Chemistry · General chemistry II · Worked example

Rule a mechanism in or out for 2NO + Br₂

Nitric oxide and bromine form nitrosyl bromide, $2 NO (g) + {Br}_{2} (g)$ $\to 2 NOBr (g)$ . Experiments give $rate = k {[NO]}^{2} [{Br}_{2}]$ . Which of two mechanisms is consistent with that rate law? Mechanism A is $NO + {Br}_{2} \to {NOBr}_{2}$ (slow), then ${NOBr}_{2} + NO \to 2 NOBr$ (fast). Mechanism B is $NO + {Br}_{2} ⇌ {NOBr}_{2}$ (fast), then ${NOBr}_{2} + NO \to 2 NOBr$ (slow).

rate = k {[NO]}^{2} [{Br}_{2}]

Write what the experiment says

Start from the overall reaction and the observed rate law. The orders, $2$ in $NO$ and $1$ in ${Br}_{2}$ , are the target every mechanism must reproduce.

2 NO + {Br}_{2} \to 2 NOBr, rate = k {[NO]}^{2} [{Br}_{2}]

\begin{array}{c} 2 NO + {Br}_{2} \to 2 NOBr \\ rate = k {[NO]}^{2} [{Br}_{2}] \end{array}

Check 1 for both mechanisms

Both mechanisms use the same two equations, and only the choice of slow step differs. Adding them, ${NOBr}_{2}$ appears on both sides and cancels, and the sum is the overall reaction. Neither fails check 1. ${NOBr}_{2}$ is the intermediate, made in the first step and used in the second.

\begin{aligned} NO + {Br}_{2} & \to {NOBr}_{2} \\ {NOBr}_{2} + NO & \to 2 NOBr \\ sum: 2 NO + {Br}_{2} & \to 2 NOBr \end{aligned}

Mechanism A, the slow step’s rate law

In A the first step is slow. It is elementary, so its rate law uses its own coefficients, one $NO$ and one ${Br}_{2}$ . It contains no intermediate, so check 3 passes at once.

rate = k_{1} [NO] [{Br}_{2}]

Mechanism A, compared and ruled out

A predicts first order in $NO$ , but the experiment shows second order. The orders disagree, so Mechanism A is ruled out, although its steps added up correctly.

k_{1} {[NO]}^{1} [{Br}_{2}] \neq k {[NO]}^{2} [{Br}_{2}]

Mechanism B, an intermediate in the slow step

In B the second step is slow, so the rate law comes from it. It contains ${NOBr}_{2}$ , an intermediate the experiment never measures, so check 3 sends the work back to the fast first step.

rate = k_{2} [{NOBr}_{2}] [NO]

Replace the intermediate using the fast equilibrium

The first step is fast and reversible, so its forward and reverse rates are equal. Solve that equality for $[{NOBr}_{2}]$ .

k_{1} [NO] [{Br}_{2}] = k_{- 1} [{NOBr}_{2}] \Rightarrow [{NOBr}_{2}] = \frac{k_{1}}{k_{- 1}} [NO] [{Br}_{2}]

\begin{array}{c} k_{1} [NO] [{Br}_{2}] = k_{- 1} [{NOBr}_{2}] \\ \Rightarrow [{NOBr}_{2}] = \frac{k_{1}}{k_{- 1}} [NO] [{Br}_{2}] \end{array}

Substitute and compare

Put that expression into the slow step’s rate law and collect the constants into one $k$ . The prediction is second order in $NO$ and first order in ${Br}_{2}$ , matching the experiment. Mechanism B passes all four checks.

rate = k_{2} \frac{k_{1}}{k_{- 1}} {[NO]}^{2} [{Br}_{2}] = k {[NO]}^{2} [{Br}_{2}]

\begin{array}{c} rate = k_{2} \frac{k_{1}}{k_{- 1}} {[NO]}^{2} [{Br}_{2}] \\ = k {[NO]}^{2} [{Br}_{2}] \end{array}

State the conclusion carefully

Mechanism B is consistent with the data, so it is possible. It is not proved, because a single collision of two $NO$ molecules with one ${Br}_{2}$ would predict the same rate law. Such three-body collisions are rare, and detecting ${NOBr}_{2}$ during the reaction would be evidence for B.

Result

Mechanism A is ruled out, because it predicts $rate = k [NO] [{Br}_{2}]$ . Mechanism B is consistent, because its fast equilibrium turns $k_{2} [{NOBr}_{2}] [NO]$ into $k {[NO]}^{2} [{Br}_{2}]$ . Consistent means possible, not proved.

Your turn

1. One like the example

For $H_{2} + 2 ICl \to I_{2} + 2 HCl$ the observed rate law is $rate = k [H_{2}] [ICl]$ . Is this mechanism consistent? The steps are $H_{2} + ICl \to HI + HCl$ (slow), then $HI + ICl \to I_{2} + HCl$ (fast).

Show the answer and explanation

Yes. It is consistent with the rate law, though not proved.

The steps add up to the overall reaction, with $HI$ made in step 1 and used in step 2, so $HI$ is the intermediate. The slow first step is elementary, so it predicts $rate = k [H_{2}] [ICl]$ , which matches. No intermediate appears in that rate law, so no substitution is needed.

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

\begin{array}{c} H_{2} + 2 ICl \to I_{2} + 2 HCl \\ H_{2} + ICl \to HI + HCl slow \\ HI + ICl \to I_{2} + HCl fast \\ rate = k [H_{2}] [ICl] \end{array}

2. A variation

For $2 NO + O_{2} \to 2 {NO}_{2}$ the observed rate law is $rate = k {[NO]}^{2} [O_{2}]$ . Show that this mechanism is consistent. The steps are $2 NO ⇌ N_{2} O_{2}$ (fast), then $N_{2} O_{2} + O_{2} \to 2 {NO}_{2}$ (slow).

Show the answer and explanation

It is consistent. The slow step gives $k_{2} [N_{2} O_{2}] [O_{2}]$ , and the fast equilibrium turns $[N_{2} O_{2}]$ into $(k_{1} / k_{- 1}) {[NO]}^{2}$ , giving $k {[NO]}^{2} [O_{2}]$ .

This time the slow step comes second, so its rate law, $k_{2} [N_{2} O_{2}] [O_{2}]$ , contains the intermediate $N_{2} O_{2}$ . The fast first step has $k_{1} {[NO]}^{2} = k_{- 1} [N_{2} O_{2}]$ , so $[N_{2} O_{2}] = (k_{1} / k_{- 1}) {[NO]}^{2}$ . Substituting gives $rate$ $= (k_{2} k_{1} / k_{- 1}) {[NO]}^{2} [O_{2}]$ , second order in $NO$ and first in $O_{2}$ , as observed.

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

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

3. Somewhere new

Iodide speeds up the decomposition of hydrogen peroxide, $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ , and the rate law is $rate = k [H_{2} O_{2}] [I^{-}]$ . In this mechanism, which species is the catalyst and which is the intermediate, and is the mechanism consistent? The steps are $H_{2} O_{2} + I^{-} \to H_{2} O + {IO}^{-}$ (slow), then $H_{2} O_{2} + {IO}^{-}$ $\to H_{2} O + O_{2} + I^{-}$ (fast).

Show the answer and explanation

$I^{-}$ is the catalyst and ${IO}^{-}$ is the intermediate. The mechanism is consistent with the rate law.

$I^{-}$ is used in step 1 and given back in step 2, so it is a catalyst. It cancels from the overall equation but appears in the rate law. ${IO}^{-}$ is made in step 1 and used in step 2, so it is the intermediate. The steps add to $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ , and the slow step predicts $k [H_{2} O_{2}] [I^{-}]$ , which matches.

\begin{array}{l} 2 H_{2} O_{2} \to 2 H_{2} O + O_{2} \\ H_{2} O_{2} + I^{-} \to H_{2} O + {IO}^{-} slow \\ H_{2} O_{2} + {IO}^{-} \to H_{2} O + O_{2} + I^{-} fast \\ rate = k [H_{2} O_{2}] [I^{-}] \end{array}

\begin{array}{c} 2 H_{2} O_{2} \to 2 H_{2} O + O_{2} \\ H_{2} O_{2} + I^{-} \to H_{2} O + {IO}^{-} slow \\ H_{2} O_{2} + {IO}^{-} \\ \to H_{2} O + O_{2} + I^{-} fast \\ rate = k [H_{2} O_{2}] [I^{-}] \end{array}

Keep exploring

Open Mechanism B in a Chemistry box and change the word fast in row 2 to slow, and slow in row 3 to fast. The rate-law row turns ✗: “The mechanism predicts rate = k[Br₂][NO], but this rate law is rate = k[Br₂][NO]².”

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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 Mechanism B in a Chemistry box See Mechanism B as an energy profile Open worked example on a board Chemistry formulas: kinetics

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Write what the experiment says

Check 1 for both mechanisms

Mechanism A, the slow step’s rate law

Mechanism A, compared and ruled out

Mechanism B, an intermediate in the slow step

Replace the intermediate using the fast equilibrium

Substitute and compare

State the conclusion carefully

Result

Your turn

1. One like the example

Yes. It is consistent with the rate law, though not proved.

2. A variation

It is consistent. The slow step gives k2[N2O2]⁢[O2]k2​[N2​O2​][O2​], and the fast equilibrium turns [N2O2][N2​O2​] into (k1/k−1)⁢[NO]2(k1​/k−1​)[NO]2, giving k⁢[NO]2[O2]k[NO]2[O2​].

3. Somewhere new

I−I− is the catalyst and IO−IO− is the intermediate. The mechanism is consistent with the rate law.

Keep exploring

Study it, then try it

It is consistent. The slow step gives $k_{2} [N_{2} O_{2}] [O_{2}]$ , and the fast equilibrium turns $[N_{2} O_{2}]$ into $(k_{1} / k_{- 1}) {[NO]}^{2}$ , giving $k {[NO]}^{2} [O_{2}]$ .

$I^{-}$ is the catalyst and ${IO}^{-}$ is the intermediate. The mechanism is consistent with the rate law.