##”1100 s”##
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 depends linearly on the concentration of the reactant, methyl isocyanide, ##”CH”_3″NC”##.
If you take the concentration of methyl isocyanide to be ##[“CH”_3″NC”]##, 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 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 . 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, ##[“CH”_3″NC”]_0##, at ##t=0##.
This will get you
##ln([“CH”_3″NC”]_0) = -k * 0 + C implies C = ln([“CH”_3″NC”]_0)##
Plug this into equation ##color(red)(“(*)”)## to get
##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, ##t_”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_”1/2″##, you will have
##[“CH”_3″NC”] = 1/2 * [“CH”_3″NC”]_0##
Plug this into the integrated rate law and solve for ##t_”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(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 , the number of sig figs you have for the rate constant.
