Exercises — Pseudo-first-order kinetics

Answer: (b) and (c). For pseudo-first-order behaviour we need the flooded reactant to be at least $\approx 50$–$100\times$ the watched one.

• (a) ratio $=0.5/0.5=1$. Both change equally → truly second-order, not pseudo.
• (b) ratio $=0.90/0.005=180$. Huge excess → $[B]\approx [B{]}_0$ holds → pseudo-first-order. ✔
• (c) water sits at $\approx 55.5\text{M}$ while the ester is dilute → excess is automatic → pseudo-first-order. ✔

A first-order constant multiplies a single concentration, so its units must be ${\text{s}}^{-1}$. A second-order constant multiplies two concentrations ($\text{M}\times \text{M}$), so to leave a rate in ${\text{M s}}^{-1}$ it must carry ${\text{M}}^{-1}{\text{s}}^{-1}$.

• $3.0\times 10^{-4}{\text{s}}^{-1}$ → this is $k^{\prime }$ (pseudo, first-order units).
• $3.0\times 10^{-4}{\text{M}}^{-1}{\text{s}}^{-1}$ → this is $k$ (true, second-order units).

Water is the flooded reactant, so $[B{]}_0=55.5\text{M}$. Rearranging $k^{\prime }=k[B{]}_0$: $k=\frac{k^{\prime }}{[B{]}_0}=\frac{2.5\times 10^{-4}}{55.5}=4.5\times 10^{-6}{\text{M}}^{-1}{\text{s}}^{-1}.$ Sanity on units: $\frac{{\text{s}}^{-1}}{\text{M}}={\text{M}}^{-1}{\text{s}}^{-1}$ ✔ — the true bimolecular units.

Convert time: $t=2\times 3600=7200\text{s}$. Use the exponential form (it hands us concentration directly, no logs needed): $[A]=[A{]}_0{e}^{-k^{\prime }t}=0.50e^{-(6.0\times 10^{-5})(7200)}=0.50e^{-0.432}.$ Now $e^{-0.432}=0.649$, so $[A]=0.50\times 0.649=0.325\text{M}.$ Fraction remaining $=0.325/0.50=0.65$, i.e. $\approx 35\%$ reacted. Reasonable for 2 h at this rate.

$t_{1/2}=\frac{0.693}{k^{\prime }}=\frac{0.693}{6.0\times 10^{-5}}=1.155\times 10^4\text{s}\approx 192.5\text{min}.$ Because $t_{1/2}$ contains only $k^{\prime }$, it is independent of $[A{]}_0$ — the signature of first-order (and hence pseudo-first-order) kinetics. See Half-life.

Look at the two plots in the figure below. What to observe: the deep-teal curve (Run 2, huge excess) is a perfectly straight descending line, whereas the burnt-orange curve (Run 1, equal concentrations) bends downward and steepens — that bend is the visual tell of genuine second-order behaviour forced onto a $\ln [A]$ axis.

• Run 1 curves because both concentrations drop, so the reaction obeys the genuine second-order integrated law, which is not linear on a $\ln [A]$ axis.
• Run 2 goes straight because $[B{]}_0/[A{]}_0=100$ freezes $[B]$; the reaction is isolated to first-order in $A$ (see Method of Isolation). The straight slope is $-k^{\prime }$, so $k^{\prime }=0.050{\text{s}}^{-1}$. $k=\frac{k^{\prime }}{[B{]}_0}=\frac{0.050}{1.0}=0.050{\text{M}}^{-1}{\text{s}}^{-1}.$ Overall order $=1$ (in $A$) $+1$ (in $B$) $=$ second order truly; observed order in Run 2 is first.

Say $[A{]}_0=1$ unit, so $[B{]}_0=10$ units. Reacting half of $A$ consumes $0.5$ units of $B$ (one $B$ per one $A$). $\text{drop in }[B]=\frac{0.5}{10}=0.05=5\%.$ A $5\%$ wander in the "constant" is already visible on a careful $\ln [A]$ plot — it produces a slight curve. So $10\times$ is marginal: fine for a rough estimate, poor for precise $k$. Aim for $\ge 50\times$.

Compare with $\ln [A]=\ln [A{]}_0-k^{\prime }t$. Matching term by term:

• Intercept $\ln [A{]}_0=-4.605\Rightarrow [A{]}_0=e^{-4.605}=0.0100\text{M}.$
• Slope $=-k^{\prime }=-0.0300\Rightarrow k^{\prime }=0.0300{\text{s}}^{-1}.$

Going from $0.020$ to $0.010\text{M}$ is exactly one half-life, so $t_{1/2}=231\text{s}$. $k^{\prime }=\frac{0.693}{t_{1/2}}=\frac{0.693}{231}=3.0\times 10^{-3}{\text{s}}^{-1}.$ True constant: $k=\frac{k^{\prime }}{[{\text{H}}_2\text{O}]}=\frac{3.0\times 10^{-3}}{55.5}=5.4\times 10^{-5}{\text{M}}^{-1}{\text{s}}^{-1}.$ Reaching $0.0025\text{M}$ from $0.020\text{M}$ is $0.0025/0.020=1/8=(1/2{)}^3$ → three half-lives: $t=3\times 231=693\text{s}.$ Cross-check with the log law: $t=\frac{1}{k^{\prime }}\ln \frac{0.020}{0.0025}=\frac{\ln 8}{3.0\times 10^{-3}}=\frac{2.079}{3.0\times 10^{-3}}=693\text{s}$ ✔.

Consuming all of $A$ removes $[A{]}_0$ worth of $B$. The fractional drift in $B$ is $\frac{[A{]}_0}{[B{]}_0}\le 0.01⟹\frac{[B{]}_0}{[A{]}_0}\ge 100.$ So a 100× excess keeps the error at the $1\%$ level — precisely the textbook threshold. For a looser $2\%$ tolerance, $50\times$ suffices.

In this low-substrate regime the whole prefactor acts as one first-order constant in $[S]$: $k^{\prime }=\frac{V_{\max }}{K_M}=\frac{2.0\times 10^{-5}}{4.0\times 10^{-3}}=5.0\times 10^{-3}{\text{s}}^{-1}.$ $t_{1/2}=\frac{0.693}{k^{\prime }}=\frac{0.693}{5.0\times 10^{-3}}=138.6\text{s}.$ Why first-order here? With $[S]\ll K_M$ the enzyme is mostly empty, so doubling $[S]$ doubles the rate — the hallmark of first-order dependence, exactly the pseudo-first-order idea in a new setting.

(a) Compute $\ln [A]$: $-4.605,-4.855,-5.104,-5.356$. Successive gaps: $-0.250,-0.249,-0.252$ — constant per equal time step, so $\ln [A]$ vs $t$ is a straight line → observed first-order. ✔

(b) Slope $=-k^{\prime }=\frac{-5.356-(-4.605)}{300-0}=\frac{-0.751}{300}=-2.50\times 10^{-3}{\text{s}}^{-1}$, so $k^{\prime }=2.50\times 10^{-3}{\text{s}}^{-1}.$ (c) $k=\frac{k^{\prime }}{[B{]}_0}=\frac{2.50\times 10^{-3}}{0.80}=3.13\times 10^{-3}{\text{M}}^{-1}{\text{s}}^{-1}.$ (d) $[A]=[A{]}_0{e}^{-k^{\prime }t}=0.0100e^{-(2.50\times 10^{-3})(500)}=0.0100e^{-1.25}=0.0100\times 0.2865=2.87\times 10^{-3}\text{M}.$

$[B{]}_0$ is essentially unchanged with temperature, so $k^{\prime }\propto k$, and the unknown factor $A$ cancels in a ratio (this is why we take a ratio — it lets us predict without ever knowing $A$): $\frac{k_2^{\prime }}{k_1^{\prime }}=\frac{k_2}{k_1}=\frac{A{e}^{-E_a/R{T}_2}}{A{e}^{-E_a/R{T}_1}}=\exp \left[-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right].$ With $T_1=298\text{K},T_2=308\text{K}$: $\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{298}-\frac{1}{308}=1.0896\times 10^{-4}{\text{K}}^{-1}.$ $\frac{k_2^{\prime }}{k_1^{\prime }}=\exp \left[\frac{50000}{8.314}\times 1.0896\times 10^{-4}\right]=\exp (0.6553)=\mathrm{1.93.}$ So $k^{\prime }$ roughly doubles for a $10^{\circ }$ rise — the classic rule-of-thumb, now derived. New $k_2^{\prime }\approx 1.0\times 10^{-3}\times 1.93=1.93\times 10^{-3}{\text{s}}^{-1}.$

Fold both "constant" species into the apparent constant: $k^{\prime }=k[{\text{H}}_2\text{O}][{\text{H}}^{+}]$. Water is genuinely fixed at $55.5\text{M}$, so it is not the culprit. The catalyst ${\text{H}}^{+}$ is regenerated each cycle (a catalyst is not consumed), but the reaction produces acetic acid; if the weak buffer is too dilute to absorb that extra acid, the solution's $[{\text{H}}^{+}]$ slowly rises. Since $k^{\prime }\propto [{\text{H}}^{+}]$, a rising $[{\text{H}}^{+}]$ makes $k^{\prime }$ grow with time, and a growing $k^{\prime }$ steepens the $\ln [\text{ester}]$ slope — exactly the observed drift.

• Diagnosis: the pseudo-first-order approximation is valid instant-by-instant, but one of the quantities hidden inside $k^{\prime }$ (the catalyst, not the solvent) is not truly clamped.
• Fix: use a stronger / more concentrated buffer so that the produced acid barely shifts $[{\text{H}}^{+}]$; then $k^{\prime }$ stays constant and the whole run collapses onto a single straight line.