11.2 Integrated rate laws

Integrated rate laws connect reactant concentrations to time, giving you equations you can actually solve rather than just differential expressions. While the differential rate law tells you the instantaneous rate, the integrated form lets you predict what concentration remains after a specific time, determine how long a reaction takes to reach a target concentration, and extract rate constants from experimental data.

Integrated Rate Laws

Deriving Integrated Rate Laws for Different Reaction Orders

Each integrated rate law comes from separating variables in the differential rate law and integrating. The results take the form of linear equations, which is what makes them so useful for data analysis.

Zero-order (rate = $k$):

$[A] = -kt + [A]_0$

This is linear in $[A]$ vs. $t$. The concentration drops at a constant rate regardless of how much reactant remains.

First-order (rate = $k[A]$):

$\ln[A] = -kt + \ln[A]_0$

This is linear in $\ln[A]$ vs. $t$. The rate slows as the reactant is consumed, since the rate depends on how much is left.

Second-order (rate = $k[A]^2$):

$\frac{1}{[A]} = kt + \frac{1}{[A]_0}$

This is linear in $\frac{1}{[A]}$ vs. $t$. The rate drops off even more steeply than first-order as concentration decreases.

In all three equations, $[A]$ is the concentration at time $t$, $[A]_0$ is the initial concentration, and $k$ is the rate constant (with units that differ by order).

Applying Integrated Rate Laws to Calculate Concentrations

To find the concentration at a given time, plug $[A]_0$, $k$, and $t$ into the appropriate equation. You need to know the reaction order first.

Zero-order example: If $k = 0.5 \text{ M/min}$ and $[A]_0 = 2.0 \text{ M}$, find $[A]$ at $t = 3 \text{ min}$:

$[A] = -kt + [A]_0 = -(0.5)(3) + 2.0 = 0.5 \text{ M}$

First-order example: If $k = 0.2 \text{ min}^{-1}$ and $[A]_0 = 1.5 \text{ M}$, find $[A]$ at $t = 5 \text{ min}$:

$[A] = [A]_0 e^{-kt} = 1.5 \cdot e^{-(0.2)(5)} = 1.5 \cdot e^{-1.0} = 0.55 \text{ M}$

The exponential form $[A] = [A]_0 e^{-kt}$ is just the rearranged version of the $\ln[A]$ equation. Use whichever is more convenient.

Second-order example: If $k = 0.1 \text{ M}^{-1}\text{min}^{-1}$ and $[A]_0 = 4.0 \text{ M}$, find $[A]$ at $t = 2 \text{ min}$:

$\frac{1}{[A]} = kt + \frac{1}{[A]_0} = (0.1)(2) + \frac{1}{4.0} = 0.45 \text{ M}^{-1}$

$[A] = \frac{1}{0.45} = 2.2 \text{ M}$

Determining Reaction Order

Analyzing Concentration-Time Data

The standard method for finding reaction order from experimental data is the graphical method: you make three plots and see which one gives a straight line.

Determining Rate Constant from the Appropriate Plot

Once you've identified which plot is linear, extract $k$ from the slope:

• Zero-order: slope = $-k$, so $k = -\text{slope}$
• First-order: slope = $-k$, so $k = -\text{slope}$
• Second-order: slope = $+k$, so $k = \text{slope}$

Notice the sign difference for second-order. The $\frac{1}{[A]}$ vs. $t$ plot has a positive slope because $\frac{1}{[A]}$ increases as $[A]$ decreases.

The y-intercept gives you the initial concentration in transformed form: $[A]_0$ for zero-order, $\ln[A]_0$ for first-order, or $\frac{1}{[A]_0}$ for second-order.

Example: You plot $\ln[A]$ vs. $t$ and get a straight line with slope $-0.15 \text{ min}^{-1}$. The reaction is first-order with $k = 0.15 \text{ min}^{-1}$.

Half-Life and Integrated Rate Laws

Relationship between Half-Life and Integrated Rate Laws

The half-life ($t_{1/2}$) is the time for $[A]$ to drop to $\frac{[A]_0}{2}$. You derive each expression by substituting $[A] = \frac{[A]_0}{2}$ into the corresponding integrated rate law.

Zero-order:

$t_{1/2} = \frac{[A]_0}{2k}$

Half-life decreases as the reaction proceeds (since $[A]_0$ effectively gets smaller with each successive half-life). Each successive half-life is shorter than the last.

First-order:

$t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$

Half-life is constant, independent of concentration. This is the hallmark of first-order kinetics and why radioactive decay (first-order) has a fixed half-life.

Second-order:

$t_{1/2} = \frac{1}{k[A]_0}$

Half-life increases as the reaction proceeds. As the concentration drops, each successive half-life gets longer.

The concentration dependence of half-life is itself a way to distinguish reaction orders experimentally.

Calculating Half-Life for Different Reaction Orders

Zero-order example: $k = 0.02 \text{ M/min}$, $[A]_0 = 1.6 \text{ M}$

$t_{1/2} = \frac{1.6}{2(0.02)} = 40 \text{ min}$

First-order example: $k = 0.1 \text{ min}^{-1}$

$t_{1/2} = \frac{0.693}{0.1} = 6.93 \text{ min}$

Second-order example: $k = 0.5 \text{ M}^{-1}\text{min}^{-1}$, $[A]_0 = 0.8 \text{ M}$

$t_{1/2} = \frac{1}{(0.5)(0.8)} = 2.5 \text{ min}$

Solving Integrated Rate Law Problems

Applying Integrated Rate Laws

These equations work in both directions: solve for $[A]$ given $t$, or solve for $t$ given a target $[A]$.

Finding concentration: For a first-order reaction with $k = 0.2 \text{ min}^{-1}$ and $[A]_0 = 2.0 \text{ M}$, find $[A]$ at $t = 10 \text{ min}$:

$\ln[A] = -(0.2)(10) + \ln(2.0) = -2.0 + 0.693 = -1.307$

$[A] = e^{-1.307} = 0.27 \text{ M}$

Finding time: For a second-order reaction with $k = 0.05 \text{ M}^{-1}\text{min}^{-1}$ and $[A]_0 = 1.2 \text{ M}$, find the time for $[A]$ to reach $0.3 \text{ M}$:

$\frac{1}{0.3} = (0.05)t + \frac{1}{1.2}$

$3.33 = 0.05t + 0.833$

$t = \frac{3.33 - 0.833}{0.05} = 50 \text{ min}$

Using Half-Life to Solve Problems

Finding $k$ from half-life: For a zero-order reaction with $[A]_0 = 2.4 \text{ M}$ and $t_{1/2} = 30 \text{ min}$:

$k = \frac{[A]_0}{2 \cdot t_{1/2}} = \frac{2.4}{2(30)} = 0.04 \text{ M/min}$

Reaching one-fourth of initial concentration: For a first-order reaction with $k = 0.08 \text{ min}^{-1}$, find the time to reach $\frac{[A]_0}{4}$. Reaching one-fourth takes exactly two half-lives (half, then half again):

$t_{1/2} = \frac{0.693}{0.08} = 8.66 \text{ min}$

$t = 2 \times t_{1/2} = 17.3 \text{ min}$

You can verify this algebraically: $\ln\frac{[A]_0/4}{[A]_0} = -kt$ gives $t = \frac{\ln 4}{k} = \frac{1.386}{0.08} = 17.3 \text{ min}$.

Analyzing Concentration-Time Data to Determine Reaction Order and Rate Constant

Here's the full workflow when you're given a data table of $[A]$ vs. $t$:

1. Calculate $\ln[A]$ and $\frac{1}{[A]}$ for each data point
2. Make three plots: $[A]$ vs. $t$, $\ln[A]$ vs. $t$, and $\frac{1}{[A]}$ vs. $t$
3. Identify which plot is linear (use $R^2$ values or visual inspection)
4. Read the slope and y-intercept from the linear plot
5. Extract $k$ from the slope (remember the sign conventions above)
6. Extract $[A]_0$ from the y-intercept by reversing the transformation

Example: A $\ln[A]$ vs. $t$ plot gives a straight line with slope $-0.12 \text{ min}^{-1}$ and y-intercept $-0.22$. The reaction is first-order with $k = 0.12 \text{ min}^{-1}$. The initial concentration is $[A]_0 = e^{-0.22} = 0.80 \text{ M}$.