2.11 · Reaction Rates and EquilibriumPractice · 反应速率与化学平衡

Start with N₂O₅: Δ[N₂O₅] = (1.00 − 0.40) mol ÷ 2.0 L = 0.30 M, so it is used up at 0.30 ÷ 5.0 = 0.060 M/min. Rates of different species are in the ratio of their coefficients, 2 : 4, so NO₂ forms at 0.060 × 2 = 0.12 M/min. Forgetting to divide by the volume gives 0.24.

A catalyst gives the reaction a different pathway with a lower activation energy. The temperature, and so the spread of molecular energies, is unchanged, but many more collisions now clear the lower barrier, so effective collisions per second rise. A catalyst doesn’t change the average kinetic energy (only heating does), doesn’t change ΔH, and doesn’t shift an equilibrium; it speeds the forward and reverse reactions equally.

Control variables: change only the factor being studied and keep everything else the same. What differs between each pair: 1 vs. 2: form; 1 vs. 3: concentration; 2 vs. 3: form and concentration; 3 vs. 4: concentration and temperature. Only trials 1 and 3 differ in concentration alone.

Look for something that changes while the reaction runs and stops changing only at equilibrium. Here the number of gas molecules doesn’t change (1 + 1 = 2), so the total pressure is constant from the start; the volume and total mass are fixed, so the density never changes either. Neither one proves anything. The color depends on [I₂], which does change as the reaction runs, so a steady color means [I₂] is constant: equilibrium. By the coefficients, at equilibrium HI is consumed twice as fast as H₂ is consumed, so the rates in D aren’t equal.

Set up an ICE table (in M): initial [CO] = [H₂O] = 0.200 and [CO₂] = [H₂] = 0; change: CO₂ and H₂ each gain 0.0889, CO and H₂O each lose 0.0889; equilibrium: [CO] = [H₂O] = 0.1111, [CO₂] = [H₂] = 0.0889. K = ([CO₂][H₂]) / ([CO][H₂O]) = (0.0889 × 0.0889) ÷ (0.1111 × 0.1111) = 0.640.

Q = (0.100 × 0.100) ÷ (0.100 × 0.100) = 1.00, which is larger than K = 0.640. Q > K means there is “too much” product, so the reaction runs in reverse until Q = K. Equal concentrations don’t mean equilibrium.

Solids (and pure liquids) are left out of the expression; only gases (and dissolved species) appear, each raised to its coefficient. So K = [H₂O]⁴ / [H₂]⁴. Products go on top and reactants on the bottom, which C reverses; D drops the exponents.

The forward reaction turns 3 gas molecules into 1. Compressing raises the pressure, so the equilibrium shifts toward fewer gas molecules, the products. Removing a product also shifts it forward. The forward reaction is exothermic, so heating shifts it backward. A catalyst only speeds things up without shifting the equilibrium. Argon added at constant volume leaves every concentration unchanged, so nothing shifts.

The added $\ce{OH-}$ combines with $\ce{H+}$ to form water, lowering [H⁺]. The equilibrium shifts to replace H⁺, which is toward $\ce{CrO4^2-}$ (the reverse direction as written), so the solution turns yellow. Adding sulfuric acid afterward raises [H⁺] and turns it orange again.

K changes only with temperature, not with concentration or pressure. Heating shifts the equilibrium in the endothermic direction, here the reverse, so at the new equilibrium there are fewer products and more reactants: K decreases. For ammonia synthesis (ΔH = −92.2 kJ/mol), K is 0.50 at 400 °C and drops to an estimated 0.059 at 500 °C.