Chemical Kinetics

Q1. (3 marks)

• (a) Average rate = change in concentration of reactant/product over a finite time interval: $\text{rate}_{avg} = -\dfrac{\Delta[A]}{\Delta t}$. (1)
• (b) Instantaneous rate = rate at a particular instant, i.e. the limit as $\Delta t \to 0$: $-\dfrac{d[A]}{dt}$. (1)
• Method: draw the tangent to the concentration–time curve at the given instant; its slope gives the instantaneous rate. (1)

Q2. (4 marks)

• Order: sum of powers of concentration terms in the experimentally determined rate law; can be zero, fractional or integer; an experimental quantity. (1)
• Molecularity: number of reacting species (atoms/molecules) colliding in an elementary step; always a whole number ($\geq 1$); a theoretical quantity. (1)
• Difference: order is experimental & may be fractional/zero, whereas molecularity is theoretical & always integral $\geq 1$. (1)
• Examples: order — decomposition of $N_2O_5$ (first order); molecularity — $2HI \to H_2 + I_2$ (bimolecular). (1)

Q3. (4 marks) $-\frac{d[A]}{dt} = k[A] \implies \frac{d[A]}{[A]} = -k\,dt$ (1) Integrate between limits $[A]_0$ (at $t=0$) and $[A]$ (at $t$): $\int_{[A]_0}^{[A]}\frac{d[A]}{[A]} = -k\int_0^t dt$ (1) $\ln\frac{[A]}{[A]_0} = -kt \implies \ln\frac{[A]_0}{[A]} = kt$ (1) Converting to base 10: $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$. (1)

Q4. (4 marks) Zero order: $-\dfrac{d[A]}{dt} = k \implies d[A] = -k\,dt$. Integrating: $[A] = [A]_0 - kt$. (2) At half-life, $[A] = \tfrac{[A]_0}{2}$, $t = t_{1/2}$: $\frac{[A]_0}{2} = [A]_0 - k\,t_{1/2} \implies k\,t_{1/2} = \frac{[A]_0}{2}$ (1) $\boxed{t_{1/2} = \frac{[A]_0}{2k}} \quad\Rightarrow\quad t_{1/2}\propto [A]_0$ (1)

Q5. (4 marks) First order: $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$. Here $[A] = \tfrac{1}{4}[A]_0$, so ratio $=4$. (1) $t = \frac{2.303}{1.15\times10^{-3}}\log 4$ (1) $\log 4 = 0.602$; $\dfrac{2.303}{1.15\times10^{-3}} = 2002.6\ \text{s}$. (1) $t = 2002.6 \times 0.602 \approx 1.206\times10^{3}\ \text{s} \approx 1205\ \text{s}$ (1)

Q6. (4 marks)

• A pseudo-first-order reaction is one that is truly higher order but behaves as first order because one reactant is present in large excess (nearly constant). (1)
• Example: $\text{CH}_3\text{COOC}_2\text{H}_5 + H_2O \xrightarrow{H^+} \text{CH}_3\text{COOH} + C_2H_5OH$. (1)
• Rate $= k[\text{ester}][H_2O]$; water is in large excess so $[H_2O]$ ≈ constant. (1)
• Thus rate $= k'[\text{ester}]$ where $k' = k[H_2O]$; the reaction appears first order. (1)

Q7. (5 marks) Arrhenius (two temperatures): $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ (1) $k_2/k_1 = 2$, $T_1=300$, $T_2=310$. $\log 2 = 0.301$. (1) $\frac{1}{300}-\frac{1}{310} = \frac{310-300}{300\times310}=\frac{10}{93000}=1.0753\times10^{-4}$ (1) $E_a = \frac{2.303\times8.314\times0.301}{1.0753\times10^{-4}}$ (1) $E_a = \frac{5.763}{1.0753\times10^{-4}} \approx 5.36\times10^{4}\ \text{J mol}^{-1} \approx 53.6\ \text{kJ mol}^{-1}$ (1)

Q8. (4 marks)

• Arrhenius equation: $k = A\,e^{-E_a/RT}$. (1)
• $A$ = frequency/pre-exponential factor; $E_a$ = activation energy; $R$ = gas constant; $T$ = absolute temperature; exponential term = fraction of effective collisions. (1)
• Plot: $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$ — straight line, negative slope. (1)
• Slope $= -E_a/R \Rightarrow E_a = -R\times(\text{slope})$; intercept $= \ln A$. (1)

Q9. (4 marks)

• Collision theory: reactant molecules must collide for a reaction; rate depends on collision frequency and the fraction of collisions that are effective. (1)
• Only collisions with energy $\geq E_a$ and proper orientation lead to product. (1)
• Frequency factor ($A$ / $Z$): number of collisions per unit time per unit volume (collision frequency). (1)
• Steric (probability) factor ($P$): fraction of collisions with correct orientation; $k = P\,Z\,e^{-E_a/RT}$. (1)

Q10. (4 marks)

• Homogeneous catalysis: catalyst and reactants in the same phase; e.g. $2SO_2 + O_2 \xrightarrow{NO} 2SO_3$ (all gases) or acid-catalysed ester hydrolysis. (1.5)
• Heterogeneous catalysis: catalyst in a different phase; e.g. Haber process $N_2 + 3H_2 \xrightarrow{Fe(s)} 2NH_3$. (1.5)
• Enzyme catalysis: highly specific, efficient biological catalyst active under mild conditions (body temperature, near-neutral pH). (1)

[
  {"claim":"Q5: time for first-order reaction to reach 1/4 concentration ≈ 1205 s","code":"k=Rational(115,100000); t=(2.303/float(k))*float(log(4,10)); result = abs(t-1205)<10"},
  {"claim":"Q4: half-life of zero order = [A]0/(2k)","code":"A0,k,t=symbols('A0 k t',positive=True); sol=solve(Eq(A0-k*t, A0/2), t)[0]; result = simplify(sol - A0/(2*k))==0"},
  {"claim":"Q7: activation energy ≈ 53.6 kJ/mol","code":"Rg=8.314; Ea=(2.303*Rg*float(log(2,10)))/((1/300)-(1/310)); result = abs(Ea/1000-53.6)<1.0"},
  {"claim":"Q3: integrated first-order gives ln([A]0/[A])=kt","code":"A0,k,t,A=symbols('A0 k t A',positive=True); expr=solve(Eq(log(A/A0), -k*t), A)[0]; result = simplify(expr - A0*exp(-k*t))==0"}
]