Thank you!!! ;0; 3.00 moles of ammonia are originally put into a 1.00-liter container and allowed to decompose (no other chemical

Question

Thank you!!! ;0;

3.00 moles of ammonia are originally put into a 1.00-liter container and allowed to decompose (no other chemicals present). At equilibrium, 0.600 moles of ammonia remained in the reaction vessel. What would be the equilibrium constant (K) for this reaction?

2 NH3(g) N2(g) + 3 H2(g(​

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Philomena 5 months 2021-08-31T06:18:51+00:00 1 Answers 0 views 0

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    2021-08-31T06:20:39+00:00

    Answer:

    Kc ≅ 100 (1 sig. fig.)

    Explanation:

                  2NH₃(g)       ⇄      N₂(g)   +   3H₂(g)

    C(i)         3.00 mol/L             0                0

    ΔC             -2x                       +x              +3x

    C(eq)     0.600 mol/L            x                3x

    3.00 – 2x = 0.600  =>  x = 3.00 – 0.600/2 = 1.2 mole/L

    At equilibrium Kc = [N₂][H₂]³/[NH₃]² = (x)(3x)³/(3·1.2)² = (1.2)(3.6)³/(0.600)² = 155.52 (calculator)

    Rounds to 1 sig. fig. based on given 3.00 moles NH₃(g)’ Kc ≅ 100.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )