Question

Write integrated rate laws for zero-order, first-order, and second-order reactions of the form \(A \longrightarrow\) products.

Short answer

Zero-order: \( [A] = [A]_0 - kt \), First-order: \( [A] = [A]_0e^{-kt} \) or \(\ln([A]) = \ln([A]_0) - kt\), Second-order: \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\).

Step-by-step solution

Understanding Zero-Order Reactions

For a zero-order reaction, the rate of reaction is independent of the concentration of reactants. It can be expressed as rate = k, where k is the rate constant. To write the integrated rate law, we integrate the differential rate equation \(\frac{d[A]}{dt} = -k\) over time.

Integrated Rate Law for Zero-Order Reaction

By integrating \(\frac{d[A]}{dt} = -k\), with initial concentration \( [A]_0 \) at \( t=0 \) and concentration [A] at time t, we get \( [A] = [A]_0 - kt \).

Understanding First-Order Reactions

For a first-order reaction, the rate is directly proportional to the concentration of the reactant. The differential rate equation is \(\frac{d[A]}{dt} = -k[A]\).

Integrated Rate Law for First-Order Reaction

Integrating \(\frac{d[A]}{dt} = -k[A]\) leads to the logarithmic form: \(\ln([A]) = \ln([A]_0) - kt\), or in its exponential form: \( [A] = [A]_0e^{-kt} \).

Understanding Second-Order Reactions

A second-order reaction rate is proportional to the square of the concentration of the reactant. The differential form of the rate equation is \(\frac{d[A]}{dt} = -k[A]^2\).

Integrated Rate Law for Second-Order Reaction

By integrating the equation \(\frac{d[A]}{dt} = -k[A]^2\), we get \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\).

Key concepts

Zero-Order Reactions

When studying chemical kinetics, zero-order reactions stand out due to their unique property: the reaction rate is constant and does not depend on the concentration of the reactant. For many students, it's surprising that such reactions exist, as we often expect higher concentrations to result in faster reactions.

To describe this behavior mathematically for a reaction where component A transforms into products, we use the integrated rate law. The formula is derived from the basic rate equation \( \frac{d[A]}{dt} = -k \) where \( [A] \) represents the concentration of A, \( t \) represents time, and \( k \) is the rate constant. To obtain the integrated law, we integrate over time from \( 0 \) to \( t \) which gives us \( [A] = [A]_0 - kt \) where \( [A]_0 \) is the initial concentration. The minus sign indicates the concentration decreases over time.

In practical terms, this means in a zero-order reaction, the reactant’s concentration decreases linearly over time. An everyday analogy might be a faucet pouring water into a sink at a constant rate; the amount of water flowing out doesn't change based on how much is in the sink.

First-Order Reactions

Moving to first-order reactions, which might be more intuitive, the rate of the reaction is directly proportional to the concentration of the reactant. To visualize this, imagine a scenario where the intensity of a glow-stick decreases as the chemicals inside are used up. The brighter it glows, the faster it fades.

The rate law for a first-order reaction is given by \( \frac{d[A]}{dt} = -k[A] \) suggesting that the rate changes as the concentration of A changes. By integrating this differential rate law, we uncover the relationship between the concentration of A at any given time \( t \) and its initial concentration \( [A]_0 \) through the formula \( \ln([A]) = \ln([A]_0) - kt \), which can also be represented as \( [A] = [A]_0e^{-kt} \).

These equations reveal that in first-order kinetics, the concentration diminishes exponentially over time, rather than linearly. This has considerable implications in processes like radioactive decay or the metabolism of drugs in the body, where the rate at which a substance changes is tied closely to its concentration at any moment.

Second-Order Reactions

Second-order reactions can be a bit more complex, as they involve the square of the concentration of the reactant. This means the rate of the reaction increases as the concentration of the reactant increases, but at a much faster pace compared to first-order reactions.

The rate of a second-order reaction is described by \( \frac{d[A]}{dt} = -k[A]^2 \) which suggests a peculiar occurrence: as reactant A is consumed, the rate of reaction doesn't simply halve if the concentration of A is halved; it becomes four times slower because of the squaring effect. Integrating this rate law is how we arrive at \( \frac{1}{[A]} = \frac{1}{[A]_0} + kt \), highlighting another distinct kinetic profile.

These kinetics are often observed in reactions where two reactant molecules must collide to form a product, like in certain types of polymerization reactions. Understandably, these reactions slow down drastically as the concentration of reactants diminish because not only is there less of them, but the likelihood of successful collisions also drops significantly. Picturing a busy dance floor clearing out over time might help conceptualize why the few remaining dancers have a harder and slower time finding a partner.