Question 1 of 12: Enzyme and Reaction-Order Kinetics

Question 1: Enzyme and Reaction-Order Kinetics (16.7 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Both alternatives offered for part (b) are worked below so the pair forms a complete study set; either alone satisfies the exam.

(a) Apparent order at S >> k_m

The Michaelis–Menten rate law is $v = V_{max}S/(k_m+S)$. When $S \gg k_m$, the denominator $k_m+S \approx S$, so $v \approx V_{max}S/S = V_{max}$: the rate collapses to a constant, independent of substrate concentration. The apparent reaction order is therefore zero order in S. Physically, at high substrate concentration every enzyme active site is already saturated with bound substrate (the enzyme is working at its maximum turnover rate), so adding more substrate cannot speed the reaction further — the rate is limited by how fast the enzyme itself can turn over, not by how often substrate molecules encounter it.

(b) — Option 1: bounding the activation energy of a two-term rate law

Given. $-r_A = k_1C_A^2/(1+k_2C_A)$, with $k_1=10^{19.39}\exp(-81.8/RT)$ and $k_2=10^{8.69}\exp(-28.4/RT)$ (activation energies in the exponents, here in units consistent with $R$ as given, kcal/mol).

Find. The minimum and maximum apparent activation energy this rate law can exhibit.

Approach. Examine the two limiting regimes of the denominator ($k_2C_A\ll 1$ and $k_2C_A\gg 1$); each limit collapses the rate to a single Arrhenius term whose exponent is the apparent activation energy.

1. Low-concentration limit ($k_2C_A \ll 1$). The denominator $\to 1$, so $$-r_A \to k_1C_A^2 = 10^{19.39}\exp\!\left(\frac{-81.8}{RT}\right)C_A^2.$$ The apparent activation energy here is simply that of $k_1$ alone: $$E_{app,\,low} = E_1 = \boxed{81.8}.$$
2. High-concentration limit ($k_2C_A \gg 1$). The denominator $\to k_2C_A$, so $$-r_A \to \frac{k_1}{k_2}C_A = 10^{19.39-8.69}\exp\!\left(\frac{-(81.8-28.4)}{RT}\right)C_A.$$ Dividing two Arrhenius terms subtracts their activation energies, giving the apparent value $$E_{app,\,high} = E_1-E_2 = 81.8-28.4 = \boxed{53.4}.$$
3. Bound the observable range. Since $E_1 > E_1-E_2$ (as $E_2>0$), every concentration in between the two limits produces an apparent $E$ that lies between these two extremes. The rate law can therefore show an apparent activation energy anywhere from $53.4$ (dilute, high-order regime) up to $81.8$ (concentrated, low-order regime), never outside that band.

(b) — Option 2: enzyme conversion at 10 min from a 2 min datum

Given. $k_m=10^{-3}$ M; $S_0=3(10^{-5})$ M; at $t=2$ min, $X=5\%$ converted.

Find. Conversion $X$ at $t=10$ min.

Approach. Since $S_0 \ll k_m$ (by more than an order of magnitude), the Michaelis–Menten rate reduces to the first-order limit $v \approx (V_{max}/k_m)S$ — use the 2 min datum to fix the pseudo-first-order constant, then integrate to 10 min.

1. Confirm the first-order regime and write the integrated law. $S_0/k_m = 3(10^{-5})/10^{-3}=0.03\ll 1$, so $k_m+S\approx k_m$ throughout the run and $-dS/dt \approx (V_{max}/k_m)S = kS$, i.e. $\ln(S_0/S)=kt$.
2. Fix $k$ from the 2 min datum. $X=0.05 \Rightarrow S/S_0 = 0.95$, so $$k = \frac{-\ln(1-0.05)}{2} = \frac{-\ln 0.95}{2} = \boxed{0.02565\ \text{min}^{-1}}.$$
3. Project to 10 min. $$X_{10} = 1-\exp(-kt) = 1-\exp[-(0.02565)(10)] = 1-\exp(-0.2565) = \boxed{22.6\%}.$$