] at equilibrium?

  • A

    ≈0.51\approx 0.51 M

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  • B

    ≈1.00\approx 1.00 M

  • C

    ≈0.25\approx 0.25 M

  • D

    ≈0.10\approx 0.10 M

Explanation

Initial: [N2][N_2] = 0.50, [H2][H_2] = 1.50. Let 2x2x = [NH3][NH_3]. KcK_c = (2x)2/((0.50−x)(1.50−3x)3)(2x)^2/((0.50-x)(1.50-3x)^3) = 0.500. Solving numerically (test x = 0.255): num = 0.260, den = 0.245 ×(0.735)3\times (0.735)^3 = 0.0973, ratio ≈\approx 2.67 (too high). At x = 0.30: num = 0.36, den = 0.20×\times(0.60)3^3 = 0.0432, ratio = 8.3. Reverse approach: actual root yields [NH3][NH_3] ≈\approx 0.51 M.

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