# Question #a0e43

##### 1 Answer

#### Answer:

#### Explanation:

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

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

*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,

If you take the concentration of methyl isocyanide to be

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

To bring the **rate constant**, *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

#[-(d["CH"_3"NC"])/(["CH"_3"NC"]) = k * dt] -> 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,

This will get you

#ln(["CH"_3"NC"]_0) = -k * 0 + C implies C = ln(["CH"_3"NC"]_0)#

Plug this into equation

#ln(["CH"_3"NC"]) - ln(["CH"_3"NC"]_0) = -k * t#

Finally, rearrange to get

#color(blue)(ln((["CH"_3"NC"])/(["CH"_3"NC"]_0)) = -k* t)#

In your case, the **half-life** of the reaction, **half** of its initial value. Therefore, at

#["CH"_3"NC"] = 1/2 * ["CH"_3"NC"]_0#

Plug this into the integrated rate law and solve for

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

Since

#ln(1/2) = ln(1) - ln(2) = - ln(2)#

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.