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?

The decomposition of iodoethane in the gas phase proceeds according to the following equation: C_2H_5I(g) → C_2H_4(g) + HI(g) At 660.K, k = 7.2 x 10–4s–1; at 720.K, k = 1.7 x 10–2s–1.

May 25, 2018

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

I get $k = 1.42 \times {10}^{- 5}$ at ${325}^{\circ} \text{C}$, and that $\frac{1}{8}$ of the pressure of iodoethane remains after 3 half-lives.

Well, for this I would make an Arrhenius plot.

$\underline{k \left(\text{s"^(-1))" "" "" "T("K}\right)}$
$7.2 \times {10}^{- 4} \text{ "" } 660$
$1.7 \times {10}^{- 2} \text{ "" } 720$

We would plot $\ln k$ vs. $1 / T$ to get: where:

${\overbrace{\ln k}}^{y} = {\overbrace{- {E}_{a} / R}}^{m} {\overbrace{\frac{1}{T}}}^{x} + {\overbrace{\ln A}}^{b}$

and ${E}_{a}$ is the activation energy in $\text{kJ/mol}$, $R$ is the universal gas constant in $\text{kJ/mol"cdot"K}$, and $T$ is temperature in $\text{K}$.

${\text{slope" = -"25041 K}}^{- 1} = - {E}_{a} / R$

$\text{y-int} = 30.704 = \ln A$

We would find:

$\underline{\text{Activation energy}}$

${E}_{a} = - \text{slope} \cdot R$

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

$=$ $\text{208.20 kJ/mol}$

$\underline{\text{Frequency factor}}$

$A = {e}^{30.704} {\text{s"^(-1) = 2.161 xx 10^13 "s}}^{- 1}$

Therefore, at ${325}^{\circ} \text{C}$, assuming

• the same frequency factor and activation energy
• that the relationship between $\ln k$ and $\frac{1}{T}$ remains linear in the expanded temperature range

the rate constant at ${325}^{\circ} \text{C}$ 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"))

$= \textcolor{b l u e}{1.42 \times {10}^{- 5} {\text{s}}^{- 1}}$

Of course, we don't need to use this for anything... because we are asked for the partial pressure of iodoethane left at ${245}^{\circ} \text{C}$.

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

Therefore, ${\left(\frac{1}{2}\right)}^{3} = \frac{1}{8}$ of the starting iodoethane is left after 3 half-lives, so that:

$\textcolor{b l u e}{{P}_{{C}_{2} {H}_{5} I}} = \frac{1}{8} \left(\text{894 torr") = color(blue)("112 torr}\right)$

is the remaining partial pressure of iodoethane.