# Question a0e43

Mar 4, 2016

$\text{1100 s}$

#### Explanation:

Yes, the half-life for a first-order reaction is

color(blue)( |bar( ul( t_"1/2" = ln(2)/k))|" ", where

$k$ - the rate constant of the reaction

Here's why that is the case.

You're dealing with a first-order reaction, so right from the start you know that the rate of the reaction depends linearly on the concentration of the reactant, methyl isocyanide, $\text{CH"_3"NC}$.

If you take the concentration of methyl isocyanide to be $\left[\text{CH"_3"NC}\right]$, you can say that rate of the reaction will be

"rate" = -(d["CH"_3"NC"])/dt

To bring the rate constant, $k$, in the mix, you need to write the differential rate law for this reaction

"rate" = -(d["CH"_3"NC"])/dt = k * ["CH"_3"NC"]

In order to be able to relate the rate of the reaction with time, you need to integrate the differential rate law. This will get you the integral rate law, which for your reaction will look like this

$\left[- \left(d \left[\text{CH"_3"NC"])/(["CH"_3"NC}\right]\right) = k \cdot \mathrm{dt}\right] \to \int$

-int(1/(["CH"_3"NC"]) * d["CH"_3"NC"]) = k * intdt

This will get you

ln(["CH"_3"NC"]) = -k * t + C" " " "color(red)("(*)")

To get rid of the integration constant, use the fact that you have an initial concentration of methyl isocyanide, ${\left[\text{CH"_3"NC}\right]}_{0}$, at $t = 0$.

This will get you

$\ln \left({\left[\text{CH"_3"NC"]_0) = -k * 0 + C implies C = ln(["CH"_3"NC}\right]}_{0}\right)$

Plug this into equation $\textcolor{red}{\text{(*)}}$ to get

$\ln \left({\left[\text{CH"_3"NC"]) - ln(["CH"_3"NC}\right]}_{0}\right) = - k \cdot t$

Finally, rearrange to get

$\textcolor{b l u e}{\ln \left(\left({\left[\text{CH"_3"NC"])/(["CH"_3"NC}\right]}_{0}\right)\right) = - k \cdot t}$

In your case, the half-life of the reaction, ${t}_{\text{1/2}}$, will be equal to the time needed for the concentration of methyl isocyanide to be reduced to half of its initial value. Therefore, at $t = {t}_{\text{1/2}}$, you will have

${\left[\text{CH"_3"NC"] = 1/2 * ["CH"_3"NC}\right]}_{0}$

Plug this into the integrated rate law and solve for ${t}_{\text{1/2}}$

ln( (1/2color(red)(cancel(color(black)(["CH"_3"NC"]))))/(color(red)(cancel(color(black)(["CH"_3"NC"]))))) = -k * t_"1/2"

Since

$\ln \left(\frac{1}{2}\right) = \ln \left(1\right) - \ln \left(2\right) = - \ln \left(2\right)$

this will get you

-ln(2) = -k * t_"1/2" = color(blue)( |bar( ul( t_"1/2" = ln(2)/k))|

Finally, plug in your values to get

t_"1/2" = ln(2)/(6.3 * 10^(-4)"s"^(-1)) = 1.1 * 10^3"s" = color(green)(| bar( ul("1100 s"))|)#

The answer is rounded to two sig figs, the number of sig figs you have for the rate constant.