AP Chemistry 5.9-5.11: Steady-State Approximation and Catalysis

Catalyst

A substance that increases the rate of a reaction without being consumed overall. It works by providing an alternative reaction pathway with a lower activation energy.

How does a catalyst affect a reaction mechanism and activation energy?

A catalyst introduces an alternative mechanism whose rate-determining step has a lower activation energy, $E_a$, than the uncatalyzed pathway. The catalyzed and uncatalyzed reactions still have the same overall reactants, products, and net enthalpy change.

How can a catalyst be identified on a reaction energy diagram?

The catalyzed pathway has a lower activation-energy barrier for the rate-determining step. If multiple steps are shown, compare the largest energy difference between a reactant or intermediate and the next transition state.

How is the activation energy for an elementary step calculated from a potential energy diagram?

$E_a=E_{\text{transition state}}-E_{\text{reactants of that step}}$. For a later step, use the energy of the preceding intermediate as the starting point.

What features remain the same for catalyzed and uncatalyzed pathways of the same overall reaction?

Both pathways begin at the same reactant energy and end at the same product energy. Therefore, a catalyst does not change the reaction's overall $\Delta H$ or the relative energies of reactants and products.

Does a catalyst necessarily change the number of elementary steps in a reaction mechanism?

No. A catalyzed mechanism may have a different number of steps, but it must provide a pathway with a faster rate-determining step, meaning a lower relevant activation energy.

Does a catalyst change the equilibrium constant or the equilibrium composition?

No. A catalyst speeds up both the forward and reverse reactions, so it does not change $K$, $\Delta G^\circ$, or the equilibrium composition. It allows equilibrium to be reached more quickly.

Intermediate

A species formed in one elementary step of a reaction mechanism and consumed in a later step. It appears in the mechanism but cancels from the overall reaction.

How can a catalyst be recognized in a proposed multistep mechanism?

A catalyst is consumed in an early step and regenerated in a later step, so it cancels when the elementary steps are added. It does not appear in the net reaction.

Homogeneous catalyst

A catalyst in the same phase as the reactants. It commonly reacts with a reactant to form an intermediate and is later regenerated.

Heterogeneous catalyst

A catalyst in a different phase from the reactants, often a solid catalyzing gas- or liquid-phase reactants. The reaction occurs at active sites on the catalyst surface.

What are the typical stages of heterogeneous catalysis?

Reactants adsorb onto the catalyst surface, their bonds are activated or weakened, the adsorbed species react, and products desorb from the surface.

Adsorption versus absorption in heterogeneous catalysis

Adsorption is the attachment of particles to a surface; absorption is penetration into the bulk of a material. Heterogeneous catalysts operate primarily through adsorption at surface active sites.

Why can a solid catalyst accelerate a reaction between gases or liquids?

The catalyst surface provides active sites that bring reactants together in favorable orientations and can weaken existing bonds or stabilize transition states, lowering the activation energy.

How does nitric oxide catalyze ozone decomposition?

A representative cycle is $\mathrm{NO+O_3\rightarrow NO_2+O_2}$, $\mathrm{O_3\rightarrow O_2+O}$, and $\mathrm{NO_2+O\rightarrow NO+O_2}$. Adding the steps gives $\mathrm{2O_3\rightarrow 3O_2}$, while NO is regenerated and therefore acts as a catalyst.

Why can a small amount of chlorine cause extensive ozone depletion?

Chlorine atoms participate in a catalytic cycle such as $\mathrm{Cl+O_3\rightarrow ClO+O_2}$ followed by $\mathrm{ClO+O\rightarrow Cl+O_2}$. Because Cl is regenerated, one chlorine atom can destroy many ozone molecules.

What is the net reaction in the chlorine-catalyzed ozone cycle described by the mechanism?

Adding the elementary steps gives $\mathrm{O_3+O\rightarrow 2O_2}$, while the chlorine species cancel from the overall reaction.

How do enzymes function as biological catalysts?

Enzymes, usually proteins, bind specific reactants called substrates at an active site and stabilize the transition state or provide a lower-energy pathway. They increase reaction rates without being consumed overall.

Active site

The region of an enzyme whose shape and chemical environment allow it to bind a particular substrate and catalyze its reaction.

Lock-and-key model of enzyme action

This model treats the active site as having a relatively fixed shape that is complementary to the substrate, like a key fitting a lock. It explains substrate specificity but is an oversimplification of enzyme flexibility.

Induced-fit model of enzyme action

This model proposes that substrate binding causes the enzyme's active site to adjust its shape, improving interactions with the substrate and facilitating the reaction. The site is flexible but not completely unstructured.

What is the role of glucose-6-phosphate dehydrogenase in cells?

Glucose-6-phosphate dehydrogenase catalyzes a rate-limiting step in a pathway that supplies NADPH. NADPH helps maintain glutathione, an antioxidant that protects red blood cells from oxidative damage.

How do catalytic converters reduce automobile emissions?

A platinum-rhodium surface catalyzes oxidation of carbon monoxide and hydrocarbons to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ while promoting reduction of nitrogen oxides to $\mathrm{N_2}$ and $\mathrm{O_2}$. These are heterogeneous catalytic processes.

Why are many automobile catalytic converters preheated?

The catalyst must reach an effective operating temperature before its active sites efficiently catalyze exhaust reactions. Preheating allows pollutant removal to begin before the engine exhaust becomes sufficiently hot.

Steady state in a chemical system

A condition in which a concentration or other state variable remains approximately constant because its rate of formation balances its rate of removal, even though reactions may continue.

How does a chemical steady state differ from chemical equilibrium?

At equilibrium, the forward and reverse reaction rates are equal and the net reaction rate is zero. At steady state, formation and consumption of an intermediate can balance while there is still a nonzero net reaction rate and ongoing conversion.

Steady-state approximation

An approximation in reaction kinetics that sets the net rate of change of an intermediate concentration to approximately zero: $d[\text{intermediate}]/dt\approx 0$. Thus, the intermediate's rate of formation is approximately equal to its rate of consumption.

For the consecutive mechanism $\mathrm{A\xrightarrow{k_1}B\xrightarrow{k_2}C}$, what is the differential rate equation for the intermediate B?

$\frac{d[B]}{dt}=k_1[A]-k_2[B]$. The first term represents formation of B from A, and the second represents consumption of B to form C.

Applying the steady-state approximation to $\mathrm{A\xrightarrow{k_1}B\xrightarrow{k_2}C}$ gives what expression for $[B]$?

Set $\frac{d[B]}{dt}=0$: $0=k_1[A]-k_2[B]$. Solving gives $[B]=\frac{k_1}{k_2}[A]$.

What condition makes the steady-state approximation valid for the intermediate in $\mathrm{A\xrightarrow{k_1}B\xrightarrow{k_2}C}$?

The intermediate must form slowly and be consumed rapidly, so $k_2\gg k_1$; a common approximate criterion is $k_2/k_1>10$. Under these conditions, $[B]$ stays low and changes relatively little.

Why is the steady-state approximation invalid when $k_1$ is greater than or comparable to $k_2$ in $\mathrm{A\rightarrow B\rightarrow C}$?

B forms rapidly but is removed slowly, so it accumulates and its concentration changes substantially. Its formation and consumption rates are not approximately balanced, violating the approximation.

For $\mathrm{A\xrightarrow{k_1}B\xrightarrow{k_2}C}$ under the steady-state approximation, what is the approximate rate of formation of C?

Because formation and consumption of B balance, the rate through the second step equals the rate of the first step: $\frac{d[C]}{dt}=k_2[B]=k_1[A]$.

What is the key mathematical advantage of the steady-state approximation?

It replaces a differential equation for an intermediate with an algebraic relationship, simplifying coupled rate equations and allowing the overall rate law to be expressed in terms of reactants.

How does the concentration of a valid steady-state intermediate typically compare with reactant and product concentrations?

The intermediate usually remains at a relatively low concentration because it is consumed nearly as quickly as it forms. Its concentration need not be exactly constant or at chemical equilibrium.

Why can the steady-state approximation be useful even though a true steady state is reached only approximately?

After a short initial period, the intermediate may change slowly enough that treating $d[\text{intermediate}]/dt$ as zero introduces little error. The approximation greatly simplifies the kinetics while preserving the dominant rate behavior.

In the mechanism for $\mathrm{H_2+Br_2\rightarrow 2HBr}$, which species are treated with the steady-state approximation?

The intermediates are atomic hydrogen and bromine, $\mathrm{H}$ and $\mathrm{Br}$. The reactants are $\mathrm{H_2}$ and $\mathrm{Br_2}$, and $\mathrm{HBr}$ is the product. Set $d[\mathrm{H}]/dt\approx0$ and $d[\mathrm{Br}]/dt\approx0$ while solving for the rate law.

What relationships result from applying the steady-state approximation to the intermediates in the hydrogen–bromine mechanism?

For $\mathrm{H}$, $0=k_2[\mathrm{Br}][\mathrm{H_2}]-k_3[\mathrm{H}][\mathrm{Br_2}]-k_4[\mathrm{H}][\mathrm{HBr}]$, so $[\mathrm{H}]=\frac{k_2[\mathrm{Br}][\mathrm{H_2}]}{k_3[\mathrm{Br_2}]+k_4[\mathrm{HBr}]}$. Combining the hydrogen and bromine steady-state equations gives $[\mathrm{Br}]=\left(\frac{k_1}{k_5}[\mathrm{Br_2}]\right)^{1/2}$.

What qualitative rate law does the steady-state approximation predict for the reaction $\mathrm{H_2+Br_2\rightarrow 2HBr}$?

Using the steady-state concentrations of H and Br gives a rate proportional to $\frac{[\mathrm{H_2}][\mathrm{Br_2}]^{1/2}}{1+K[\mathrm{HBr}]/[\mathrm{Br_2}]}$, where $K$ is a combination of rate constants. The denominator accounts for inhibition by HBr.