13.2 Integrated Rate Laws, Graphical Methods & Half-Lives

13.2 Integrated Rate Laws, Graphical Methods & Half-Lives

Quick Summary: While differential rate laws express velocity as a function of concentration, integrated rate laws express reactant concentration as an explicit function of elapsed time ($t$). For zero-order reactions, $[\text{A}]_t = -kt + [\text{A}]_0$ and $t_{1/2} = [\text{A}]_0 / (2k)$. For first-order reactions, $\ln[\text{A}]_t = -kt + \ln[\text{A}]_0$ and $t_{1/2} = 0.693 / k$, which remains invariant regardless of initial concentration. For second-order reactions, $1/[\text{A}]_t = kt + 1/[\text{A}]_0$ and $t_{1/2} = 1 / (k[\text{A}]_0)$. Graphical identification of reaction order relies on determining which of three plots ($[\text{A}]$ vs $t$, $\ln[\text{A}]$ vs $t$, or $1/[\text{A}]$ vs $t$) produces a straight line.

---

1. Operational Need for Integrated Rate Laws

Differential rate laws relate instantaneous rates to concentrations, but laboratory experiments monitor concentration changes over discrete time intervals. Integrating differential rate laws using calculus produces mathematical equations that relate concentration directly to time, enabling chemists to:

1. Calculate the concentration of a reactant remaining after any specified duration.
2. Determine the time required for a reactant concentration to drop to a specified threshold.
3. Compute the half-life ($t_{1/2}$) of the reaction.

---

2. Zero-Order Kinetics

In a zero-order reaction, the rate is completely independent of reactant concentration: $-\frac{d[\text{A}]}{dt} = k[\text{A}]^0 = k$ Integrating both sides from $t = 0$ (where $[\text{A}] = [\text{A}]_0$) to time $t$ (where $[\text{A}] = [\text{A}]_t$): $\int_{[\text{A}]_0}^{[\text{A}]_t} d[\text{A}] = -k \int_0^t dt \implies [\text{A}]_t - [\text{A}]_0 = -kt$ $[\text{A}]_t = -kt + [\text{A}]_0$ This equation matches the standard slope-intercept linear form ($y = mx + b$):

• Linear Plot: A plot of $[\text{A}]$ on the y-axis versus $t$ on the x-axis yields a straight line.
• Slope: $\text{Slope} = -k$.
• y-Intercept: $\text{Intercept} = [\text{A}]_0$.

Half-Life Derivation for Zero Order

The half-life ($t_{1/2}$) is the time required for the reactant concentration to decrease to half its initial value ($[\text{A}]_{t_{1/2}} = \frac{1}{2}[\text{A}]_0$): $\frac{1}{2}[\text{A}]_0 = -k t_{1/2} + [\text{A}]_0 \implies k t_{1/2} = [\text{A}]_0 - \frac{1}{2}[\text{A}]_0 = \frac{1}{2}[\text{A}]_0$ $t_{1/2} = \frac{[\text{A}]_0}{2k}$ Key Feature: For zero-order reactions, the half-life is directly proportional to initial concentration. As the reaction proceeds, $[\text{A}]_0$ becomes smaller for each successive period, so successive half-lives become progressively shorter. Zero-order kinetics typically occurs in surface-catalyzed reactions where active catalyst sites are fully saturated (e.g., decomposition of gaseous $\text{NH}_3$ on hot platinum wire).

---

3. First-Order Kinetics

In a first-order reaction, the rate is directly proportional to reactant concentration: $-\frac{d[\text{A}]}{dt} = k[\text{A}]$ Separating variables and integrating: $\int_{[\text{A}]_0}^{[\text{A}]_t} \frac{d[\text{A}]}{[\text{A}]} = -k \int_0^t dt \implies \ln[\text{A}]_t - \ln[\text{A}]_0 = -kt$ $\ln[\text{A}]_t = -kt + \ln[\text{A}]_0$ Rearranging into logarithmic quotient and exponential forms yields: $\ln\left(\frac{[\text{A}]_t}{[\text{A}]_0}\right) = -kt \quad \Longleftrightarrow \quad [\text{A}]_t = [\text{A}]_0 e^{-kt}$

• Linear Plot: A plot of $\ln[\text{A}]$ on the y-axis versus $t$ on the x-axis produces a straight line.
• Slope: $\text{Slope} = -k$.
• y-Intercept: $\text{Intercept} = \ln[\text{A}]_0$.

Half-Life Derivation for First Order

Substitute $[\text{A}]_{t_{1/2}} = \frac{1}{2}[\text{A}]_0$ into the quotient form: $\ln\left(\frac{\frac{1}{2}[\text{A}]_0}{[\text{A}]_0}\right) = -k t_{1/2} \implies \ln\left(\frac{1}{2}\right) = -\ln(2) = -k t_{1/2}$ $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ Critical Exam Insight: The half-life of a first-order process is strictly constant and completely independent of the starting concentration $[\text{A}]_0$. Regardless of whether the initial concentration is $1.0\text{ M}$ or $0.001\text{ M}$, the time required to consume $50\%$ of the remaining reactant is identical. All nuclear radioactive decays (such as carbon-14 dating and uranium decay series) follow first-order kinetics.

---

4. Second-Order Kinetics

For a reaction that is second order with respect to a single reactant: $-\frac{d[\text{A}]}{dt} = k[\text{A}]^2$ Separating variables and integrating: $\int_{[\text{A}]_0}^{[\text{A}]_t} -\frac{d[\text{A}]}{[\text{A}]^2} = k \int_0^t dt \implies \left[\frac{1}{[\text{A}]}\right]_{[\text{A}]_0}^{[\text{A}]_t} = kt$ $\frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0}$

• Linear Plot: A plot of the reciprocal concentration $\frac{1}{[\text{A}]}$ on the y-axis versus $t$ on the x-axis produces a straight line.
• Slope: $\text{Slope} = +k$ (Note the positive slope!).
• y-Intercept: $\text{Intercept} = \frac{1}{[\text{A}]_0}$.

Half-Life Derivation for Second Order

Substitute $[\text{A}]_{t_{1/2}} = \frac{1}{2}[\text{A}]_0$: $\frac{1}{\frac{1}{2}[\text{A}]_0} = k t_{1/2} + \frac{1}{[\text{A}]_0} \implies \frac{2}{[\text{A}]_0} - \frac{1}{[\text{A}]_0} = k t_{1/2}$ $t_{1/2} = \frac{1}{k[\text{A}]_0}$ Key Feature: For second-order reactions, the half-life is inversely proportional to initial concentration. As the reactant is consumed, $[\text{A}]$ drops, causing each subsequent half-life to double in length: $t_{1/2}(2) = 2 \times t_{1/2}(1)$, and $t_{1/2}(3) = 4 \times t_{1/2}(1)$.

---

5. Master Comparison Table of Integrated Rate Laws

---

6. Worked Quantitative Example: First-Order Decay Calculations

Problem: The thermal decomposition of dinitrogen pentoxide in the gas phase: $2\text{ N}_2\text{O}_5(g) \longrightarrow 4\text{ NO}_2(g) + \text{O}_2(g)$ is first order. Suppose that at the temperature of the experiment $k = 6.20 \times 10^{-4}\text{ s}^{-1}$.

Part A: Calculate the reaction half-life in seconds and minutes. $t_{1/2} = \frac{0.69315}{k} = \frac{0.69315}{6.20 \times 10^{-4}\text{ s}^{-1}} = 1118\text{ s} = \frac{1118\text{ s}}{60\text{ s/min}} = 18.6\text{ min}$

Part B: If the initial concentration of $\text{N}_2\text{O}_5$ is $0.250\text{ M}$, calculate the concentration remaining after $3600\text{ s}$ (1.00 hour). Substitute into the first-order integrated rate law: $\ln[\text{N}_2\text{O}_5]_t = -kt + \ln[\text{N}_2\text{O}_5]_0$ $\ln[\text{N}_2\text{O}_5]_t = -(6.20 \times 10^{-4}\text{ s}^{-1})(3600\text{ s}) + \ln(0.250)$ $\ln[\text{N}_2\text{O}_5]_t = -2.232 + (-1.3863) = -3.6183$ Take the natural antilogarithm ($e^x$): $[\text{N}_2\text{O}_5]_t = e^{-3.6183} = 0.0268\text{ M}$

Part C: Determine the percentage of $\text{N}_2\text{O}_5$ decomposed after 1.00 hour. $\%\text{ Decomposed} = \frac{[\text{A}]_0 - [\text{A}]_t}{[\text{A}]_0} \times 100\% = \frac{0.250\text{ M} - 0.0268\text{ M}}{0.250\text{ M}} \times 100\% = 89.3\%$