# A certain first-order reaction has a rate of "0.04 M/s" after 10 seconds and "0.03 M/s" after 20 seconds. What is the half-life for this reaction?

Aug 14, 2017

${t}_{\text{1/2" = "24.1 s}}$.

Interesting question. The rates $r \left(t\right)$ defined at each point in a reaction shall be given by the first order rate law:

$r \left(10\right) = k {\left[A\right]}_{10} = {\text{0.04 M"cdot"s}}^{- 1}$

$r \left(20\right) = k {\left[A\right]}_{20} = {\text{0.03 M"cdot"s}}^{- 1}$

where the subscript in front of $\left[A\right]$ indicates the time in seconds. The rate constant $k$ is the same for the same reaction at the same temperature.

Each would follow the same integrated rate law for first-order one-reactant reactions:

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

where ${\left[A\right]}_{0}$ is the initial concentration of $A$.

Since the rate constants $k$ are the same, we can write a ratio of the concentrations and relate that to the ratio of the rates:

$\frac{{\left[A\right]}_{20}}{{\left[A\right]}_{10}} = \frac{{r}_{20}}{{r}_{10}}$

So, using the integrated rate law for each time:

$\ln {\left[A\right]}_{10} = - 10 k + \ln {\left[A\right]}_{0}$

$\implies \ln {\left[A\right]}_{10} + 10 k = \ln {\left[A\right]}_{0}$

$\ln {\left[A\right]}_{20} = - 20 k + \ln {\left[A\right]}_{0}$

$\implies \ln {\left[A\right]}_{20} + 20 k = \ln {\left[A\right]}_{0}$

Since $\ln {\left[A\right]}_{0}$ for two times in a given reaction is the same for all time (i.e. the we start at the same initial concentration), we can thus say that:

$\ln {\left[A\right]}_{20} + 20 k = \ln {\left[A\right]}_{10} + 10 k$

$\ln {\left[A\right]}_{20} - \ln {\left[A\right]}_{10} = \ln \left(\frac{{\left[A\right]}_{20}}{{\left[A\right]}_{10}}\right) = 10 k - 20 k$

Plug in the ratio of the rates:

$\ln \left(\frac{{r}_{20}}{{r}_{10}}\right) = - 10 k$

Lastly, the half-life ${t}_{\text{1/2}}$ of a first-order reaction is given by

$\boldsymbol{{t}_{\text{1/2}} = \frac{\ln 2}{k}}$,

so $k = \frac{\ln 2}{t} _ \text{1/2}$. Thus:

$\ln \left(\frac{{r}_{20}}{{r}_{10}}\right) = - \frac{10 \ln 2}{t} _ \text{1/2}$

$\implies \textcolor{b l u e}{{t}_{\text{1/2}}} = - \frac{10 \ln 2}{\ln \left({r}_{20} / {r}_{10}\right)} = - \frac{10 \ln 2}{\ln \left(0.03 / 0.04\right)}$

$=$ $\underline{\textcolor{b l u e}{\text{24.1 s}}}$