Exercises — Catalysis — homogeneous, heterogeneous, enzyme catalysis

• (a) Mo — a promoter. It is not the catalyst itself; it boosts the activity of the Fe catalyst.
• (b) As on Pt — a poison. Arsenic clings to the platinum active sites and destroys its catalytic activity.
• (c) aqueous $H^{+}$ — a homogeneous catalyst. The ester, water, and $H^{+}$ are all in the same liquid phase (see Adsorption is not needed here — no surface).
• (d) $V_2{O}_5$ — a heterogeneous catalyst. A solid acting on gaseous $S{O}_2+O_2$: different phase, surface reaction.

1. False. A catalyst is regenerated — it comes out unchanged in amount.
2. False. A catalyst lowers only the hill (the transition-state peak), never the valleys. $\Delta H=H_{prod}-H_{react}$ is fixed, so exo/endo character never flips. ("Lower the hill, not the house.")
3. True. A negative catalyst (inhibitor) blocks the easy path, so molecules must climb a higher effective $E_a$ → slower.

What we do: use the speed-up ratio, because it isolates the catalyst's effect (same $A$, same $T$ cancel out). Step 1 — get $\Delta E_a$ in joules. $\Delta E_a=E_{a,un}-E_{a,cat}=95-60=35\text{kJ/mol}=35000\text{J/mol}$. Step 2 — compute $RT$ (the "thermal energy scale" we measure the hill against): $RT=8.314\times 310=2577.3\text{J/mol}.$ Step 3 — divide (the drop measured in units of $RT$): $\frac{\Delta E_a}{RT}=\frac{35000}{2577.3}=\mathrm{13.58.}$ Step 4 — exponentiate: $\frac{k_{cat}}{k_{un}}=e^{13.58}\approx 7.9\times 10^5.$ The reaction runs about 790,000× faster — a 35 kJ cut, magnified by the exponential.

Tool: the Michaelis–Menten law from the parent note, $v=\frac{V_{max}[S]}{K_M+[S]}.$ (a) $[S]=K_M=2.0\text{mM}$. When substrate equals $K_M$, the denominator is $K_M+K_M=2K_M$, so $v=\frac{V_{max}{K}_M}{2K_M}=\frac{V_{max}}{2}=2.0\mu {\text{M s}}^{-1}.$ This is why $K_M$ is defined as "the $[S]$ giving half of $V_{max}$." (b) $[S]=18.0\text{mM}$. $v=\frac{4.0\times 18.0}{2.0+18.0}=\frac{72.0}{20.0}=3.6\mu {\text{M s}}^{-1}.$ Notice we are already at $90\%$ of $V_{max}$ — approaching saturation.

(1) $\Delta H$ is the vertical gap between the reactant valley and the product valley (the pink arrow). The catalyst only pulls the peak down; the valleys stay put, so $\Delta H$ is unchanged. (2) Forward $E_a$ = (peak) − (reactant level). Uncatalysed peak sits at $100\text{kJ/mol}$, catalysed peak at $60\text{kJ/mol}$, reactants at $0$. So forward $E_a$ drops from $100$ to $60$: a fall of $40\text{kJ/mol}$. The product valley sits at $-20\text{kJ/mol}$. Backward $E_a$ = (peak) − (product level):

• uncatalysed: $100-(-20)=120\text{kJ/mol}$,
• catalysed: $60-(-20)=80\text{kJ/mol}$. Backward $E_a$ also falls by $40\text{kJ/mol}$ — the same amount. (3) Since $k_f$ and $k_b$ each scale by the identical factor $e^{40000/RT}$, $K_{eq}=\frac{k_f}{k_b}\Rightarrow K_{eq}^{cat}=\frac{k_f\cdot e^{40000/RT}}{k_b\cdot e^{40000/RT}}=\frac{k_f}{k_b}=K_{eq}.$ The equal factors cancel. See Chemical Equilibrium.

Both share the same uncatalysed baseline, so compare them directly: $\frac{k_Y}{k_X}=e^{(50000-45000)/(8.314\times 300)}=e^{5000/2494.2}=e^{2.005}\approx \mathrm{7.4.}$ A mere $5\text{kJ/mol}$ makes Y about 7.4× faster. Because $E_a$ is inside the exponential, even "small" differences in the hill height are amplified into large speed differences — this is why catalyst screening chases every kJ.

Set up $RT$: $RT=8.314\times 298=2477.6\text{J/mol}$. (1) Uncatalysed: $k_{un}=10^{13}{e}^{-90000/2477.6}=10^{13}{e}^{-36.32}=10^{13}\times 1.68\times 10^{-16}\approx 1.68\times 10^{-3}{\text{s}}^{-1}.$ Catalysed: $k_{cat}=10^{13}{e}^{-55000/2477.6}=10^{13}{e}^{-22.20}=10^{13}\times 2.27\times 10^{-10}\approx 2.27\times 10^3{\text{s}}^{-1}.$ (2) For a fixed reaction, time to reach a given conversion is inversely proportional to $k$ (faster $k$ → shorter time). So $t_{cat}=t_{un}\times \frac{k_{un}}{k_{cat}}=5\text{h}\times \frac{1.68\times 10^{-3}}{2.27\times 10^3}=5\times 7.40\times 10^{-7}\text{h}.$ That is $3.70\times 10^{-6}\text{h}\approx 0.0133\text{s}$. A five-hour job collapses to roughly 13 milliseconds — the power of a $35\text{kJ}$ cut. (See Rate Law and Order of Reaction for why $t\propto 1/k$ holds for a fixed order.)

Set up: we want $v=0.80V_{max}$. Plug into the law and cancel $V_{max}$: $0.80=\frac{[S]}{K_M+[S]}.$ Solve (cross-multiply): $0.80(K_M+[S])=[S]\Rightarrow 0.80K_M=[S](1-0.80)=0.20[S].$ $[S]=\frac{0.80}{0.20}{K}_M=4K_M=4\times 5.0=20.0\text{mM}.$ To go from $50\%$ (at $[S]=K_M$) up to $80\%$ takes four times $K_M$ — the curve flattens, so each extra percent of $V_{max}$ costs disproportionately more substrate. That is saturation in action; see Enzymes and Proteins (Biomolecules).

Which tool and why? We know the ratio and want the energy drop — so invert the speed-up formula. The inverse of an exponential is the natural logarithm $\ln$ (it answers "$e$ to what power gives this ratio?"). Start: $\frac{k_{cat}}{k_{un}}=e^{\Delta E_a/RT}\Rightarrow \ln \left(\frac{k_{cat}}{k_{un}}\right)=\frac{\Delta E_a}{RT}.$ Take the log: $\ln (5.0\times 10^4)=10.82$. Multiply by $RT$: $RT=8.314\times 350=2909.9\text{J/mol}$, so $\Delta E_a=10.82\times 2909.9=3.148\times 10^4\text{J/mol}\approx 31.5\text{kJ/mol}.$ A $50,000\times$ boost required only a $\sim 31.5\text{kJ/mol}$ cut — again the exponential doing the heavy lifting.

(1) From the parent's definition $V_{max}=k_2[E{]}_0$: $V_{max}=300{\text{s}}^{-1}\times 0.010\text{mM}=3.0{\text{mM s}}^{-1}.$ (2) Apply Michaelis–Menten: $v=\frac{V_{max}[S]}{K_M+[S]}=\frac{3.0\times 1.0}{4.0+1.0}=\frac{3.0}{5.0}=0.60{\text{mM s}}^{-1}.$ (3) Fraction $=v/V_{max}=0.60/3.0=0.20$, i.e. $20\%$ of $V_{max}$. Because the fraction $[S]/(K_M+[S])$ is also the fraction of enzyme bound as $ES$, only about $20\%$ of active sites are occupied at any instant here — we are on the low-$[S]$, near-first-order part of the curve, far from saturation. Compare Figure s02.

• Convert $E_a$ to joules before dividing by $RT$? ::: Always — $R$ is in J.
• Inverse operation to pull $E_a$ out of $e^{E_a/RT}$? ::: Natural logarithm $\ln$.
• Time vs rate constant relationship (fixed order)? ::: $t\propto 1/k$ — bigger $k$, shorter time.
• $[S]$ giving half $V_{max}$? ::: $[S]=K_M$.
• $[S]$ giving $80\%$ of $V_{max}$? ::: $[S]=4K_M$.
• Does a catalyst change the two valleys of the energy diagram? ::: No — only the peak.