What is the rate constant for this first-order decomposition at 325°C? If the initial pressure of iodoethane is 894 torr at 245°C, what is the pressure of iodoethane after three half-lives?

Well, apparently the second part of the question has nothing to do with the first part...

I get `k = 1.42 xx 10^(-5)` at `325^@ "C"`, and that `1/8` of the pressure of iodoethane remains after 3 half-lives.

Well, for this I would make an Arrhenius plot.

`ul(k("s"^(-1))" "" "" "T("K"))`
`7.2 xx 10^(-4)" "" "660`
`1.7 xx 10^(-2)" "" "720`

We would plot `lnk` vs. `1//T` to get:

where:

`overbrace(ln k)^(y) = overbrace(-E_a/R)^(m) overbrace(1/T)^(x) + overbrace(ln A)^(b)`

and `E_a` is the activation energy in `"kJ/mol"`, `R` is the universal gas constant in `"kJ/mol"cdot"K"`, and `T` is temperature in `"K"`.

`"slope" = -"25041 K"^(-1) = -E_a/R`

`"y-int" = 30.704 = ln A`

We would find:

`ul"Activation energy"`

`E_a = -"slope"cdotR`

`= -(-"25041 K"^(-1)) cdot "0.008314472 kJ/mol"cdot"K"`

`=` `"208.20 kJ/mol"`

`ul"Frequency factor"`

`A = e^(30.704) "s"^(-1) = 2.161 xx 10^13 "s"^(-1)`

Therefore, at `325^@ "C"`, assuming

• the same frequency factor and activation energy
• that the relationship between `lnk` and `1/T` remains linear in the expanded temperature range

the rate constant at `325^@ "C"` is:

`color(blue)(k_(598)) = 2.161 xx 10^13 "s"^(-1) cdot e^(-("208.20 kJ/mol")//("0.008314472 J/mol"cdot"K"//"598.15 K"))`

`= color(blue)(1.42 xx 10^(-5) "s"^(-1))`

Of course, we don't need to use this for anything... because we are asked for the partial pressure of iodoethane left at `245^@ "C"`.

We know that three half-lives have passed, and that the half-life is independent of starting concentration, just like radioactive decay.

Therefore, `(1/2)^(3) = 1/8` of the starting iodoethane is left after 3 half-lives, so that:

`color(blue)(P_(C_2H_5I)) = 1/8("894 torr") = color(blue)("112 torr")`

is the remaining partial pressure of iodoethane.