Unit 5 · KineticsPractice · 动力学

Rates of change are in the ratio of the coefficients: $\ce{N2O5}$ : $\ce{NO2}$ = 2 : 4. So $\ce{N2O5}$ is consumed at 0.032 × 2/4 = 0.016 M/s. Flipping the ratio gives 0.064.

Compare trials 1 and 2: [B] is constant and doubling [A] doubles the rate (×2.0), so it is first order in A. Compare trials 1 and 3: [A] is constant and doubling [B] multiplies the rate by 4.0 = 2², so it is second order in B. So rate = $k[\ce{A}][\ce{B}]^{2}$.

k = rate ÷ ([A][B]²) = $0.0036$ ÷ (0.15 × 0.20²) = 0.600 M⁻² s⁻¹. The overall order is 3, so k has units of M⁻² s⁻¹.

For a first-order reaction $t_{1/2}$ = ln 2 ÷ k = 0.693 ÷ $0.00231$ = 3.00 × 10² s, independent of concentration. 87.5% decomposed leaves 12.5% = 1/8 = (1/2)³, so 3 half-lives: 3 × 3.00 × 10² = 9.00 × 10² s.

Compute 1/[A]: 2.0, 6.0, 10.0, 14.0, 18.0. It rises by about 4.0 every 20 s, so 1/[A] vs. t is linear: the reaction is second order, and the slope is k (0.20 M⁻¹ s⁻¹). [A] itself falls more and more slowly, and ln[A] doesn’t fall by equal steps either, so it isn’t zero or first order.

An elementary step is a single collision, so its rate law follows directly from the reactant coefficients: one NO and one NO₃ collide, a bimolecular step with rate = k[NO][NO₃]. “Only by experiment” applies to overall reactions, and a product like NO₂ doesn’t appear in the rate law.

k depends only on temperature (and on a catalyst). Raising the temperature raises the average kinetic energy, so a much larger fraction of collisions reach the activation energy and k increases. Changing concentrations changes the rate (how often molecules collide) but not k, and removing product doesn’t change k either.

The reverse reaction climbs from the products (−15) to the transition state (125): $E_a$(reverse) = 125 − (−15) = 140 kJ/mol. The forward activation energy is 125 − 40 = 85 kJ/mol and ΔH = −55 kJ/mol; the two activation energies differ by exactly ΔH.

Add the three steps and cancel what appears on both sides: step 1 makes 2 Cl and steps 2 and 3 each use 1, so Cl cancels; $\ce{CCl3}$ is made in step 2 and used in step 3, so it cancels too. What remains is $\ce{CHCl3 + Cl2 -> HCl + CCl4}$. Cl and $\ce{CCl3}$ are intermediates and don’t appear in the overall reaction.

The rate-determining step is step 2: rate = $k_2$[Cl][$\ce{CHCl3}$]. Cl is an intermediate, so use the fast equilibrium in step 1: $K_1 = \dfrac{[\ce{Cl}]^2}{[\ce{Cl2}]}$, so [Cl] = $K_1^{1/2}[\ce{Cl2}]^{1/2}$. Substituting gives rate = $k_2K_1^{1/2}$[$\ce{CHCl3}$][$\ce{Cl2}$]$^{1/2}$. The intermediate Cl can’t appear in the rate law, and the order in $\ce{Cl2}$ is 1/2, not 1.

Step 1’s activation energy is 60 − 0 = 60; step 2 starts from the intermediate (20): 95 − 20 = 75. Step 2 is larger (and has the highest transition state), so it is rate-determining. The products (−30) are below the reactants (0), so ΔH = −30 kJ/mol: exothermic.

The dashed curve has two peaks and a valley: the catalyst changes the pathway into two steps with an intermediate, and its highest peak (70) is far below the original (110). The reactants and products are unchanged, so ΔH is the same; the forward and reverse activation energies drop by the same amount, so the reverse reaction speeds up too.