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?


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


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


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