The equilibrium constant, KC, for the following reaction is 6 × 10-2 at 500 °C.N2(g) + 3H2(g) ⇌ 2NH3(g)Given:[N2]Initial = 3.0 × 10-5 M[H2]Initial =4.0 × 10-3 M[NH3]Initial = 3.0 × 10-3 MCalculate the reaction quotient and determine which way the reaction is shifting at this point.
Question
The equilibrium constant, KC, for the following reaction is 6 × 10-2 at 500 °C.N2(g) + 3H2(g) ⇌ 2NH3(g)Given:[N2]Initial = 3.0 × 10-5 M[H2]Initial =4.0 × 10-3 M[NH3]Initial = 3.0 × 10-3 MCalculate the reaction quotient and determine which way the reaction is shifting at this point.
Solution
The reaction quotient, Q, is calculated in the same way as the equilibrium constant, K, but uses the initial concentrations instead of the equilibrium concentrations.
For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), the reaction quotient Q is given by:
Q = [NH3]^2 / ([N2] * [H2]^3)
Substituting the given initial concentrations into this equation gives:
Q = (3.0 × 10^-3)^2 / ((3.0 × 10^-5) * (4.0 × 10^-3)^3) = 0.5625
Comparing Q to the given equilibrium constant Kc = 6 × 10^-2, we see that Q > Kc.
When Q > Kc, the reaction will shift to the left to reach equilibrium. This means the concentrations of the reactants N2 and H2 will increase, while the concentration of the product NH3 will decrease.
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