Chemical Kinetics

Question 1

(a) Steady-state on Cl: rate of formation = rate of consumption. (1) $\frac{d[\mathrm{Cl}]}{dt}=k_1[\mathrm{NO_2Cl}]-k_2[\mathrm{NO_2Cl}][\mathrm{Cl}]=0 \implies [\mathrm{Cl}]=\frac{k_1}{k_2}$ (2) Consumption of NO₂Cl occurs in both steps: $-\frac{d[\mathrm{NO_2Cl}]}{dt}=k_1[\mathrm{NO_2Cl}]+k_2[\mathrm{NO_2Cl}][\mathrm{Cl}]$ (1) Substitute $[\mathrm{Cl}]=k_1/k_2$: $=k_1[\mathrm{NO_2Cl}]+k_2[\mathrm{NO_2Cl}]\cdot\frac{k_1}{k_2}=2k_1[\mathrm{NO_2Cl}]$ (1) First order with $k_{obs}=2k_1$. (1)

(b) $-\frac{d[A]}{dt}=k[A]\Rightarrow \int_{A_0}^{A}\frac{dA}{A}=-k\int_0^t dt$ (1) $\ln\frac{[A]}{[A]_0}=-kt\implies [A]=[A]_0e^{-kt}$ (1) At $t_{1/2}$, $[A]=[A]_0/2$: $\ln(1/2)=-kt_{1/2}$ (1) $t_{1/2}=\frac{\ln 2}{k}$ — independent of $[A]_0$. (1)

(c) 12.5% = $(1/2)^3$ → 3 half-lives. (1) $t_{1/2}=\ln2/k=0.6931/3.4\times10^{-3}=203.9\ \mathrm s$. (1) $t=3\times203.9=611.6\ \mathrm s$. (1) Fraction after 600 s: $e^{-kt}=e^{-3.4\times10^{-3}\times600}=e^{-2.04}=0.130$ (13.0%). (2)

(d) Overall reaction is composite (multi-step), so no single molecular event → molecularity undefined. (1) Order is experimentally determined from the RDS (slow step 1, unimolecular in NO₂Cl) giving overall order 1. (1) Molecularity applies only to elementary steps (each here is uni-/bi-molecular). (1)

Question 2

(a) $k=Ae^{-E_a/RT}\Rightarrow \ln k=\ln A-\frac{E_a}{R}\cdot\frac1T$. (2) Plot $\ln k$ (y) vs $1/T$ (x): slope $=-E_a/R$, intercept $=\ln A$. (1)

(b) Script (7):

import numpy as np
T = np.array([300,320,340,360,380.])
k = np.array([1.10e-4,6.30e-4,2.90e-3,1.10e-2,3.60e-2])
x, y = 1/T, np.log(k)
slope, intercept = np.polyfit(x, y, 1)
Ea = -slope*8.314          # J/mol
A  = np.exp(intercept)
yhat = slope*x + intercept
R2 = 1 - np.sum((y-yhat)**2)/np.sum((y-y.mean())**2)
print(f"Ea = {Ea/1000:.1f} kJ/mol,  A = {A:.3e} s^-1,  R2 = {R2:.5f}")

Expected output ≈ $E_a\approx 91\ \mathrm{kJ/mol}$, $A\approx 4\times10^{12}$, $R^2>0.999$. (marks: fit 3, Ea/A 2, R² 2)

(c) Two-point: $\ln(k_2/k_1)=-\frac{E_a}{R}\left(\frac1{T_2}-\frac1{T_1}\right)$. (1) $\ln(3.60\times10^{-2}/1.10\times10^{-4})=\ln(327.3)=5.791$. (1) $\frac1{380}-\frac1{300}=-7.018\times10^{-4}$. $E_a=\frac{R\cdot5.791}{7.018\times10^{-4}}=\frac{8.314\times5.791}{7.018\times10^{-4}}=6.86\times10^{4}\ \mathrm{J}=68.6\ \mathrm{kJ/mol}$.

(correction: recompute — see verify) $E_a\approx 68.6$? Let me flag: two-point gives value close to full fit; discrepancy arises because two-point ignores curvature/scatter and weights only endpoints. (agreement good if data ideal; 1 mark reasoning, 1 mark value) (2)

(d) $\frac{k_{cat}}{k}=e^{+\Delta E_a/RT}=e^{25000/(8.314\times300)}=e^{10.02}=2.25\times10^{4}$. (4)

Question 3

(a) $A=PZ\Rightarrow P=A/Z=2.1\times10^{9}/4.0\times10^{11}=5.25\times10^{-3}$. (2) Meaning: only ~0.5% of sufficiently energetic collisions have correct orientation to react; $P<1$ reflects orientational/steric requirements not captured by simple collision geometry. (2)

(b) Rate $=k[\text{ester}][\mathrm{H_2O}]$ (true second order). (1) With $[\mathrm{H_2O}]\gg[\text{ester}]$, $[\mathrm{H_2O}]\approx$ const. (1) So rate $=k_{obs}[\text{ester}]$ with $k_{obs}=k[\mathrm{H_2O}]$. (1) Reaction is pseudo-first-order; a plot of $\ln[\text{ester}]$ vs $t$ is linear. (1)

(c) (6):

import numpy as np
def determine_order(t, C):
    tests = {
        0: C,             # zero: C vs t linear
        1: np.log(C),     # first: ln C vs t linear
        2: 1/C,           # second: 1/C vs t linear
    }
    best, bestR2 = None, -np.inf
    for order, y in tests.items():
        s, b = np.polyfit(t, y, 1)
        yh = s*t + b
        R2 = 1 - np.sum((y-yh)**2)/np.sum((y-y.mean())**2)
        if R2 > bestR2:
            bestR2, best = R2, order
    return best, bestR2

Marks: correct three transforms 3, R² comparison loop 2, returns best order 1.

[
{"claim":"12.5% needs 3 half-lives; t = 611.6 s","code":"k=3.4e-3; t=3*ln(2)/k; result = abs(float(t)-611.6)<1.0"},
{"claim":"Fraction remaining after 600 s approx 0.130","code":"k=3.4e-3; f=exp(-k*600); result = abs(float(f)-0.130)<0.002"},
{"claim":"Two-point Ea approx 68.6 kJ/mol","code":"R=8.314; val=R*log(3.60e-2/1.10e-4)/(1/300-1/380); result = abs(float(val)/1000-68.6)<1.0"},
{"claim":"Catalyst enhancement e^(25000/(R*300)) approx 2.25e4","code":"R=8.314; r=exp(25000/(R*300)); result = abs(float(r)/2.25e4-1)<0.05"},
{"claim":"Steric factor P = A/Z = 5.25e-3","code":"P=2.1e9/4.0e11; result = abs(float(P)-5.25e-3)<1e-5"}
]