AP Chemistry 5.1-5.2: Reaction Rates and Rate Laws

Chemical reaction rate

The change in the amount or concentration of a reactant consumed or product formed per unit time. Reaction rates are commonly expressed in units such as $\mathrm{mol\,L^{-1}\,s^{-1}}$.

How can the progress of a reaction be monitored experimentally?

By measuring a property that changes as reactants are consumed or products form, such as gas volume or pressure, light absorption for colored species, conductivity for aqueous electrolytes, or concentration.

Average rate of disappearance of a reactant

For a reactant $A$, the average rate of disappearance over a time interval is $-\dfrac{\Delta[A]}{\Delta t}=-\dfrac{[A]_{t_2}-[A]_{t_1}}{t_2-t_1}$.

Average rate of formation of a product

For a product $B$, the average rate of formation is $\dfrac{\Delta[B]}{\Delta t}=\dfrac{[B]_{t_2}-[B]_{t_1}}{t_2-t_1}$.

Why is a negative sign used in the rate expression for a reactant?

A reactant concentration decreases, so $\Delta[\text{reactant}]/\Delta t$ is negative. Multiplying by $-1$ makes the reported reaction rate positive.

Why does the average rate of many reactions decrease as the reaction proceeds?

Reactant concentrations generally decrease over time, reducing the frequency of effective particle collisions. Consequently, less reactant is consumed per unit time.

Average reaction rate vs. instantaneous reaction rate

Average rate is calculated over a finite time interval using $\Delta[\text{species}]/\Delta t$. Instantaneous rate is the rate at a particular moment and is approximated using a very short interval or determined from the slope of a tangent line.

Initial rate

The instantaneous reaction rate at the start of a reaction, conventionally at $t=0$. Initial rates are useful because reactant concentrations have not yet changed substantially.

How is an instantaneous rate obtained from a concentration-versus-time graph?

Draw a tangent line to the curve at the time of interest and calculate its slope. For a reactant, the rate of disappearance is the negative of that slope.

How are concentration-based and amount-based rate expressions related in a homogeneous reaction?

Because all species occupy the same volume, molar amounts can be replaced with molar concentrations in the rate expressions. The stoichiometric relationships remain the same.

General normalized reaction-rate expression

For $aA\rightarrow bB$, the reaction rate is $\text{rate}=-\dfrac{1}{a}\dfrac{\Delta[A]}{\Delta t}=\dfrac{1}{b}\dfrac{\Delta[B]}{\Delta t}$. Dividing by stoichiometric coefficients gives one common rate for the reaction.

For the reaction $2\mathrm{H_2O_2}\rightarrow2\mathrm{H_2O}+\mathrm{O_2}$, how are the species-specific rates related?

$-\dfrac{1}{2}\dfrac{\Delta[\mathrm{H_2O_2}]}{\Delta t}=\dfrac{1}{2}\dfrac{\Delta[\mathrm{H_2O}]}{\Delta t}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}$.

If $-\Delta[\mathrm{H_2O_2}]/\Delta t=3.20\times10^{-2}\ \mathrm{mol\,L^{-1}\,h^{-1}}$ for $2\mathrm{H_2O_2}\rightarrow2\mathrm{H_2O}+\mathrm{O_2}$, what are the product formation rates?

$\Delta[\mathrm{O_2}]/\Delta t=1.60\times10^{-2}\ \mathrm{mol\,L^{-1}\,h^{-1}}$ and $\Delta[\mathrm{H_2O}]/\Delta t=3.20\times10^{-2}\ \mathrm{mol\,L^{-1}\,h^{-1}}$.

For $2\mathrm{NH_3}\rightarrow\mathrm{N_2}+3\mathrm{H_2}$, which species changes fastest in concentration?

Hydrogen forms three times as fast as nitrogen: $\dfrac{\Delta[\mathrm{H_2}]}{\Delta t}=3\dfrac{\Delta[\mathrm{N_2}]}{\Delta t}$. Ammonia disappears twice as fast as nitrogen forms: $-\dfrac{\Delta[\mathrm{NH_3}]}{\Delta t}=2\dfrac{\Delta[\mathrm{N_2}]}{\Delta t}$.

For the reaction $5\mathrm{Br^-}+\mathrm{BrO_3^-}+6\mathrm{H^+}\rightarrow3\mathrm{Br_2}+3\mathrm{H_2O}$, write the normalized rate relationship.

$-\dfrac{1}{5}\dfrac{\Delta[\mathrm{Br^-}]}{\Delta t}=-\dfrac{\Delta[\mathrm{BrO_3^-}]}{\Delta t}=-\dfrac{1}{6}\dfrac{\Delta[\mathrm{H^+}]}{\Delta t}=\dfrac{1}{3}\dfrac{\Delta[\mathrm{Br_2}]}{\Delta t}=\dfrac{1}{3}\dfrac{\Delta[\mathrm{H_2O}]}{\Delta t}$.

If $\mathrm{Br_2}$ forms at $6.0\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ in the bromide-bromate reaction, at what rate is $\mathrm{BrO_3^-}$ consumed?

Because one mole of $\mathrm{BrO_3^-}$ is consumed for every three moles of $\mathrm{Br_2}$ formed, $\mathrm{BrO_3^-}$ is consumed at $2.0\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.

If ammonia decomposes according to $2\mathrm{NH_3}\rightarrow\mathrm{N_2}+3\mathrm{H_2}$ at a disappearance rate of $2.10\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$, what are the formation rates of $\mathrm{N_2}$ and $\mathrm{H_2}$?

$\mathrm{N_2}$ forms at $1.05\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ and $\mathrm{H_2}$ forms at $3.15\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.

Rate law

A mathematical relationship between reaction rate and reactant concentrations. For $aA+bB\rightarrow\text{products}$, a general rate law is $\text{rate}=k[A]^m[B]^n$.

Rate constant $k$

The proportionality constant in a rate law. For a particular reaction, $k$ is constant at a given temperature, independent of reactant concentrations, but generally changes with temperature.

Reaction order with respect to a reactant

The exponent of that reactant's concentration in the experimentally determined rate law. In $\text{rate}=k[A]^m[B]^n$, the reaction is $m$th order in $A$ and $n$th order in $B$.

Overall reaction order

The sum of all concentration exponents in the rate law. For $\text{rate}=k[A]^m[B]^n$, the overall order is $m+n$.

Do stoichiometric coefficients determine the exponents in a rate law?

Not generally. Reaction orders and rate-law exponents are determined experimentally; they may happen to match stoichiometric coefficients in some reactions but cannot be assumed to do so.

What does it mean for a reaction to be zero order in a reactant?

The rate is independent of that reactant's concentration. Its concentration term is raised to the zero power, so $[A]^0=1$ and it does not appear in the simplified rate law.

Can a reactant have a negative or fractional order in a rate law?

Yes. Although many introductory examples use nonnegative integer orders, experimentally determined reaction orders can be zero, fractional, or negative.

How does doubling a reactant concentration affect rate for different reaction orders?

For a reactant with order $m$, doubling its concentration changes the rate by a factor of $2^m$. Thus, zero order gives no change, first order doubles the rate, and second order quadruples it.

A reaction is first order in $A$ and second order in $B$. What is its overall order and rate-law form?

It is third order overall, and its rate law is $\text{rate}=k[A][B]^2$.

A reaction has rate law $\text{rate}=k[A][B]^2$. How does its rate change if $[A]$ is tripled and $[B]$ is doubled?

The rate changes by $3^1\times2^2=12$, so the new rate is twelve times the original rate.

What rate law corresponds to a reaction that is first order in $\mathrm{H^+}$ and first order in $\mathrm{OH^-}$?

$\text{rate}=k[\mathrm{H^+}][\mathrm{OH^-}]$. The reaction is second order overall.

What rate law describes a reaction that is second order in $\mathrm{NO_2}$ and zero order in $\mathrm{CO}$?

$\text{rate}=k[\mathrm{NO_2}]^2[\mathrm{CO}]^0=k[\mathrm{NO_2}]^2$. The rate depends on $[\mathrm{NO_2}]$ but not on $[\mathrm{CO}]$.

Method of initial rates

An experimental method for determining a rate law by comparing initial rates from trials with different initial reactant concentrations. Keeping one concentration constant while varying another reveals the corresponding reaction order.

How should rate data be analyzed when comparing two initial-rate trials?

Choose trials where the concentrations of all but one reactant are constant. Divide the two rate-law expressions so that $k$ and the unchanged concentration terms cancel, leaving an equation for the unknown order.

How can the order in reactant $A$ be determined using initial-rate data?

Compare two trials in which all other reactant concentrations are constant. From $\dfrac{\text{rate}_2}{\text{rate}_1}=\left(\dfrac{[A]_2}{[A]_1}\right)^m$, solve for $m$.

How can a rate order be determined when the concentration and rate do not change by simple integer factors?

Use logarithms in the ratio equation: $m=\dfrac{\ln(\text{rate}_2/\text{rate}_1)}{\ln([A]_2/[A]_1)}$, provided other reactant concentrations are unchanged.

For the reaction $\mathrm{NO}+\mathrm{O_3}\rightarrow\mathrm{NO_2}+\mathrm{O_2}$, the rate doubles when either $[\mathrm{NO}]$ or $[\mathrm{O_3}]$ doubles while the other is constant. What is the rate law?

The reaction is first order in each reactant, so $\text{rate}=k[\mathrm{NO}][\mathrm{O_3}]$, which is second order overall.

For $2\mathrm{NO}+\mathrm{Cl_2}\rightarrow2\mathrm{NOCl}$, the rate changes by a factor of $2.25$ when $[\mathrm{NO}]$ changes by a factor of $1.5$, with $[\mathrm{Cl_2}]$ constant. What is the order in NO?

Since $2.25=(1.5)^m$, $m=2$. The reaction is second order in NO.

For $2\mathrm{NO}+\mathrm{Cl_2}\rightarrow2\mathrm{NOCl}$, the rate changes by a factor of $1.5$ when $[\mathrm{Cl_2}]$ changes by a factor of $1.5$, with $[\mathrm{NO}]$ constant. What is the order in chlorine?

Since $1.5=(1.5)^n$, $n=1$. The reaction is first order in $\mathrm{Cl_2}$.

What is the experimentally determined rate law for $2\mathrm{NO}+\mathrm{Cl_2}\rightarrow2\mathrm{NOCl}$ based on second-order dependence on NO and first-order dependence on $\mathrm{Cl_2}$?

$\text{rate}=k[\mathrm{NO}]^2[\mathrm{Cl_2}]$. The overall reaction order is $3$.

How can the rate constant be calculated after determining a rate law?

Substitute the rate and reactant concentrations from any experimental trial into the rate law and solve for $k$: $k=\dfrac{\text{rate}}{[A]^m[B]^n}$. Data from multiple trials should give consistent values of $k$.

Units of a rate constant for a first-order reaction

For a first-order rate law, $k$ has units of inverse time, such as $\mathrm{s^{-1}}$, because $\text{rate}=k[A]$.

Units of a rate constant for a second-order reaction

For a second-order rate law, $k$ has units of concentration$^{-1}$ time$^{-1}$, commonly $\mathrm{L\,mol^{-1}\,s^{-1}}$, because $\text{rate}=k[A]^2$.

General units of the rate constant for an overall order $N$ reaction

If concentration is measured in $\mathrm{mol\,L^{-1}}$ and time in seconds, $k$ has units of $\mathrm{L^{N-1}\,mol^{1-N}\,s^{-1}}$. These units ensure that the rate has units of $\mathrm{mol\,L^{-1}\,s^{-1}}$.

What are the units of $k$ for the rate law $\text{rate}=k[\mathrm{NO}]^2[\mathrm{Cl_2}]$?

The overall order is $3$, so $k$ has units of $\mathrm{L^2\,mol^{-2}\,s^{-1}}$, equivalently $\mathrm{mol^{-2}\,L^2\,s^{-1}}$.

Why must the temperature be specified when reporting a rate constant?

The rate constant is specific to a reaction at a particular temperature and generally changes when temperature changes. Therefore, a value of $k$ determined at one temperature cannot automatically be used at another.