Chemistry · General chemistry II · Worked example

Explain a negative order with a mechanism

Ozone decomposes, $2 O_{3} (g) \to 3 O_{2} (g)$ , with the observed rate law $rate = k {[O_{3}]}^{2} {[O_{2}]}^{- 1}$ . Show that this mechanism is consistent with it. In a fast, reversible first step an ozone molecule splits into $O_{2}$ and a free oxygen atom. In a slow second step the oxygen atom attacks another $O_{3}$ , giving two $O_{2}$ .

2 O_{3} \to 3 O_{2}, rate = k \frac{{[O_{3}]}^{2}}{[O_{2}]}

Write the fast first step

The $NOBr$ example had a positive order in every reactant. Here a product, $O_{2}$ , slows the reaction, an order of $- 1$ . The overall equation is the given. Its first step splits an ozone molecule into $O_{2}$ and an oxygen atom, and the double arrow marks it fast and reversible.

O_{3} ⇌ O_{2} + O fast

Write the slow second step

The oxygen atom then attacks a second ozone. This step is slow, so it sets the rate.

O + O_{3} \to 2 O_{2} slow

Check that the steps add up

Adding the two steps puts an oxygen atom on both sides, and it cancels, leaving $2 O_{3} \to 3 O_{2}$ , the overall equation. The free oxygen atom is the intermediate.

Write the slow step’s rate law

The slow step is elementary, so its rate law uses its own coefficients. It contains the concentration of oxygen atoms, an intermediate the experiment never measures.

rate = k_{2} [O] [O_{3}]

Replace the intermediate

The fast first step is at equilibrium, so its forward rate, $k_{1} [O_{3}]$ , equals its reverse rate, $k_{- 1} [O_{2}]$ times the oxygen-atom concentration. Solving gives that concentration as $(k_{1} / k_{- 1}) [O_{3}] / [O_{2}]$ . More $O_{2}$ pushes the equilibrium back and leaves fewer oxygen atoms for the slow step.

[O] = \frac{k_{1}}{k_{- 1}} \cdot \frac{[O_{3}]}{[O_{2}]}

Compare with the experiment

Substituting gives $rate$ $= (k_{2} k_{1} / k_{- 1}) {[O_{3}]}^{2} / [O_{2}]$ , second order in $O_{3}$ and order $- 1$ in $O_{2}$ , the observed rate law. In a Chemistry box the order $- 1$ is written as the power $- 1$ on $[O_{2}]$ .

rate = \frac{k_{2} k_{1}}{k_{- 1}} \cdot \frac{{[O_{3}]}^{2}}{[O_{2}]} = k {[O_{3}]}^{2} {[O_{2}]}^{- 1}

\begin{array}{c} rate = \frac{k_{2} k_{1}}{k_{- 1}} \cdot \frac{{[O_{3}]}^{2}}{[O_{2}]} \\ = k {[O_{3}]}^{2} {[O_{2}]}^{- 1} \end{array}

State what it shows

The mechanism is consistent, so it is possible. A negative order in a product is a strong hint of a fast equilibrium that releases that product before the slow step.

Result

The mechanism is consistent. The slow step’s rate is $k_{2}$ times $[O_{3}]$ times the oxygen-atom concentration, and the fast equilibrium makes that concentration $(k_{1} / k_{- 1}) [O_{3}] / [O_{2}]$ , so $rate = k {[O_{3}]}^{2} {[O_{2}]}^{- 1}$ .

Your turn

1. One like the example

Chloroform reacts with chlorine, ${CHCl}_{3} + {Cl}_{2} \to {CCl}_{4} + HCl$ , with $rate = k [{CHCl}_{3}] {[{Cl}_{2}]}^{\frac{1}{2}}$ . Show that this mechanism is consistent. The steps are ${Cl}_{2} ⇌ 2 Cl$ (fast), then $Cl + {CHCl}_{3} \to HCl + {CCl}_{3}$ (slow), then ${CCl}_{3} + Cl \to {CCl}_{4}$ (fast).

Show the answer and explanation

It is consistent. The slow step gives $k_{2} [Cl] [{CHCl}_{3}]$ , and the fast equilibrium makes $[Cl]$ proportional to ${[{Cl}_{2}]}^{\frac{1}{2}}$ .

The steps add up with $Cl$ and ${CCl}_{3}$ as intermediates. The slow step contains $[Cl]$ . The fast first step gives $k_{1} [{Cl}_{2}] = k_{- 1} {[Cl]}^{2}$ , so $[Cl] = {(k_{1} / k_{- 1})}^{\frac{1}{2}} {[{Cl}_{2}]}^{\frac{1}{2}}$ . Substituting gives $rate = k [{CHCl}_{3}] {[{Cl}_{2}]}^{\frac{1}{2}}$ . A half order points to a molecule that splits in two before the slow step.

\begin{array}{l} {CHCl}_{3} + {Cl}_{2} \to {CCl}_{4} + HCl \\ {Cl}_{2} ⇌ 2 Cl fast \\ Cl + {CHCl}_{3} \to HCl + {CCl}_{3} slow \\ {CCl}_{3} + Cl \to {CCl}_{4} fast \\ rate = k [{CHCl}_{3}] {[{Cl}_{2}]}^{1 / 2} \end{array}

\begin{array}{c} {CHCl}_{3} + {Cl}_{2} \to {CCl}_{4} + HCl \\ {Cl}_{2} ⇌ 2 Cl fast \\ Cl + {CHCl}_{3} \to HCl + {CCl}_{3} slow \\ {CCl}_{3} + Cl \to {CCl}_{4} fast \\ rate = k [{CHCl}_{3}] {[{Cl}_{2}]}^{1 / 2} \end{array}

2. A variation

For $2 NO + 2 H_{2} \to N_{2} + 2 H_{2} O$ the observed rate law is $rate = k {[NO]}^{2} [H_{2}]$ . Test this three-step mechanism. The steps are $2 NO ⇌ N_{2} O_{2}$ (fast), then $N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O$ (slow), then $N_{2} O + H_{2} \to N_{2} + H_{2} O$ (fast).

Show the answer and explanation

It is consistent. The rate law is first order in $H_{2}$ , although the overall equation uses two $H_{2}$ .

Three steps this time, with two intermediates, $N_{2} O_{2}$ and $N_{2} O$ . They add to $2 NO + 2 H_{2} \to N_{2} + 2 H_{2} O$ . The slow step gives $k_{2} [N_{2} O_{2}] [H_{2}]$ , and the fast equilibrium gives $[N_{2} O_{2}] = (k_{1} / k_{- 1}) {[NO]}^{2}$ . The prediction, $k {[NO]}^{2} [H_{2}]$ , matches. The second $H_{2}$ enters only after the slow step, so it does not appear in the rate law.

\begin{array}{l} 2 NO + 2 H_{2} \to N_{2} + 2 H_{2} O \\ 2 NO ⇌ N_{2} O_{2} fast \\ N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O slow \\ N_{2} O + H_{2} \to N_{2} + H_{2} O fast \\ rate = k {[NO]}^{2} [H_{2}] \end{array}

\begin{array}{c} 2 NO + 2 H_{2} \to N_{2} + 2 H_{2} O \\ 2 NO ⇌ N_{2} O_{2} fast \\ N_{2} O_{2} + H_{2} \\ \to N_{2} O + H_{2} O slow \\ N_{2} O + H_{2} \to N_{2} + H_{2} O fast \\ rate = k {[NO]}^{2} [H_{2}] \end{array}

3. Somewhere new

For $H_{2} + I_{2} \to 2 HI$ the rate law is $rate = k [H_{2}] [I_{2}]$ , which looks like one collision of $H_{2}$ with $I_{2}$ . Show that a different mechanism gives the same rate law. The steps are $I_{2} ⇌ 2 I$ (fast), then $H_{2} + 2 I \to 2 HI$ (slow).

Show the answer and explanation

The slow step gives $k_{2} [H_{2}] {[I]}^{2}$ , and ${[I]}^{2} = (k_{1} / k_{- 1}) [I_{2}]$ , so $rate = k [H_{2}] [I_{2}]$ . Two mechanisms predict the same rate law.

The steps add up with I as the intermediate. The fast equilibrium gives $k_{1} [I_{2}] = k_{- 1} {[I]}^{2}$ , so ${[I]}^{2} = (k_{1} / k_{- 1}) [I_{2}]$ . Substituting into the slow step gives $k [H_{2}] [I_{2}]$ , exactly what a single collision would predict. The rate law alone cannot tell the two apart, which is why a matching rate law makes a mechanism possible, not proved.

\begin{array}{l} H_{2} + I_{2} \to 2 HI \\ I_{2} ⇌ 2 I fast \\ H_{2} + 2 I \to 2 HI slow \\ rate = k [H_{2}] [I_{2}] \end{array}

Keep exploring

In the Chemistry box, change the last row to $rate = k {[O_{3}]}^{2}$ . The row turns ✗: “The mechanism predicts rate = k[O₂]⁻¹[O₃]², but this rate law is rate = k[O₃]².”

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 9 key terms Test the ozone mechanism in a Chemistry box See the ozone mechanism as an energy profile Open worked example on a board Chemistry formulas: kinetics

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Write the fast first step

Write the slow second step

Check that the steps add up

Write the slow step’s rate law

Replace the intermediate

Compare with the experiment

State what it shows

Result

Your turn

1. One like the example

It is consistent. The slow step gives k2[Cl]⁢[CHCl3]k2​[Cl][CHCl3​], and the fast equilibrium makes [Cl][Cl] proportional to [Cl2]12[Cl2​]21​.

2. A variation

It is consistent. The rate law is first order in H2H2​, although the overall equation uses two H2H2​.

3. Somewhere new

The slow step gives k2[H2]⁢[I]2k2​[H2​][I]2, and [I]2=(k1/k−1)⁢[I2][I]2=(k1​/k−1​)[I2​], so rate=k⁢[H2]⁢[I2]rate=k[H2​][I2​]. Two mechanisms predict the same rate law.

Keep exploring

Study it, then try it

It is consistent. The slow step gives $k_{2} [Cl] [{CHCl}_{3}]$ , and the fast equilibrium makes $[Cl]$ proportional to ${[{Cl}_{2}]}^{\frac{1}{2}}$ .

It is consistent. The rate law is first order in $H_{2}$ , although the overall equation uses two $H_{2}$ .

The slow step gives $k_{2} [H_{2}] {[I]}^{2}$ , and ${[I]}^{2} = (k_{1} / k_{- 1}) [I_{2}]$ , so $rate = k [H_{2}] [I_{2}]$ . Two mechanisms predict the same rate law.