AP Chemistry - Equilibrium

Chemical Equilibrium is a state where the rate of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products. The Equilibrium Position refers to the relative concentrations of reactants and products at equilibrium, indicating the extent to which a reaction proceeds to completion. The Equilibrium Expression is a mathematical expression that relates the concentrations of reactants and products at equilibrium to a constant value (K), while the Reaction Quotient (Q) measures the relative amounts of products and reactants present at any given time, predicting the direction in which a reaction must shift to reach equilibrium.

1) Characterizing a System at Chemical Equilibrium

At equilibrium, the rates of the forward and reverse reactions are equal, meaning reactants convert to products at the same rate products convert back to reactants. The overall composition of the reaction mixture remains constant at equilibrium, indicating that the ratio of reactants and products remains constant, though not necessarily equal.

2) Reaction Quotient (Q) Expressions for Gas-Phase Reactions

NO(g) + O3(g) \rightleftharpoons NO2(g) + O2(g): Q = \frac{[NO2][O2]}{[NO][O3]}

2 O3(g) \rightleftharpoons 3 O2(g): Q = \frac{[O*2]^3}{[O_3]^2}

3) Calculating the Reaction Quotient

For the reaction O2(g) + 2 SO2(g) \rightleftharpoons 2 SO3(g), given [O2] = 3.70 \ M, [SO2] = 4.50 \ M, and [SO3] = 2.50 \ M, the reaction quotient Q is calculated as follows: Q = \frac{[SO3]^2}{[O2][SO_2]^2} = \frac{(2.50)^2}{(3.70)(4.50)^2} = 0.083

4) Calculating the Reaction Quotient (Reversed Reaction)

For the reversed reaction 2 SO3(g) \rightleftharpoons O2(g) + 2 SO2(g), with the same given concentrations, the reaction quotient Q is: Q = \frac{[O2][SO2]^2}{[SO3]^2} = \frac{(3.70)(4.50)^2}{(2.50)^2} = 11.99

5) Predicting the Direction of Shift to Reach Equilibrium

For the reaction N2(g) + 3 H2(g) \rightleftharpoons 2 NH_3(g), where K = 6.0 \times 10^{-2}, we analyze three scenarios:

a) If [NH3]0 = 1.0 \times 10^{-3} \ M, [N2]0 = 1.0 \times 10^{-5} \ M, and [H2]0 = 2.0 \times 10^{-3} \ M, then Q = \frac{[NH3]^2}{[N2][H*2]^3} = 1.25 \times 10^{7}. Since Q > K, the system shifts left.

b) If [NH3]0 = 2.00 \times 10^{-4} \ M, [N2]0 = 1.50 \times 10^{-5} \ M, and [H2]0 = 3.54 \times 10^{-1} \ M, then Q = \frac{[NH3]^2}{[N2][H*2]^3} = 0.06. Since Q < K, the system shifts right.

c) If [NH3]0 = 1.00 \times 10^{-4} \ M, [N2]0 = 5.0 \ M, and [H2]0 = 1.0 \times 10^{-2} \ M, then Q = \frac{[NH3]^2}{[N2][H*2]^3} = 0.002. Since Q < K, the system shifts right.

6) Equilibrium Constant and Reaction Quotient

For the reaction N2O4(g) \rightleftharpoons 2 NO2(g), where K{eq} = 8.00, and given [N2O*4] = 3.75 \times 10^{-2} \ M and $$[NO_2] = 5.88