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

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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

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

##### 1 Answer

Since the data wasn't linear, I got around

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

#"H"_3"C""O""CH"_3(g) -> "CH"_4(g) + "CO"(g) + "H"_2(g)# ,which has a

rate lawassumed to be (where for non-gases, we'd use#[H_3COCH_3]# instead of#P_(H_3COCH_3# ):

#bb(r(t) = kP_(H_3COCH_3) = -(DeltaP_(H_3COCH_3))/(Deltat))#

#= (DeltaP_(CH_4))/(Deltat) = (DeltaP_(CO))/(Deltat) = (DeltaP_(H_2))/(Deltat)# The rate for

consumptionof dimethyl ether is numerically negative (cancelled out by the negative sign out front), and the rates ofproductionfor 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

(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**,

#kP = (dP)/(dt)#

Separation of variables gives:

#kdt = 1/(P)dP#

The left side is integrated from *isn't* necessarily the total pressure of the products):

#int_(0)^(t) kdt = int_(P_i)^(P)1/PdP#

The integral of *current* pressure

#kt = lnP - lnP_i#

This gives us our **integrated rate law** for *this* process, plugging in

#=> bb(stackrel(y)overbrace(lnP) = stackrel(m)overbrace(k)stackrel(x)overbrace(t) + stackrel(b)overbrace(lnP_(H_3COCH_3)))#

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

#ln[A] = -kt + ln[A]_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 *is* linear (the

In the integrated rate law, the y-intercept was given at

#lnP_i = lnP_(H_3COCH_3) = ln(60000)#

(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

#lnP = ln(106123)#

#= 11.57#

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

#ln(106123) = k("0.6 hrs") + ln(60000)#

#11.572 = k("0.6 hrs") + 11.002#

#=> color(blue)(k) = (11.572 - 11.002)/("0.6 hrs")#

#~~# #color(blue)("0.95/hr")#

If you use "trial" 3, you'd get

(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.)