# What is the rate constant for dimethyl ether after a couple of hours?

## While heating dimethylether is decomposing according to the equation: CH3OCH3 --> CH4+ CO+ H2 Pressure is going to increase because number of moles is going to increase (closed vessel). During time pressure is given in the next way: Time: 0 hours-60 000Pa, After 0,6hours-106 123Pa, after 0,7 hours-119 188Pa, after 1,0 hour 134 653Pa. In the beginning only dimethyl ether was present, by knowing that, find the rate constant for this first order reaction. Solution: 0,972/h

Jan 4, 2017

Since the data wasn't linear, I got around $\text{0.95/hr}$ to $\text{0.98/hr}$. Basically, we were supposed to use the integrated rate law, and find $k$ as the slope.

You're basically given a table of time vs. pressure, of certain "trials" of the same process where the experimenter checks the pressure after a given amount of time passes:

color(white)([(color(black)("Trial"),color(black)("time"("hr")),color(black)("Pressure" ("Pa"))),(color(black)(1),color(black)(0.000),color(black)(60000.)),(color(black)(2),color(black)(0.600),color(black)(106123)),(color(black)(3),color(black)(0.700),color(black)(119188)),(color(black)(4),color(black)(1.000),color(black)(134653))])

This is for the reaction

${\text{H"_3"C""O""CH"_3(g) -> "CH"_4(g) + "CO"(g) + "H}}_{2} \left(g\right)$,

which has a rate law assumed to be (where for non-gases, we'd use $\left[{H}_{3} C O C {H}_{3}\right]$ instead of P_(H_3COCH_3):

$\boldsymbol{r \left(t\right) = k {P}_{{H}_{3} C O C {H}_{3}} = - \frac{\Delta {P}_{{H}_{3} C O C {H}_{3}}}{\Delta t}}$

$= \frac{\Delta {P}_{C {H}_{4}}}{\Delta t} = \frac{\Delta {P}_{C O}}{\Delta t} = \frac{\Delta {P}_{{H}_{2}}}{\Delta t}$

The rate for consumption of dimethyl ether is numerically negative (cancelled out by the negative sign out front), and the rates of production for the remaining gases are all numerically positive.

You can kind of predict that we're going to have to consider this data as a plot, so let's derive the integrated rate law for this process, for the rate $\frac{\mathrm{dP}}{\mathrm{dt}}$, where $d = \Delta$ for incrementally small changes in pressure.

(If you don't know Calculus, you can feel free to skip past this to the result, but some textbooks hand-wave the math because not every general chemistry student knows Calculus.)

Since the pressure increases, $\frac{\mathrm{dP}}{\mathrm{dt}} > 0$:

$k P = \frac{\mathrm{dP}}{\mathrm{dt}}$

Separation of variables gives:

$k \mathrm{dt} = \frac{1}{P} \mathrm{dP}$

The left side is integrated from $t = 0$ to some time $t$, and the right side is integrated from the initial pressure ${P}_{i} = {P}_{{H}_{3} C O C {H}_{3}}$ to the current pressure $P$ (which isn't necessarily the total pressure of the products):

${\int}_{0}^{t} k \mathrm{dt} = {\int}_{{P}_{i}}^{P} \frac{1}{P} \mathrm{dP}$

The integral of $\frac{1}{P}$ is $\ln P$, evaluated from ${P}_{i}$ to the current pressure $P$ (since we don't know when the reaction ended), and the integral on the left of $1$ is just $t$, evaluated from $0$ to $t$ hours:

$k t = \ln P - \ln {P}_{i}$

This gives us our integrated rate law for this process, plugging in ${P}_{i} = {P}_{{H}_{3} C O C {H}_{3}}$:

$\implies \boldsymbol{\stackrel{y}{\overbrace{\ln P}} = \stackrel{m}{\overbrace{k}} \stackrel{x}{\overbrace{t}} + \stackrel{b}{\overbrace{\ln {P}_{{H}_{3} C O C {H}_{3}}}}}$

compared to the one for general concentrations in which the initial concentration decreases instead of increasing:

$\ln \left[A\right] = - k t + \ln {\left[A\right]}_{0}$

I plotted this in Excel to get:

For some reason, whereas this graph is supposed to be linear, it turned out to taper off farther away from $t = 0$, so we should find the earliest points where it is linear (the ${R}^{2}$ would have been $0.972$).

In the integrated rate law, the y-intercept was given at $t = \text{0 hrs}$:

$\ln {P}_{i} = \ln {P}_{{H}_{3} C O C {H}_{3}} = \ln \left(60000\right)$
(since we can't take the $\ln$ of something with units, we have implicitly divided by a "standard pressure" that takes out the units)

$= 11.00$

Supposedly, the rate constant is the same for the same reaction no matter what trial we look at. So, let us choose $t = \text{0.600 hrs}$ and $P = \text{106123 Pa}$ ("trial" 2). Then:

$\ln P = \ln \left(106123\right)$

$= 11.57$

The rate constant is then found from manipulating what we now have:

$\ln \left(106123\right) = k \left(\text{0.6 hrs}\right) + \ln \left(60000\right)$

$11.572 = k \left(\text{0.6 hrs}\right) + 11.002$

$\implies \textcolor{b l u e}{k} = \frac{11.572 - 11.002}{\text{0.6 hrs}}$

$\approx$ $\textcolor{b l u e}{\text{0.95/hr}}$

If you use "trial" 3, you'd get $k = \text{0.98/hr}$ (though it shouldn't be the case if the graph was linear, since $k$ IS the slope).

(Note that it'd take a very small change in the initial pressure or any of the current pressures to alter the rate constant by a few tenths, so we may be a bit off if the textbook author used more decimal places than us behind the scenes.)