3
$\begingroup$

For the reaction: $$\ce{F + Br_2 -> FBr + Br}$$ The gas concentrations as a function of time are given by: $$kt=\ce{\frac{1}{[Br2]_0-[F]_0}}\ce{\ln(\frac{[Br2][F]_0}{[F][Br_2]_0})}$$ Given that $\ce{[F]_0=4 x 10^{-9}}$ and $\ce{[Br2]=1x 10^{-10}}$ I am meant to show that the rate constant is first order with respect to $\ce{Br2}$. How do I do this?

Fluorine atoms are in excess but is it a large enough excess to assume that the isolation method is valid? I could then cancel terms but would I the have to differentiate to find the rate law?

$\endgroup$
4
$\begingroup$

Maybe it would be best to reorganise that to

$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right] \frac{\ce{[Br_2]_0}\ce{[F]}}{\ce{[F]_0}}$

and as there is 40 times more $\ce{F}$ than $\ce{Br_2}$, it is safe to assume that $\ce{[F]}\approx\ce{[F]_0}$. I.e. even if all the $\ce{Br_2}$ is gone there is still $\ce{[F]}=3.9\times10^{-9}$. This simplifies the above equation to

$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right]\ce{[Br_2]_0}$

which is the familiar form of the expression of concentration. It is also easy to see what the pseudo first-order rate constant is.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.