Time for Given Percentage Completion

CHEIntegrated Rate Laws & Calculations

These numericals ask for the time to reach a stated percentage completion, or the concentration left after a given time, almost always for first-order kinetics. Speed comes from plugging the right concentration ratio into the integrated equation.

Master first-order time formula

t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}

set

[A]_0=100

[A]=100-(\%\,\text{completed})

Standard completion times (first order)

t_{50\%} = \dfrac{0.693}{k} \qquad t_{75\%} = \dfrac{1.386}{k} \qquad t_{99\%} = \dfrac{4.606}{k}

derived from

\log 2,\log 4,\log 100

Concentration after time t

[A] = [A]_0\times 10^{-kt/2.303} = \dfrac{[A]_0}{2^{\,t/t_{1/2}}}

second form when

t

is a whole multiple of

t_{1/2}

Master formula: $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$ ; set $[A]_0 = 100$ and $[A] = 100 - (\%\text{ completed})$ .
Standard milestones for first order: $50\%$ at $t = \dfrac{0.693}{k}$ , $75\%$ at $\dfrac{1.386}{k}$ , $99\%$ at $\dfrac{4.606}{k}$ .
Half-life ladder: $25\%$ remaining $= 2$ half-lives, $12.5\%$ remaining $= 3$ half-lives, since each $t_{1/2}$ halves the amount left.
To find concentration after time $t$ , rearrange to $[A] = [A]_0 \times 10^{-kt/2.303}$ , or count half-lives if $t$ is a whole multiple of $t_{1/2}$ .
Useful logs to keep handy: $\log 2 = 0.301$ , $\log 4 = 0.602$ , $\log 100 = 2$ — combine these for most percentage problems.
Trap: ' $99\%$ complete' means $\dfrac{[A]_0}{[A]} = 100$ (not $99$ ); always work with concentration remaining, not consumed.
Keep $k$ and $t$ in the same time unit (s, min, h) throughout, since first-order $k$ has units of inverse time.
For $90\%$ completion, $\dfrac{[A]_0}{[A]}=\dfrac{100}{10}=10$ , so $t_{90\%}=\dfrac{2.303}{k}\log 10=\dfrac{2.303}{k}$ .
Since $t_{75\%}=2t_{1/2}$ and $t_{99.9\%}=10t_{1/2}$ , you can answer many parts just by counting half-lives without re-deriving $k$ .
If $t_{1/2}$ is given instead of $k$ , first compute $k=0.693/t_{1/2}$ , then substitute into the master formula.
$25\%$ remaining means $75\%$ done; $12.5\%$ remaining means $87.5\%$ done — convert carefully between 'remaining' and 'completed' before applying the formula.
The required time grows logarithmically: going from $99\%$ to $99.9\%$ adds one more half-life-scale chunk, not a tenfold jump.

Where the marks go

Reading ' $99\%$ complete' as $[A]_0/[A]=99$ instead of $100$ — use concentration remaining, $[A]=1$ .
Confusing 'percentage remaining' with 'percentage completed' when setting $[A]$ in the ratio.
Mismatched units: leaving $k$ in $s^{-1}$ while $t$ comes out in minutes (or vice versa).
Using $0.693/k$ for a non-first-order reaction — these milestone formulae are valid only for first order.
Forgetting that the answer for $50\%$ equals one half-life, then double-counting it when stacking half-lives.

How the board asks it

Numericalmaster integrated first-order equation
A first-order reaction has a rate constant $k = 2.303 \times 10^{-3}\ \text{s}^{-1}$ . Calculate the time required for the reaction to be $80\%$ complete. $(\log 5 = 0.699)$
Numericalfinding k from percentage data, then a second milestone
A first-order reaction is found to be $30\%$ complete in $30\ \text{minutes}$ . Calculate the rate constant $k$ and hence the time required for the reaction to be $90\%$ complete.
Numericalhalf-life ladder for first order
The half-life of a first-order reaction is $20\ \text{min}$ . Calculate the time taken for $75\%$ of the reactant to be consumed.
Numericalconcentration remaining after time t
For a first-order reaction with $k = 1.15 \times 10^{-3}\ \text{s}^{-1}$ , calculate the fraction of the initial concentration of reactant that remains after $1000\ \text{s}$ .
Give reasonsthe '99% complete' trap where $[A]_0/[A]=100$
For a first-order reaction, show that the time required for $99\%$ completion is twice the time required for $90\%$ completion.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.