AP Chemistry - Equilibrium Notes

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in reactant and product concentrations. The equilibrium position indicates whether product formation or reactant retention is favored. The equilibrium expression mathematically connects reactant and product concentrations at equilibrium to the equilibrium constant KK. The reaction quotient QQ measures the relative amounts of products and reactants at any given time, helping determine if a reaction is at equilibrium and, if not, the direction it must shift to reach equilibrium.

At chemical equilibrium, the forward and reverse reaction rates are equal, and the concentrations of reactants and products remain constant, though not necessarily equal.

For gas-phase reactions in the atmosphere, the reaction quotient QQ can be expressed as follows: For the reaction NO(g)+O3(g)⇌NO2(g)+O2(g)NO(g) + O3(g) \rightleftharpoons NO2(g) + O2(g), Q=[NO2][O2][NO][O3]Q = \frac{[NO2][O2]}{[NO][O3]}. For the reaction 2O3(g)⇌3O2(g)2O3(g) \rightleftharpoons 3O2(g), Q=[O2]3[O3]2Q = \frac{[O2]^3}{[O3]^2}.

To calculate the reaction quotient QQ, consider the reaction O2(g)+2SO2(g)⇌2SO3(g)O2(g) + 2SO2(g) \rightleftharpoons 2SO3(g) with given concentrations [O2]=3.70 M[O2] = 3.70 \text{ M}, [SO2]=4.50 M[SO2] = 4.50 \text{ M}, and [SO3]=2.50 M[SO3] = 2.50 \text{ M}. The reaction quotient QQ is calculated as Q=[SO3]2[O2][SO2]2=(2.50)2(3.70)(4.50)2=6.253.70×20.25=6.2574.925≈0.0834Q = \frac{[SO3]^2}{[O2][SO2]^2} = \frac{(2.50)^2}{(3.70)(4.50)^2} = \frac{6.25}{3.70 \times 20.25} = \frac{6.25}{74.925} \approx 0.0834. Conversely, for the reaction 2SO3(g)⇌O2(g)+2SO2(g)2SO3(g) \rightleftharpoons O2(g) + 2SO2(g) with the same concentrations, Q=[O2][SO2]2[SO3]2=(3.70)(4.50)2(2.50)2=3.70×20.256.25=74.9256.25≈11.99Q = \frac{[O2][SO2]^2}{[SO3]^2} = \frac{(3.70)(4.50)^2}{(2.50)^2} = \frac{3.70 \times 20.25}{6.25} = \frac{74.925}{6.25} \approx 11.99.

To predict shifts in equilibrium, consider the ammonia synthesis reaction at 5000°C: N2(g)+3H2(g)⇌2NH3(g)N2(g) + 3H2(g) \rightleftharpoons 2NH3(g), where K=6.0×10−2K = 6.0 \times 10^{-2}. Given [NH3]0=1.0×10−3 M[NH3]0 = 1.0 \times 10^{-3} \text{ M}, [N2]0=1.0×10−5 M[N2]0 = 1.0 \times 10^{-5} \text{ M}, and [H2]0=2.0×10−3 M[H2]0 = 2.0 \times 10^{-3} \text{ M}, Q=[NH3]02[N2]0[H2]03=(1.0×10−3)2(1.0×10−5)(2.0×10−3)3=1.0×10−6(1.0×10−5)(8.0×10−9)=1.0×10−68.0×10−14=1.25×107Q = \frac{[NH3]0^2}{[N2]0[H2]0^3} = \frac{(1.0 \times 10^{-3})^2}{(1.0 \times 10^{-5})(2.0 \times 10^{-3})^3} = \frac{1.0 \times 10^{-6}}{(1.0 \times 10^{-5})(8.0 \times 10^{-9})} = \frac{1.0 \times 10^{-6}}{8.0 \times 10^{-14}} = 1.25 \times 10^7. Since Q > K, the system will shift to the left. If [NH3]0=2.00×10−4 M[NH3]0 = 2.00 \times 10^{-4} \text{ M}, [N2]0=1.50×10−5 M[N2]0 = 1.50 \times 10^{-5} \text{ M}, and [H2]0=3.54×10−1 M[H2]0 = 3.54 \times 10^{-1} \text{ M}, Q=[NH3]02[N2]0[H2]03=(2.00×10−4)2(1.50×10−5)(3.54×10−1)3=4.00×10−8(1.50×10−5)(0.04447)=4.00×10−86.67×10−7≈0.06Q = \frac{[NH3]0^2}{[N2]0[H2]0^3} = \frac{(2.00 \times 10^{-4})^2}{(1.50 \times 10^{-5})(3.54 \times 10^{-1})^3} = \frac{4.00 \times 10^{-8}}{(1.50 \times 10^{-5})(0.04447)} = \frac{4.00 \times 10^{-8}}{6.67 \times 10^{-7}} \approx 0.06. Since Q < K, the system will shift to the right. If [NH3]0=1.00×10−4 M[NH3]0 = 1.00 \times 10^{-4} \text{ M}, [N2]0=5.0 M[N2]0 = 5.0 \text{ M}, and [H2]0=1.0×10−2 M[H2]0 = 1.0 \times 10^{-2} \text{ M}, Q=[NH3]02[N2]0[H2]03=(1.00×10−4)2(5.0)(1.0×10−2)3=1.00×10−8(5.0)(1.0×10−6)=1.00×10−85.0×10−6=0.002Q = \frac{[NH3]0^2}{[N2]0[H2]0^3} = \frac{(1.00 \times 10^{-4})^2}{(5.0)(1.0 \times 10^{-2})^3} = \frac{1.00 \times 10^{-8}}{(5.0)(1.0 \times 10^{-6})} = \frac{1.00 \times 10^{-8}}{5.0 \times 10^{-6}} = 0.002. Again, since Q < K, the system will shift to the right.

For the gas-phase reaction $$N2O4(g) \rightleftharpoons 2