Chemical Kinetics

Q1

(a) [5] Rate law: $-\dfrac{d[A]}{dt} = k[A]$ (1) Separate: $\dfrac{d[A]}{[A]} = -k\,dt$ (1) Integrate both sides from $[A]_0\to[A]$, $0\to t$: $\int_{[A]_0}^{[A]} \dfrac{d[A]}{[A]} = -k\int_0^t dt$ (1) $\ln[A] - \ln[A]_0 = -kt$ (1) $\Rightarrow \ln[A] = \ln[A]_0 - kt$ (1)

(b) [4] At $t_{1/2}$, $[A] = [A]_0/2$: $\ln\frac{[A]_0/2}{[A]_0} = -kt_{1/2}$ → $\ln(1/2) = -kt_{1/2}$ (2) $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$ (1) Why: Because $[A]_0$ cancels in the log ratio, $t_{1/2}$ depends only on $k$ — the fractional decay per unit time is constant, so the time to halve is always the same regardless of where you start. (1)

(c) [3] 40% complete → $[A] = 0.60[A]_0$. $k = \dfrac{1}{t}\ln\frac{[A]_0}{[A]} = \dfrac{1}{25}\ln\frac{1}{0.6}$ (1) $= \dfrac{0.5108}{25} = 0.02043\ \text{min}^{-1}$ (1) $t_{1/2} = 0.693/0.02043 = 33.9\ \text{min}$ (1)

Q2

(a) [4] $-\dfrac{d[A]}{dt} = k[A]^2$ (1); $\dfrac{-d[A]}{[A]^2} = k\,dt$ (1) Integrate: $\left[\dfrac{1}{[A]}\right]_{[A]_0}^{[A]} = kt$ (1) $\dfrac{1}{[A]} - \dfrac{1}{[A]_0} = kt \Rightarrow \dfrac{1}{[A]} = \dfrac{1}{[A]_0} + kt$ (1)

(b) [2] $[A]=[A]_0/2$: $\dfrac{2}{[A]_0} - \dfrac{1}{[A]_0} = kt_{1/2}$ → $t_{1/2} = \dfrac{1}{k[A]_0}$ (2) (inversely proportional to $[A]_0$)

(c) [4]

• Zero order: plot $[A]$ vs $t$ → straight line, slope $= -k$. (1)
• First order: plot $\ln[A]$ vs $t$ → straight line, slope $= -k$. (1)
• Second order: plot $1/[A]$ vs $t$ → straight line, slope $= +k$. (1) Whichever plot gives the best straight line reveals the order (only one linearises). (1)

Q3

(a) [6] Two-point: $\ln\dfrac{k_2}{k_1} = -\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right) = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ (2) $\ln(8/2) = \ln 4 = 1.386$ (1) $\frac{1}{300}-\frac{1}{320} = 3.3333\times10^{-3} - 3.125\times10^{-3} = 2.0833\times10^{-4}$ (1) $E_a = \dfrac{R\ln 4}{(1/T_1 - 1/T_2)} = \dfrac{8.314\times1.386}{2.0833\times10^{-4}}$ (1) $= 55{,}300\ \text{J/mol} \approx 55.3\ \text{kJ/mol}$ (1)

(b) [4]

import math
def Ea(k1, k2, T1, T2):
    R = 8.314
    ea = R * math.log(k2/k1) / (1/T1 - 1/T2)
    return ea / 1000  # kJ/mol
print(Ea(2e-3, 8e-3, 300, 320))

(1 mark import/log ratio; 1 mark correct $1/T_1-1/T_2$; 1 mark /R factor; 1 mark units conversion & print) (4)

(c) [2] Catalyst lowers $E_a$ by providing an alternative path → increases $k$ (larger factor $e^{-E_a/RT}$). On the Arrhenius plot ($\ln k$ vs $1/T$) the slope $-E_a/R$ becomes less steep (smaller magnitude); intercept $\ln A$ may also shift. (2)

Q4

(a) [4] True rate $= k[\text{ester}][\text{H}_2\text{O}]$. Water is the solvent, in vast excess, so $[\text{H}_2\text{O}]$ is effectively constant. (2) Then rate $= k_{obs}[\text{ester}]$ with $k_{obs} = k[\text{H}_2\text{O}]$ — the reaction appears first order (pseudo-first-order). (2)

(b) [6] Compare Expt 1 & 2 ($[B]$ fixed): $[A]$ doubles, rate ×4 → $2^m=4$, $m=2$ (2nd order in A). (2) Compare Expt 2 & 3 ($[A]$ fixed): $[B]$ doubles, rate unchanged → $2^n=1$, $n=0$ (0 order in B). (2) Overall order $= m+n = 2$. Rate $= k[A]^2$. (2)

Q5

(a) [5] RDS is step 2: rate $= k_2[N_2O_2][O_2]$. (1) Pre-equilibrium: $K = \dfrac{[N_2O_2]}{[NO]^2}$ → $[N_2O_2] = K[NO]^2$. (2) Substitute: rate $= k_2 K [NO]^2[O_2]$. (1) Define $k = k_2K$: rate $= k[NO]^2[O_2]$. (1)

(b) [3] Overall order $= 2 + 1 = 3$ (third order). (1) The overall reaction $2NO+O_2$ would formally have molecularity 3 (termolecular), which is extremely rare; the mechanism shows it actually proceeds via bimolecular elementary steps. (1) Molecularity is defined only for elementary steps (integer count of species colliding), whereas the observed order comes from the composite mechanism — hence they can differ. (1)

Q6

(a) [4] $k = P\,Z\,e^{-E_a/RT}$ (or $k = PA e^{-E_a/RT}$), where $Z/A$ = collision frequency factor. (1) $A$ (frequency factor): number of collisions per unit time per unit concentration — the maximum possible rate if every sufficiently energetic collision reacted. (1.5) $P$ (steric factor): fraction accounting for correct orientation of colliding molecules; $P\le1$ typically, correcting for the fact that only properly-aligned collisions lead to reaction. (1.5)

(b) [4] Collision theory: reaction occurs when molecules collide with energy $\ge E_a$ and correct orientation; treats molecules as hard spheres, no explicit complex. (1) TST: reactants pass through a high-energy activated complex (transition state) in quasi-equilibrium at the energy barrier top; rate governed by decomposition of this complex. (2) Advantage of TST: incorporates molecular structure/thermodynamics ($\Delta G^{\ddagger}$) and predicts $A$ and $P$ better than the crude hard-sphere collision picture. (1)

[
{"claim":"Q1c: k = ln(1/0.6)/25 ≈ 0.02043 /min","code":"k=ln(1/Rational(6,10))/25; result=abs(float(k)-0.020431)<1e-4"},
{"claim":"Q1c: t_half = 0.693/k ≈ 33.9 min","code":"k=ln(1/Rational(6,10))/25; th=ln(2)/k; result=abs(float(th)-33.92)<0.1"},
{"claim":"Q3a: Ea ≈ 55.3 kJ/mol","code":"R=8.314; Ea=R*ln(4)/(Rational(1,300)-Rational(1,320)); result=abs(float(Ea)/1000-55.3)<0.5"},
{"claim":"Q4b: order in A is 2 (rate x4 when A doubled)","code":"m=log(4,2); result=m==2"}
]