Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# Methyl isonitrile, CH3NC, isomerizes when heated to give acetonitrile (methyl cyanide), CH3CN CH3NC (g) -&gt; CH3CN (g) The reaction is first order. At 230 ºC, the rate constant for the isomerization is 6.3 x 10-4 /s. a) What is the half-life?

"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.

LEARN MORE EFFECTIVELY AND GET BETTER GRADES!