If the rate constant is 3.5 xx 10^(-3) "s"^(-1) for a first order decay of reactant A, how long does it take the reaction to get to 24% completion?

A) $\text{285.7 s}$ B) $\text{407.7 s}$ C) $7.84 \times {10}^{- 5} \text{s}$ D) $4.08 \times {10}^{- 4} \text{s}$ E) $\text{78.41 s}$

Mar 21, 2017

I got about $\text{78 s}$.

A) is incorrect because it is merely one over $k$.
B) is incorrect because it results from taking 24% completion to mean 24% of reactant $A$ leftover instead of 76% of $A$ leftover.
C) is incorrect because it results from an exponent error due to not using parentheses around $k$ in evaluating the time, even though the rest was correct.
D) is incorrect for the same reasons as C), except that it also takes the reaction to have 24% of $A$ leftover instead of 76%.

A first-order reaction

$A \to B$

follows the rate law:

$r \left(t\right) = k \left[A\right] = - \frac{\Delta \left[A\right]}{\Delta t} = \frac{\Delta \left[B\right]}{\Delta t}$

For an infinitesimally small time $\mathrm{dt}$, we can rewrite this as:

$r \left(t\right) = k \left[A\right] = - \frac{d \left[A\right]}{\mathrm{dt}} = \frac{d \left[B\right]}{\mathrm{dt}}$

Rearranging, we get:

$- k \mathrm{dt} = \frac{1}{\left[A\right]} d \left[A\right]$

What we can do is "integrate" from time $t = 0$ to time $t$ on the left and from initial concentration ${\left[A\right]}_{0}$ to current concentration $\left[A\right]$ on the right.

In simple words, this describes the evolution of time as compared to how the concentration changes, in a "smooth" manner (rather than "stepwise").

$- {\int}_{0}^{t} k \mathrm{dt} = {\int}_{{\left[A\right]}_{0}}^{\left[A\right]} \frac{1}{\left[A\right]} d \left[A\right]$

$- \left(k \cdot t - k \cdot 0\right) = \ln \left[A\right] - \ln {\left[A\right]}_{0}$

Therefore, we have the integrated rate law for first-order kinetics:

$\boldsymbol{\ln \left[A\right] = - k t + \ln {\left[A\right]}_{0}}$

Manipulating this rate law, we can determine at what time this reaction is at 24% completion, i.e. when 24% of $A$ is gone, or when 76% of $A$ is left. In other words, $\left[A\right] = 0.76 {\left[A\right]}_{0}$.

This means:

$\ln \left[A\right] - \ln {\left[A\right]}_{0} = - k t$

$\ln \setminus \frac{\left[A\right]}{{\left[A\right]}_{0}} = - k t$

$\ln \setminus \frac{0.76 \cancel{{\left[A\right]}_{0}}}{\cancel{{\left[A\right]}_{0}}} = - k t$

$\ln \left(0.76\right) = - k t$

Therefore:

$\textcolor{b l u e}{t} = - \ln \frac{0.76}{k}$

$= - \ln \frac{0.76}{3.5 \times {10}^{- 3} {\text{s}}^{- 1}}$

$=$ $\textcolor{b l u e}{\text{78.41 s}}$

This is closest to answer E.