Chemistry · General chemistry II · Worked example

Find a rate law from initial rates

For $2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$ at a fixed temperature, three trials give these initial rates. Find the order in $NO$ , the order in $O_{2}$ , the rate law and $k$ with units.

rate = k {[NO]}^{m} {[O_{2}]}^{n}

Lay out the data

Look for pairs of trials that differ in one concentration only. Trials 1 and 2 differ only in $[NO]$ , and trials 1 and 3 only in $[O_{2}]$ . Trials 2 and 3 differ in both, so they cannot give either order on their own.

Initial-rate data (illustrative)
Trial	$[NO]$ $(M)$	$[O_{2}]$ $(M)$	Initial rate $(M / s)$
1	$0.020$	$0.010$	$0.028$
2	$0.040$	$0.010$	$0.112$
3	$0.020$	$0.020$	$0.056$

Order in $NO$ from trials 1 and 2

Dividing trial 2’s rate law by trial 1’s cancels $k$ and $[O_{2}]$ , which are the same in both. $[NO]$ doubled and the rate went from $0.028$ to $0.112 M / s$ , four times as fast. $2^{m} = 4$ gives $m = 2$ .

\frac{0.112}{0.028} = {(\frac{0.040}{0.020})}^{m} \Rightarrow 4 = 2^{m} \Rightarrow m = 2

\begin{array}{c} \frac{0.112}{0.028} = {(\frac{0.040}{0.020})}^{m} \\ 4 = 2^{m} \Rightarrow m = 2 \end{array}

Order in $O_{2}$ from trials 1 and 3

This time $k$ and $[NO]$ cancel. $[O_{2}]$ doubled and the rate doubled, from $0.028$ to $0.056 M / s$ . $2^{n} = 2$ gives $n = 1$ .

\frac{0.056}{0.028} = {(\frac{0.020}{0.010})}^{n} \Rightarrow 2 = 2^{n} \Rightarrow n = 1

\begin{array}{c} \frac{0.056}{0.028} = {(\frac{0.020}{0.010})}^{n} \\ 2 = 2^{n} \Rightarrow n = 1 \end{array}

Write the rate law

Second order in $NO$ and first order in $O_{2}$ gives a third-order reaction overall. Here the orders happen to equal the coefficients, $2$ and $1$ . That is a result of the experiment, not something the equation could have told you.

rate = k {[NO]}^{2} [O_{2}]

Solve for $k$ with trial 1

Divide trial 1’s rate by ${[NO]}^{2} [O_{2}]$ . The units are $(M / s) \div M^{3} = M^{- 2} s^{- 1}$ , the units of a third-order rate constant. The arithmetic is $0.028 \div (0.00040 \times 0.010)$ $= 0.028 \div (4.0 \times 10^{- 6})$ $= 7.0 \times 10^{3}$ .

k = \frac{0.028 M s^{- 1}}{{(0.020 M)}^{2} (0.010 M)} = 7.0 \times 10^{3} M^{- 2} s^{- 1}

\begin{array}{c} k = \frac{0.028 M s^{- 1}}{{(0.020 M)}^{2} (0.010 M)} \\ = 7.0 \times 10^{3} M^{- 2} s^{- 1} \end{array}

Check with the other trials

Trial 2 gives $0.112 \div ({0.040}^{2} \times 0.010)$ $= 7.0 \times 10^{3}$ as well, so one $k$ fits both. The rate law with this $k$ predicts trial 3 exactly, $(7.0 \times 10^{3}) {(0.020)}^{2} (0.020)$ $= 0.056 M / s$ , a run that was used only for the order in $O_{2}$ .

{rate}_{3} = (7.0 \times 10^{3}) {(0.020)}^{2} (0.020) = 0.056 M/s

\begin{array}{c} {rate}_{3} = (7.0 \times 10^{3}) \\ \times {(0.020)}^{2} (0.020) \\ = 0.056 M/s \end{array}

Result

$Rate = k {[NO]}^{2} [O_{2}]$ , third order overall, with $k = 7.0 \times 10^{3} M^{- 2} s^{- 1}$ .

rate = k {[NO]}^{2} [O_{2}]

Your turn

1. One like the example

For $A + B \to C$ , trial 1 has $[A] = 0.10 M$ , $[B] = 0.10 M$ and a rate of $3.0 \times 10^{- 4} M / s$ . Trial 2 has $[A] = 0.30 M$ , $[B] = 0.10 M$ and $9.0 \times 10^{- 4} M / s$ . Trial 3 has $[A] = 0.10 M$ , $[B] = 0.30 M$ and $3.0 \times 10^{- 4} M / s$ . Find the rate law and $k$ .

Show the answer and explanation

$Rate = k [A]$ , with $k = 3.0 \times 10^{- 3} s^{- 1}$ .

Trials 1 and 2 differ only in $[A]$ . Tripling $[A]$ tripled the rate, so the order in A is $1$ . Trials 1 and 3 differ only in $[B]$ , and tripling $[B]$ left the rate unchanged, so the order in B is $0$ . Then $k = 3.0 \times 10^{- 4} \div 0.10$ $= 3.0 \times 10^{- 3}$ , in $s^{- 1}$ because the reaction is first order overall.

\begin{array}{l} \frac{9.0 \times 10^{- 4} M/s}{3.0 \times 10^{- 4} M/s} = 3.0 \\ \frac{0.30 M}{0.10 M} = 3.0 \\ rate = k [A] \\ \frac{3.0 \times 10^{- 4} M/s}{0.10 M} = 3.0 \times 10^{- 3} s^{- 1} \end{array}

2. A variation

For $A + 2 B \to products$ , trial 1 has $[A] = 0.10 M$ , $[B] = 0.20 M$ and a rate of $2.0 \times 10^{- 3} M / s$ . Trial 2 has $[A] = 0.30 M$ , $[B] = 0.20 M$ and $1.8 \times 10^{- 2} M / s$ . Trial 3 has $[A] = 0.10 M$ , $[B] = 0.40 M$ and $4.0 \times 10^{- 3} M / s$ . Find the rate law and $k$ .

Show the answer and explanation

$Rate = k {[A]}^{2} [B]$ , with $k = 1.0 M^{- 2} s^{- 1}$ .

Tripling $[A]$ multiplied the rate by $9.0$ , and $3^{2} = 9$ , so the order in A is $2$ . Doubling $[B]$ doubled the rate, so the order in B is $1$ , although B’s coefficient is $2$ . The reaction is third order overall. $k = 2.0 \times 10^{- 3}$ $\div ({0.10}^{2} \times 0.20)$ $= 2.0 \times 10^{- 3} \div 0.0020$ $= 1.0 M^{- 2} s^{- 1}$ .

\begin{array}{l} \frac{1.8 \times 10^{- 2} M/s}{2.0 \times 10^{- 3} M/s} = 9.0 \\ \frac{0.30 M}{0.10 M} = 3.0 \\ \frac{4.0 \times 10^{- 3} M/s}{2.0 \times 10^{- 3} M/s} = 2.0 \\ \frac{0.40 M}{0.20 M} = 2.0 \\ rate = k {[A]}^{2} [B] \\ \frac{2.0 \times 10^{- 3} M/s}{{(0.10 M)}^{2} (0.20 M)} = 1.0 M^{- 2} s^{- 1} \end{array}

3. Somewhere new

Use the worked rate law, $rate = k {[NO]}^{2} [O_{2}]$ with $k = 7.0 \times 10^{3} M^{- 2} s^{- 1}$ . Starting from trial 1, $[NO]$ is halved to $0.010 M$ . What $[O_{2}]$ keeps the rate at $0.028 M / s$ ?

Show the answer and explanation

$[O_{2}] = 0.040 M$ , four times trial 1’s $0.010 M$ .

Halving $[NO]$ multiplies the rate by ${(\frac{1}{2})}^{2} = \frac{1}{4}$ , because the order in $NO$ is $2$ . To win the rate back, $[O_{2}]$ must rise by a factor of $4$ , since the order in $O_{2}$ is $1$ . Putting $0.010 M$ and $0.040 M$ into the rate law gives $(7.0 \times 10^{3}) {(0.010)}^{2} (0.040)$ $= 0.028 M / s$ , the rate of trial 1.

\begin{array}{l} rate = k {[NO]}^{2} [O_{2}] \\ k = 7.0 \times 10^{3} M^{- 2} s^{- 1} \\ [NO] = 0.010 M \\ [O_{2}] = 0.040 M \\ rate = (7.0 \times 10^{3} M^{- 2} s^{- 1}) {(0.010 M)}^{2} (0.040 M) = 0.028 M/s \end{array}

\begin{array}{c} rate = k {[NO]}^{2} [O_{2}] \\ k = 7.0 \times 10^{3} M^{- 2} s^{- 1} \\ [NO] = 0.010 M \\ [O_{2}] = 0.040 M \\ rate = (7.0 \times 10^{3} M^{- 2} s^{- 1}) \\ \times {(0.010 M)}^{2} (0.040 M) \\ = 0.028 M/s \end{array}

Keep exploring

Open Kinetics & Arrhenius with trials 1 and 2 and change the second rate to $0.056 M / s$ . The order it estimates falls from $2$ to $1$ , because the rate now only doubles.

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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 7 key terms Check the orders and k in a Chemistry box Compare trials 1 and 2 in Kinetics & Arrhenius Open worked example on a board Chemistry formulas: experimental analysis

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Lay out the data

Order in NONO from trials 1 and 2

Order in O2O2​ from trials 1 and 3

Write the rate law

Solve for kk with trial 1

Check with the other trials

Result

Your turn

1. One like the example

Rate=k⁢[A]Rate=k[A], with k=3.0×10−3 s−1k=3.0×10−3 s−1.

2. A variation

Rate=k⁢[A]2[B]Rate=k[A]2[B], with k=1.0 M−2 s−1k=1.0 M−2 s−1.

3. Somewhere new

[O2]=0.040 M[O2​]=0.040 M, four times trial 1’s 0.010 M0.010 M.

Keep exploring

Study it, then try it

Order in $NO$ from trials 1 and 2

Order in $O_{2}$ from trials 1 and 3

Solve for $k$ with trial 1

$Rate = k [A]$ , with $k = 3.0 \times 10^{- 3} s^{- 1}$ .

$Rate = k {[A]}^{2} [B]$ , with $k = 1.0 M^{- 2} s^{- 1}$ .

$[O_{2}] = 0.040 M$ , four times trial 1’s $0.010 M$ .