AP Chemistry 5.3: Concentration Changes Over Time

Rate law for a reaction involving multiple reactants

A common empirical form is $\text{rate}=k[A]^m[B]^n$. The exponents are the partial orders with respect to each reactant, and their sum is the overall reaction order.

What is the meaning of the rate constant $k$?

The rate constant is the proportionality constant in a rate law; its numerical value depends on conditions such as temperature. Its units depend on the overall reaction order.

Can reaction orders usually be determined from the balanced chemical equation?

No. For an overall reaction, reaction orders and the rate law must generally be determined experimentally because the mechanism may involve multiple elementary steps. Stoichiometric coefficients equal reaction orders only for an elementary step.

How does reaction order affect the units of $k$?

Since rate has units of $\mathrm{M\,time^{-1}}$ and $\text{rate}=k[A]^n$, the units of $k$ are $\mathrm{M^{1-n}\,time^{-1}}$. Therefore, zero-, first-, and second-order units are respectively $\mathrm{M\,time^{-1}}$, $\mathrm{time^{-1}}$, and $\mathrm{M^{-1}\,time^{-1}}$.

Integrated rate law

An integrated rate law relates a reactant's concentration to elapsed time. It can be used to calculate concentration after a given time or the time required for a specified amount of reaction to occur.

How does a differential rate law differ from an integrated rate law?

A differential rate law relates reaction rate to reactant concentrations, such as $\text{rate}=k[A]^n$. An integrated rate law relates concentration to time.

Zero-order differential rate law

For a zero-order reaction, $\text{rate}=k$. The rate is constant with respect to the reactant concentration under the specified conditions.

Zero-order integrated rate law

For a zero-order reaction, $[A]_t=[A]_0-kt$. A plot of $[A]$ versus $t$ is linear with slope $-k$ and y-intercept $[A]_0$.

How can the concentration of a reactant be calculated at a particular time for a zero-order reaction?

Use $[A]_t=[A]_0-kt$. Rearranging gives $t=\frac{[A]_0-[A]_t}{k}$.

Units of the rate constant for a zero-order reaction

For $\text{rate}=k$, $k$ has units of concentration per time, commonly $\mathrm{M\,s^{-1}}$.

First-order differential rate law

For a first-order reaction involving one reactant, $\text{rate}=k[A]$. The rate is directly proportional to $[A]$.

First-order integrated rate law in exponential form

The concentration-time relationship is $[A]_t=[A]_0e^{-kt}$. This describes exponential decay of the reactant concentration.

First-order integrated rate law in logarithmic forms

Equivalent forms are $\ln\left(\frac{[A]_t}{[A]_0}\right)=-kt$, $\ln\left(\frac{[A]_0}{[A]_t}\right)=kt$, and $\ln[A]_t=-kt+\ln[A]_0$.

What plot identifies first-order kinetics, and what do its slope and intercept represent?

A plot of $\ln[A]$ versus $t$ is linear for a first-order reaction. Its slope is $-k$ and its y-intercept is $\ln[A]_0$.

How can the time for a specified fraction of a first-order reactant to remain be calculated?

Use $t=\frac{1}{k}\ln\left(\frac{[A]_0}{[A]_t}\right)$. For example, if 20.0% remains, $t=\frac{\ln(5)}{k}$.

Units of the rate constant for a first-order reaction

For $\text{rate}=k[A]$, $k$ has units of inverse time, such as $\mathrm{s^{-1}}$ or $\mathrm{d^{-1}}$.

Why does reactant concentration generally decrease nonlinearly with time for a first-order reaction?

The rate decreases as the reactant is consumed because $\text{rate}=k[A]$. Integration gives exponential decay, rather than the linear concentration-time relationship characteristic of zero-order kinetics.

Second-order differential rate law for one reactant

For the simple second-order case, $\text{rate}=k[A]^2$. The rate is proportional to the square of the reactant concentration.

Second-order integrated rate law

For a reaction with $\text{rate}=k[A]^2$, $\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}$.

What plot identifies second-order kinetics, and what do its slope and intercept represent?

A plot of $1/[A]$ versus $t$ is linear for this second-order reaction. Its slope is $k$ and its y-intercept is $1/[A]_0$.

How can the concentration at time $t$ be calculated for a second-order reaction?

Substitute the known values into $\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}$, then take the reciprocal: $[A]_t=\frac{1}{kt+1/[A]_0}$.

Units of the rate constant for a second-order reaction

For $\text{rate}=k[A]^2$, $k$ has units of inverse concentration-time, commonly $\mathrm{M^{-1}\,s^{-1}}$.

How can concentration-time data be used to determine reaction order?

Test the linearity of three plots: $[A]$ versus $t$ for zero order, $\ln[A]$ versus $t$ for first order, and $1/[A]$ versus $t$ for second order. The plot that is most linear identifies the corresponding order.

How is the rate constant obtained from integrated-law plots?

For a zero-order plot, $k=-\text{slope}$; for a first-order plot, $k=-\text{slope}$; and for a second-order plot, $k=\text{slope}$.

What does it mean if a plot used to test a particular integrated rate law is not linear?

The reaction is not consistent with that reaction order under the conditions studied. Another integrated-law plot should be tested.

How can a rate constant be estimated from two concentration-time data points?

For first order, use $k=\frac{\ln([A]_0/[A]_t)}{t}$; for second order, use $k=\frac{1/[A]_t-1/[A]_0}{t}$; for zero order, use $k=\frac{[A]_0-[A]_t}{t}$. The time and concentration units must be consistent.

Half-life of a reaction

The half-life, $t_{1/2}$, is the time required for the reactant concentration or amount to decrease to one-half its initial value. During each subsequent half-life, half of the amount remaining is consumed.

Zero-order half-life relationship

For a zero-order reaction, $t_{1/2}=\frac{[A]_0}{2k}$. The half-life increases with initial concentration and decreases as $k$ increases.

First-order half-life relationship

For a first-order reaction, $t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{k}$. It is independent of the initial reactant concentration.

Why does a first-order reactant decrease by equal fractions during equal half-lives?

First-order kinetics produce exponential decay, so the time for a concentration to change by a given ratio is constant. Thus, each half-life reduces the remaining concentration by 50%, regardless of the starting concentration.

Second-order half-life relationship

For the simple second-order reaction, $t_{1/2}=\frac{1}{k[A]_0}$. The half-life depends on the initial concentration, so $k$ cannot be found from the half-life alone unless $[A]_0$ is known.

How does the half-life of a second-order reaction change as the reaction proceeds?

It increases as the reactant concentration decreases. More generally, the time required for a further halving depends on the concentration at the beginning of that interval.

How can the half-life formulas distinguish reaction orders?

A concentration-independent half-life indicates first-order behavior. A half-life proportional to $[A]_0$ indicates zero-order behavior, while a half-life inversely proportional to $[A]_0$ indicates second-order behavior.

How does increasing the rate constant affect half-life?

For zero-, first-, and second-order reactions, the half-life is inversely proportional to $k$. A larger $k$ corresponds to a faster reaction and a shorter half-life.

What happens to a first-order reactant concentration after three half-lives?

The fraction remaining is $(1/2)^3=1/8$, or 12.5% of the initial concentration. The fraction consumed is therefore 87.5%.

How can a first-order decay problem be solved when the initial concentration is not given?

If the fraction remaining is specified, use the ratio form $t=\frac{1}{k}\ln([A]_0/[A]_t)$. The unknown initial concentration cancels when $[A]_t$ is expressed as a fraction of $[A]_0$.