Unit 7 · EquilibriumPractice · 平衡

At first there is only A, so the reverse rate is zero and the forward rate is at its largest. As A is used up the forward rate falls; as B builds up the reverse rate rises. Equilibrium is reached when the two rates are equal. After that, [A] and [B] stay constant, but both reactions keep going: the equilibrium is dynamic. The reverse rate never overtakes the forward rate, since then B would decrease and the system would move back to equilibrium.

Pure solids have constant concentrations and are left out of the equilibrium expression. Dropping both solids leaves the two gases, H₂O and CO₂, both products, so Kc = [H₂O][CO₂].

Q = [H₂][I₂] ÷ [HI]² = 0.10 × 0.050 ÷ 0.50² = 0.020, exactly K, so the system is already at equilibrium: the forward and reverse rates are equal and the concentrations no longer change. Forgetting to square [HI] gives 0.010; flipping the fraction gives 50.

$K_p = \dfrac{P_{\ce{NO}}^{2}P_{\ce{Br2}}}{P_{\ce{NOBr}}^{2}}$; substituting, 0.12² × 0.060 ÷ 0.42² = 8.64 × 10⁻⁴ ÷ 0.1764 = 4.9 × 10⁻³. Each coefficient of 2 becomes an exponent of 2.

K = [B]/[A] = 1.0 × 10⁻¹², so at equilibrium there is a trillionth as much B as A: reactants dominate. K describes the equilibrium composition, not the speed. A catalyst speeds up the forward and reverse reactions equally, so it changes neither K nor the equilibrium composition.

① + ② gives A(g) ⇌ C(g), K = K₁ × K₂ = 0.0040 × 50 = 0.20. The target is that reversed (take the reciprocal) and doubled (square it): K = (1 ÷ 0.20)² = 5.0² = 25.

ICE: PCl₅ 0.100 − x, PCl₃ and Cl₂ both x. x² ÷ (0.100 − x) = 0.042. The “x is small” shortcut gives x ≈ √(0.042 × 0.1) = 0.065, which is 65% of the starting concentration, far above 5%, so it fails. Rearranging gives x² + 0.042x − 0.0042 = 0, and the positive root is x = 0.0471, so [Cl₂] = 0.0471 M.

Count the molecules: 2 H₂, 1 I₂, 6 HI, so the concentrations are 0.20, 0.10, and 0.60 M. Q = 0.60² ÷ (0.20 × 0.10) = 18 < 50: too little product, so the reaction proceeds forward and makes more HI.

There are 3 mol of gas on the left and 1 mol on the right. Compressing shifts the equilibrium toward fewer gas molecules, the forward direction, so CH₃OH increases. The forward reaction is exothermic, so heating shifts it in reverse; a catalyst only reaches equilibrium faster; and argon at constant volume leaves each gas’s concentration (partial pressure) unchanged, so nothing shifts.

Doubling the volume halves every concentration. Q = [NOBr]² ÷ ([NO]²[Br₂]): the top is multiplied by (½)² and the bottom by (½)³, so Q doubles. Q = 2K > K, and the equilibrium shifts in reverse, toward the side with more gas molecules, partly making up for the drop in pressure.

Mixing equal volumes halves each concentration: [Ag⁺] = [Br⁻] = 1.0 × 10⁻⁴ M. Q = [Ag⁺][Br⁻] = 1.0 × 10⁻⁸, far above 5.0 × 10⁻¹³, so a precipitate forms. Using the concentrations before mixing gives 4.0 × 10⁻⁸ (the same conclusion by luck, but the wrong Q); Q is the product of the two concentrations, not their sum.

CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). F⁻ is the conjugate base of the weak acid HF, so added H⁺ combines with F⁻ to form HF. That lowers [F⁻], making Q < K_sp, and more CaF₂ dissolves. NaF or Ca(NO₃)₂ adds a common ion and lowers the solubility. More water dissolves more CaF₂ in total, but the moles dissolved per liter (the molar solubility) stay the same.