N2(g) + 3 H2(g) <=> 2 NH3 (g)?
Consider the gas-phase reaction, N2(g) + 3 H2(g) <=> 2 NH3 (g), for which Kp = 43 at 400 K. If the mixture is analyzed and found to contain 1.5 bar of N2, 0.42 bar of H2 and 2.1 bar of NH3, describe the situation:
1. Q > K and more products will be made to reach equilibrium.
2. Q < K and more reactants will be made to reach equilibrium.
3. Q < K and more products will be made to reach equilibrium.
4. Q > K and more reactants will be made to reach equilibrium.
5. Within 1 decimal place, Q = K and the reaction is at equilibrium
The answer is 3, I just need to know how to get it. thanks!
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your answer is:
3. Q < K and more products will be made to reach equilibrium.
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N2(g) + 3 H2(g) <=> 2 NH3 (g),
Kp = 43 = [NH3]^2 / [N2] [H2]^3
Q = [NH3]^2 / [N2] [H2]^3
Q = [2.1]^2 / [1.5] [0.42]^3
Q = (4.41) / [1.5] (0.07409)
Q = (4.41) / (0.1111)
Q = 39.7
the present level "Q" at 40. is less than the equilb level "K" at 43, ....
indicating a smaller numerator amount of products than at equilib....
the equilib will shift to the right, to increase the products to reach equilib
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