Chemistry · General chemistry II · Worked example

Use the first-order integrated rate law

A compound A decomposes in solution by first-order kinetics with $k = 0.0410 \min^{- 1}$ . A solution starts at ${[A]}_{0} = 0.880 M$ . Find $[A]$ after $30.0 \min$ , the half-life, and the time at which $[A]$ has fallen to $0.110 M$ .

Pick the law

The reaction is first order, so the concentration decays exponentially. $k$ is in $\min^{- 1}$ and the times are in minutes, so the units already agree and nothing needs converting.

[A] = {[A]}_{0} e^{- k t}

Find $[A]$ after $30.0 \min$

The exponent is $k t$ $= (0.0410 \min^{- 1}) (30.0 \min)$ $= 1.23$ , a pure number, because minutes cancel. Then $e^{- 1.23} = 0.2923$ , so $29 %$ of A is left.

[A] = (0.880 M) e^{- 1.23} = 0.257 M

Find the half-life

For a first-order reaction the half-life depends only on $k$ , so the starting $0.880 M$ plays no part.

t_{1 / 2} = \frac{0.693}{0.0410 \min^{- 1}} = 16.9 \min

\begin{array}{c} t_{1 / 2} = \frac{0.693}{0.0410 \min^{- 1}} \\ = 16.9 \min \end{array}

Find the time to reach $0.110 M$

Now the unknown is $t$ , so use the logarithmic form and solve it for $t$ . $ln (0.880) - ln (0.110)$ $= ln 8.00$ $= 2.079$ , and $2.079 \div 0.0410 \min^{- 1}$ $= 50.7 \min$ .

t = \frac{ln (0.880) - ln (0.110)}{0.0410 \min^{- 1}} = 50.7 \min

\begin{array}{c} t = \frac{ln (0.880) - ln (0.110)}{0.0410 \min^{- 1}} \\ = 50.7 \min \end{array}

Check with half-lives

$0.110 M$ is one eighth of $0.880 M$ , and one eighth is three halvings, $0.880$ $\to 0.440$ $\to 0.220$ $\to 0.110$ . Three half-lives is $3 \times 16.9 \min = 50.7 \min$ , the same answer by a second route.

Equal half-lives for a first-order reaction
Time $(\min)$	Half-lives	$[A]$ $(M)$
$0$	$0$	$0.880$
$16.9$	$1$	$0.440$
$33.8$	$2$	$0.220$
$50.7$	$3$	$0.110$

Result

$[A] = 0.257 M$ after $30.0 \min$ . The half-life is $16.9 \min$ , and $[A]$ reaches $0.110 M$ at $50.7 \min$ .

Your turn

1. One like the example

A first-order reaction has $k = 0.0150 \min^{- 1}$ and starts at ${[A]}_{0} = 0.500 M$ . What is $[A]$ after $100 \min$ ?

Show the answer and explanation

$0.112 M$ .

$k t = (0.0150 \min^{- 1}) (100 \min)$ $= 1.50$ , a pure number. $e^{- 1.50} = 0.2231$ , so $22.3 %$ of A is left, and $0.500 M \times 0.2231$ $= 0.112 M$ . The half-life is $0.693 \div 0.0150$ $= 46.2 \min$ , and $100 \min$ is a little over two half-lives, so a little under a quarter should be left.

\begin{array}{l} [A] = {[A]}_{0} e^{- k t} \\ {[A]}_{0} = 0.500 M \\ k = 0.0150 \min^{- 1} \\ t = 100 \min \\ [A] = (0.500 M) e^{- (0.0150 \min^{- 1}) (100 \min)} = 0.112 M \end{array}

\begin{array}{c} [A] = {[A]}_{0} e^{- k t} \\ {[A]}_{0} = 0.500 M \\ k = 0.0150 \min^{- 1} \\ t = 100 \min \\ [A] = (0.500 M) \\ \times e^{- (0.0150 \min^{- 1}) (100 \min)} \\ = 0.112 M \end{array}

2. A variation

For the same reaction, how long does $[A]$ take to fall from $0.500 M$ to $0.100 M$ ?

Show the answer and explanation

$107 \min$ .

This time the unknown is $t$ , so use $ln [A] = ln {[A]}_{0} - k t$ and solve it for $t = (ln {[A]}_{0} - ln [A]) / k$ . $ln 0.500 - ln 0.100$ $= ln 5.00$ $= 1.609$ , and $1.609 \div 0.0150 \min^{- 1}$ $= 107 \min$ . That is $2.3$ half-lives, and one fifth is a little less than a quarter, which agrees.

\begin{array}{l} ln [A] = ln {[A]}_{0} - k t \\ {[A]}_{0} = 0.500 M \\ [A] = 0.100 M \\ k = 0.0150 \min^{- 1} \\ t = \frac{ln (0.500) - ln (0.100)}{0.0150 \min^{- 1}} = 107 \min \end{array}

\begin{array}{c} ln [A] = ln {[A]}_{0} - k t \\ {[A]}_{0} = 0.500 M \\ [A] = 0.100 M \\ k = 0.0150 \min^{- 1} \\ t = \frac{ln (0.500) - ln (0.100)}{0.0150 \min^{- 1}} \\ = 107 \min \end{array}

3. Somewhere new

A first-order decomposition has a half-life of $35.0 \min$ . Starting at $0.200 M$ , how long until only $10.0 %$ of the reactant is left?

Show the answer and explanation

$116 \min$ , about $3.3$ half-lives.

First turn the half-life into $k$ , $k = ln 2 \div t_{1 / 2}$ $= 0.693 \div 35.0 \min$ $= 0.0198 \min^{- 1}$ . $10.0 %$ of $0.200 M$ is $0.0200 M$ , and $ln 0.200 - ln 0.0200$ $= ln 10.0$ $= 2.303$ . $t = 2.303 \div 0.0198 \min^{- 1}$ $= 116 \min$ . The starting concentration cancels, so any first-order reaction with this half-life takes $116 \min$ to fall to a tenth.

\begin{array}{l} k = \frac{ln 2}{t_{1 / 2}} \\ t_{1 / 2} = 35.0 \min \\ k = \frac{0.693}{35.0 \min} = 0.0198 \min^{- 1} \\ ln [A] = ln {[A]}_{0} - k t \\ {[A]}_{0} = 0.200 M \\ [A] = 0.0200 M \\ t = \frac{ln (0.200) - ln (0.0200)}{0.0198 \min^{- 1}} = 116 \min \end{array}

\begin{array}{c} k = \frac{ln 2}{t_{1 / 2}} \\ t_{1 / 2} = 35.0 \min \\ k = \frac{0.693}{35.0 \min} \\ = 0.0198 \min^{- 1} \\ ln [A] = ln {[A]}_{0} - k t \\ {[A]}_{0} = 0.200 M \\ [A] = 0.0200 M \\ t = \frac{ln (0.200) - ln (0.0200)}{0.0198 \min^{- 1}} \\ = 116 \min \end{array}

Keep exploring

In the Chemistry box for the concentration, change the exponent to $+ 1.23$ , a sign slip. The row turns to ✗: “This working doesn’t give what row 4 gives with the values defined above.” Tap the ✗ and choose Show the value to see what row 4 gives. In Kinetics & Arrhenius, set the selected time to $50.7$ and the concentration comes back as $0.110 mol / L$ .

Return to the concept →

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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 6 key terms Check [A] in a Chemistry box Check the half-life in a Chemistry box Check the time in a Chemistry box Model it in Kinetics & Arrhenius See the half-lives in Graph Open worked example on a board Chemistry formulas: kinetics

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Pick the law

Find [A][A] after 30.0 min30.0 min

Find the half-life

Find the time to reach 0.110 M0.110 M

Check with half-lives

Result

Your turn

1. One like the example

0.112 M0.112 M.

2. A variation

107 min107 min.

3. Somewhere new

116 min116 min, about 3.33.3 half-lives.

Keep exploring

Study it, then try it

Find $[A]$ after $30.0 \min$

Find the time to reach $0.110 M$

$0.112 M$ .

$107 \min$ .

$116 \min$ , about $3.3$ half-lives.