4.8 Integrated Rate Laws, Half-Life, Arrhenius Behaviour & Catalysis

4.8 Integrated Rate Laws, Half-Life, Arrhenius Behaviour & Catalysis

3. Integrated Rate Laws and Half-Life Kinetics

Integrated rate laws express reactant concentration directly as a continuous function of elapsed time ($t$):

Zero-Order Kinetics

• Differential Rate Law: $\text{Rate} = -\frac{d[\text{A}]}{dt} = k$
• Integrated Rate Law: $$[\text{A}]_t = [\text{A}]_0 - k t$$
• Linear Graphical Plot: A plot of $[\text{A}]$ versus time ($t$) yields a straight line with slope $= -k$ and y-intercept $= [\text{A}]_0$.
• Half-Life ($t_{1/2}$): $$t_{1/2} = \frac{[\text{A}]_0}{2 k}$$ Notice: The zero-order half-life is directly proportional to initial concentration. As the reaction progresses and reactant is depleted, successive half-lives become shorter.
• Clinical Application: Zero-order kinetics characterizes saturated enzyme systems where substrate concentration vastly exceeds enzyme active site capacity ($[S] \gg K_m$, Michaelis-Menten $V_{\text{max}}$ regime). A classic clinical example is ethanol metabolism by hepatic alcohol dehydrogenase: the liver eliminates a constant mass of alcohol per hour (approximately $7-10\text{ g/hour}$), regardless of blood alcohol concentration.

First-Order Kinetics

• Differential Rate Law: $\text{Rate} = -\frac{d[\text{A}]}{dt} = k [\text{A}]$
• Integrated Rate Law: $$\ln[\text{A}]_t = \ln[\text{A}]_0 - k t \iff [\text{A}]_t = [\text{A}]_0 e^{-k t}$$
• Linear Graphical Plot: A plot of $\ln[\text{A}]$ versus time ($t$) yields a straight line with slope $= -k$ and y-intercept $= \ln[\text{A}]_0$.
• Half-Life ($t_{1/2}$): $$t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$$ Critical Feature: The first-order half-life is completely independent of initial reactant concentration. The time required to eliminate $50\%$ of the drug is constant regardless of whether initial plasma concentration is $100\text{ mg/L}$ or $10\text{ mg/L}$.
• Clinical / Physical Application: Radioactive decay of radioisotopes (e.g., $^{14}\text{C}, ^{131}\text{I}, ^{99m}\text{Tc}$) and linear pharmacokinetic drug clearance (most therapeutic pharmaceuticals follow first-order elimination where clearance rate is proportional to circulating plasma concentration).

Second-Order Kinetics

• Differential Rate Law: $\text{Rate} = -\frac{d[\text{A}]}{dt} = k [\text{A}]^2$
• Integrated Rate Law: $$\frac{1}{[\text{A}]_t} = \frac{1}{[\text{A}]_0} + k t$$
• Linear Graphical Plot: A plot of $\frac{1}{[\text{A}]}$ versus time ($t$) yields a straight line with positive slope $= +k$ and y-intercept $= \frac{1}{[\text{A}]_0}$.
• Half-Life ($t_{1/2}$): $$t_{1/2} = \frac{1}{k [\text{A}]_0}$$ Notice: The second-order half-life is inversely proportional to initial concentration. As the reaction proceeds and $[\text{A}]_0$ declines, each successive half-life doubles in duration ($t_{1/2, 2} = 2 \times t_{1/2, 1}$).

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4. The Arrhenius Equation and Temperature Dependence

The rate constant $k$ is not a static constant; it is exquisitely sensitive to temperature, governed by the Arrhenius equation:

$$k = A e^{-E_a / RT}$$

Where:

• $A$ is the pre-exponential frequency factor, representing the frequency of collisions with proper steric alignment.
• $E_a$ is the activation energy (in $\text{J/mol}$).
• $R$ is the ideal gas constant ($8.314\text{ J}/(\text{mol}\cdot\text{K})$).
• $T$ is absolute temperature in Kelvin.

Linear Form of the Arrhenius Equation

Taking the natural logarithm of both sides transforms the exponential curve into a linear slope-intercept equation:

$$\ln(k) = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln(A)$$

• An Arrhenius plot of $\ln(k)$ on the y-axis versus $\frac{1}{T}$ on the x-axis produces a straight line with: $$\text{Slope} = -\frac{E_a}{R} \implies E_a = -R \times (\text{Slope})$$ $$\text{y-intercept} = \ln(A)$$

Two-Point Form for Temperature Shifts

To calculate how a rate constant shifts between temperature $T_1$ and temperature $T_2$:

$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

• MCAT Rule of Thumb: For many biological and chemical reactions near room temperature ($300\text{ K}$), every $10^\circ\text{C}$ increase in temperature roughly doubles the reaction rate ($k_2 / k_1 \approx 2$).

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5. Catalysis and Biological Enzymes

A catalyst is a chemical agent that accelerates the velocity of a chemical reaction without being permanently consumed or chemically altered in the process.

Mechanism of Action

Catalysts accelerate reactions by stabilizing the transition state and providing an alternative reaction pathway possessing a substantially lower activation energy ($E_a$):

$$\text{Rate Constant Ratio: } \frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{(E_{a,\text{uncat}} - E_{a,\text{cat}})/RT} = e^{\Delta E_a / RT}$$

What Catalysts DO versus What They CANNOT DO

• What Catalysts DO:
  1. Lower activation energy for both the forward pathway ($E_{a,\text{fwd}}$) and the reverse pathway ($E_{a,\text{rev}}$) by the exact same numerical amount.
  2. Accelerate both the forward rate constant ($k_f$) and reverse rate constant ($k_r$) by the identical multiplicative factor.
  3. Accelerate the rate at which dynamic equilibrium is reached.
• What Catalysts CANNOT DO:
  1. Cannot alter the equilibrium constant ($K_{\text{eq}}$): Because $K_{\text{eq}} = k_f / k_r$, increasing both rate constants by the same factor leaves their ratio unchanged.
  2. Cannot shift the equilibrium position or change equilibrium concentrations.
  3. Cannot alter thermodynamic state functions: Catalysts have zero effect on $\Delta H_{\text{rxn}}, \Delta S_{\text{rxn}}$, or $\Delta G_{\text{rxn}}$. If a reaction is thermodynamically non-spontaneous ($\Delta G > 0$), adding a catalyst cannot make it spontaneous.

Biological Enzymes

Enzymes are specialized biological macromolecular catalysts (almost always proteins, occasionally catalytic RNA ribozymes) that accelerate biochemical transformations by factors of $10^6$ to $10^{14}$ under physiological conditions ($37^\circ\text{C}, 1\text{ atm}, \text{pH } 7.4$). Enzymes achieve their catalytic power by:

• Transition State Complementarity: The enzyme active site is structurally and electrostatically complementary to the transition state, rather than to the substrate or product.
• Induced Fit: Substrate binding triggers conformational changes that align catalytic residues, strain substrate bonds, and exclude water to maximize electrostatic interactions.
• Catalytic Strategies: General acid-base catalysis (proton transfer via His, Asp, Glu, Lys), covalent catalysis (nucleophilic attack forming transient covalent intermediates via Ser, Cys, Lys), and metal ion cofactor catalysis (polarization of substrate bonds by $\text{Mg}^{2+}, \text{Zn}^{2+}, \text{Fe}^{2+}$).

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6. Worked Numerical Examples

Example 1: Rate Law and Rate Constant Determination

A physical chemist investigates the gas-phase reaction: $2\text{NO}_{(g)} + \text{O}_{2(g)} \longrightarrow 2\text{NO}_{2(g)}$. Initial rate measurements at $25^\circ\text{C}$ yield the following experimental data:

Determine the partial reaction orders ($m, n$), the overall rate law, and the value and units of the rate constant $k$.

Solution:

1. Determine the order with respect to $\text{NO}$ ($m$):
   • Compare Trial 1 and Trial 2 where $[\text{O}_2]$ is held constant at $0.010\text{ M}$: $$\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{NO}]_2}{[\text{NO}]_1}\right)^m \implies \frac{1.0 \times 10^{-4}}{2.5 \times 10^{-5}} = \left(\frac{0.020}{0.010}\right)^m$$ $$4.0 = (2.0)^m \implies m = 2 \quad (\text{Second order with respect to NO})$$
2. Determine the order with respect to $\text{O}_2$ ($n$):
   • Compare Trial 1 and Trial 3 where $[\text{NO}]$ is held constant at $0.010\text{ M}$: $$\frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[\text{O}_2]_3}{[\text{O}_2]_1}\right)^n \implies \frac{5.0 \times 10^{-5}}{2.5 \times 10^{-5}} = \left(\frac{0.020}{0.010}\right)^n$$ $$2.0 = (2.0)^n \implies n = 1 \quad (\text{First order with respect to } \text{O}_2)$$
3. Write the complete rate law: $$\text{Rate} = k [\text{NO}]^2 [\text{O}_2]^1$$ The overall order is $2 + 1 = 3$ (third-order reaction).
4. Calculate the rate constant $k$ using data from Trial 1: $$k = \frac{\text{Rate}}{[\text{NO}]^2 [\text{O}_2]} = \frac{2.5 \times 10^{-5}\text{ M/s}}{(0.010\text{ M})^2 (0.010\text{ M})} = \frac{2.5 \times 10^{-5}}{1.0 \times 10^{-6}} = 25\text{ M}^{-2}\cdot\text{s}^{-1}$$ Units: $\text{M}^{1-3}\cdot\text{s}^{-1} = \text{M}^{-2}\cdot\text{s}^{-1}$ (or $\text{L}^2/(\text{mol}^2\cdot\text{s})$).

Example 2: First-Order Pharmacokinetic Half-Life

A patient receives an intravenous bolus of a cardioprotective drug, achieving an initial peak plasma concentration of $80.0\text{ mg/L}$. The drug undergoes first-order renal clearance with an elimination rate constant $k = 0.231\text{ h}^{-1}$. Determine the elimination half-life ($t_{1/2}$) and calculate the plasma concentration remaining after $9.0\text{ hours}$.

Solution:

1. Calculate the half-life ($t_{1/2}$): $$t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{0.231\text{ h}^{-1}} = 3.0\text{ hours}$$
2. Calculate the number of elapsed half-lives in $9.0\text{ hours}$: $$n_{\text{half-lives}} = \frac{9.0\text{ h}}{3.0\text{ h}} = 3.0\text{ half-lives}$$
3. Calculate remaining concentration after 3 successive half-lives:
   • After 1 half-life ($3.0\text{ h}$): $[\text{Drug}] = 80.0 / 2 = 40.0\text{ mg/L}$
   • After 2 half-lives ($6.0\text{ h}$): $[\text{Drug}] = 40.0 / 2 = 20.0\text{ mg/L}$
   • After 3 half-lives ($9.0\text{ h}$): $[\text{Drug}] = 20.0 / 2 = 10.0\text{ mg/L}$
4. Confirmation using the integrated first-order rate equation: $$[\text{Drug}]_t = [\text{Drug}]_0 e^{-k t} = (80.0\text{ mg/L}) e^{-(0.231 \times 9.0)} = 80.0 \times e^{-2.079} = 80.0 \times 0.125 = 10.0\text{ mg/L}$$ Exactly $10.0\text{ mg/L}$ of the drug remains in plasma after 9 hours.