Question

A reaction of the form $$aA \longrightarrow Products$$gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

Short answer

The time required for the reaction to reach \(22.9\%\) completion is \(36.1 \,\mathrm{s}\).

Step-by-step solution

Calculate the reaction rate constant

The slope of the given plot equals the negative value of the reaction rate constant, so: \[k = -(-7.35 \times 10^{-3})\] \[k = 7.35 \times 10^{-3} \,\mathrm{s^{-1}}\]

Determine the concentration of A at 22.9% completion

To find the concentration of A at \(22.9\%\) completion, we'll multiply the initial concentration with the remaining percentage: \[[\mathrm{A}] = [\mathrm{A}]_{0} \times \left(1 - \frac{22.9}{100}\right)\] \[[\mathrm{A}] = 0.0100 \,\mathrm{M} \times \left(1- 0.229\right)\] \[[\mathrm{A}] = 0.0100 \,\mathrm{M} \times 0.771\] \[[\mathrm{A}] = 0.00771 \,\mathrm{M}\]

Calculate the time required for 22.9% completion

Now, we can insert the values of \(k, [\mathrm{A}]_{0}\), and \([\mathrm{A}]\) into the formula for the first-order reaction, and solve for the time, \(t\): \[\ln\frac{[\mathrm{A}]_{0}}{[\mathrm{A}]} = kt\] \[\ln\frac{0.0100}{0.00771} = (7.35 \times 10^{-3}) t\] \[0.265 = 7.35 \times 10^{-3}t\] Now we'll solve for \(t\): \[t = \frac{0.265}{7.35 \times 10^{-3}}\] \[t = 36.1 \,\mathrm{s}\] So, it takes 36.1 seconds for the reaction to reach \(22.9\%\) completion.