Question #a0e43
1 Answer
Explanation:
Yes, the half-life for a first-order reaction is
#color(blue)( |bar( ul( t_"1/2" = ln(2)/k))|" "# , where
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,
#"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,
#["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.