AP Chemistry: Integrated Rate Laws

In chemistry, predicting how long a reaction takes or what concentration remains is a practical necessity, whether you're a pharmacist determining a drug's shelf life, an engineer modeling pollutant decay, or a doctor understanding medication dosage intervals. While differential rate laws tell you the rate at a specific concentration, integrated rate laws are the mathematical tools that connect concentration directly to time, allowing you to forecast the future of a reaction. Mastering these laws transforms you from a passive observer into an active predictor of chemical change.

The Foundation: From Rate to Concentration

A rate law expresses the relationship between the reaction rate and the concentrations of reactants. For a generic reaction $aA\to products$, the rate law is often written as $Rate=k[A{]}^n$, where $k$ is the rate constant and $n$ is the reaction order. This is a differential rate law. An integrated rate law, however, is derived by integrating this differential equation, resulting in a formula that expresses $[A]$ as a function of time $t$. The form of this equation depends entirely on the reaction order ($n=0,1,or2$). Determining which integrated rate law to use is the first step in solving any concentration-time problem.

Zero-Order Reactions: A Constant Rate Process

A zero-order reaction has a rate that is independent of the concentration of the reactant. Its rate law is $Rate=k[A{]}^0=k$. The rate is constant. Integrating this yields the zero-order integrated rate law:

$$[A{]}_t=-kt+[A{]}_0$$

Here, $[A{]}_t$ is the concentration at time $t$, $[A{]}_0$ is the initial concentration, and $k$ is the rate constant. Notice this equation has the form of a straight line ($y=mx+b$), where $y=[A{]}_t$, $m=-k$, $x=t$, and $b=[A{]}_0$.

Example: Imagine a catalytic decomposition on a saturated metal surface. If the initial concentration of a gas is 2.0 M and $k=0.025M/s$, what is the concentration after 30 seconds? $$[A{]}_{30}=-(0.025M/s)(30s)+2.0M=-0.75M+2.0M=1.25M$$

A plot of $[A]$ versus $t$ will yield a straight line with a slope of $-k$. The half-life ($t_{1/2}$), the time required for the concentration to drop to half its initial value, for a zero-order reaction is concentration-dependent: $t_{1/2}=[A{]}_0/(2k)$. This means the higher the starting concentration, the longer the half-life.

First-Order Reactions: The Exponential Decay

First-order reactions are extremely common (e.g., radioactive decay, many decomposition reactions). The rate is directly proportional to the concentration: $Rate=k[A]$. Integration gives the first-order integrated rate law, most commonly used in its linear form:

$$\ln [A{]}_t=-kt+\ln [A{]}_0$$

A plot of $\ln [A]$ versus $t$ gives a straight line with a slope of $-k$. This is the most important graphical test for first-order behavior.

Example: A first-order reaction has a rate constant $k=0.015s^{-1}$. If $[A{]}_0=0.80M$, what is $[A]$ after 2.0 minutes (120 s)? $$\ln [A{]}_t=-(0.015s^{-1})(120s)+\ln (0.80)$$ $$\ln [A{]}_t=-1.8+(-0.223)=-2.023$$ $$[A{]}_t=e^{-2.023}=0.132M$$

The half-life for a first-order reaction is a constant, independent of initial concentration: $t_{1/2}=\frac{\ln (2)}{k}\approx \frac{0.693}{k}$. This constant half-life is a unique fingerprint of first-order kinetics. If you know the half-life, you can instantly find $k$, and vice-versa.

Second-Order Reactions: When Two Particles Must Meet

A reaction is second-order overall if its rate is proportional to the square of a single reactant's concentration ($Rate=k[A{]}^2$) or to the product of two reactants' first-order concentrations ($Rate=k[A][B]$ where $[A]=[B]$). For the simpler $Rate=k[A{]}^2$ case, integration yields the linear form:

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

A plot of $\frac{1}{[A]}$ versus $t$ yields a straight line with a positive slope equal to $k$.

Example: For a second-order reaction with $k=0.055M^{-1}{s}^{-1}$ and $[A{]}_0=0.65M$, find $[A]$ at $t=75s$. $$\frac{1}{[A{]}_t}=(0.055M^{-1}{s}^{-1})(75s)+\frac{1}{0.65M}$$ $$\frac{1}{[A{]}_t}=4.125+1.538=5.663M^{-1}$$ $$[A{]}_t=\frac{1}{5.663M^{-1}}=0.177M$$

The half-life for a second-order reaction depends on the initial concentration: $t_{1/2}=\frac{1}{k[A{]}_0}$. Notice the inverse relationship—a higher initial concentration leads to a shorter half-life, which makes sense intuitively: with more particles, collisions happen more frequently.

Graphical Determination of Reaction Order

This is the primary application of the linear forms of the integrated rate laws. You don't need to know the order in advance; you let the data tell you.

Step-by-Step Process:

1. Collect concentration vs. time data for a reactant.
2. Make three plots:

• $[A]$ vs. $t$
• $\ln [A]$ vs. $t$
• $\frac{1}{[A]}$ vs. $t$

1. Identify the linear plot. The one that gives the straight line reveals the reaction order.

• Zero-order: $[A]$ vs. $t$ is linear.
• First-order: $\ln [A]$ vs. $t$ is linear.
• Second-order: $\frac{1}{[A]}$ vs. $t$ is linear.

1. Extract the rate constant ($k$) from the slope of the linear plot (remembering the sign: slope = $-k$ for zero and first order, slope = $+k$ for second order).

This graphical method is a powerful, model-agnostic way to determine kinetics from experimental data.

Common Pitfalls

1. Using the Wrong Integrated Law Without Checking Order: The most frequent error is plugging numbers into the first-order equation simply because it's the most familiar. Correction: Always check if the problem states the order or provides data that lets you determine it graphically. If data is given, test for linearity as described above before choosing your equation.

1. Misinterpreting Plots and Slopes: Students often forget the sign or units of the slope. Correction: Memorize the three linear forms and what the slope represents:

• Zero-order: Plot $[A]$ vs $t$, slope = $-k$ (units: $M/s$).
• First-order: Plot $\ln [A]$ vs $t$, slope = $-k$ (units: $s^{-1}$).
• Second-order: Plot $1/[A]$ vs $t$, slope = $+k$ (units: $M^{-1}{s}^{-1}$).

1. Confusing Half-Life Equations: Applying the first-order half-life formula ($t_{1/2}=0.693/k$) to a zero- or second-order reaction will give a wrong answer. Correction: Associate each half-life formula with its order:

• Zero-order: $t_{1/2}=[A{]}_0/(2k)$
• First-order: $t_{1/2}=0.693/k$
• Second-order: $t_{1/2}=1/(k[A{]}_0)$

1. Algebraic Mistakes with Logarithms and Reciprocals: Solving $\ln [A{]}_t=-kt+\ln [A{]}_0$ requires careful algebra to isolate $[A{]}_t$. Correction: Practice the steps: 1) Calculate the right side as a single number. 2) Use the $e^x$ function to undo the natural log: $[A{]}_t=e^{\text{(calculated number)}}$.

Summary

• Integrated rate laws link reactant concentration directly to time, enabling prediction. The correct law to use depends solely on the reaction order.
• Each order has a distinct linear plot: $[A]$ vs. $t$ for zero-order, $\ln [A]$ vs. $t$ for first-order, and $1/[A]$ vs. $t$ for second-order. The linear plot identifies the order, and its slope gives the rate constant $k$.
• Half-life behavior is a key differentiator: constant for first-order, dependent on initial concentration for zero-order ($t_{1/2}$ increases with $[A{]}_0$) and second-order ($t_{1/2}$ decreases with $[A{]}_0$).
• The graphical method is an essential experimental technique for determining reaction order and the rate constant from raw concentration-time data.
• Always confirm the reaction order before selecting an equation, and pay meticulous attention to the units and signs when working with slopes and constants.