Kinetics problem

[hermanntrude]

This is one that i causing me some serious greif. I wrote it on the board in class, thinking it was simple. Unfortunately I didn't look at the last part of the question. Now i'm stumped and my students are feeling crappy because if their teacher can't do it then how are they supposed to?

 

here goes:

 

Pure DTBP at 147°C and 975 mmHg decomposes to acetone and ethane in a reaction flask at constant volume and constant temperature. the rate constant is [math]8.66 * 10^-3 min^-1[/math].

 

[ce] C8H18O2(g) -> 2C3H6O(g) + C2H6(g) [/ce]

 

a) calculate the hlaf-life of the reaction. I've done this using t½ = ln2/k, which gives 80.04 min

b) calculate the pressure of DTBP after 95.0 minutes. Ive done this using the integrated rate law for a first-order reaction, subsituting partial pressures for concentrations. This gives 428 mmHg

c) Calculate the total pressure of gas in the reaction vessel after 95.0 minutes. I've done this too, using the relationship between the rates of appearance and disappearance of each species. This gives 2068mmHg.

 

The first three parts are all checked against the published answers and are correct.

 

part d) is giving me trouble:

 

d) calculate the time required to produce a total pressure of 1225 mmHg.

 

I'm totally flummoxed. I have pages and pages of scribbles none of which give me the right answer (which should be 15.8 minutes).

 

I'm thinking graphing in excel would probably work but it's not what's expected, i'm sure. I need a simple method which can be presented in class as the right method, but i'm going crazy.

[NeonBlack]

If you can do part c, it looks like part d is almost the exact same thing. Have you considered the possibility that the textbook (or wherever you got the problem) contains a mistake? It's certainly not unheard of.

[hermanntrude]

This is from a worksheet from another instructor. He has been known to be wrong before but only once. I, however get stuff wrong all the time

[D H]

Assume the DTBP and decomposition products are ideal gases. The quantity (# moles) of the decomposition products is three times the quantity of DTBP that has decayed. Since the decomposition occurs under constant volume/temperature conditions, the partial pressure of the decomposition products will be 3 times the decrease in DTBP partial pressure.

 

Denote [math]P_0[/math] as the pressure at time t=0 (pure DTBP). The partial pressure of the DTBP at some later time t will be

 

[math]P_{DTBP}(t) = P_0e^{-kt}[/math]

 

The partial pressure of the decomposition products at that time will be

 

[math]P_{decomp}(t) = 3(P_0-P_{DTBP}(t)) = 3P_0(1-e^{-kt})[/math]

 

The total pressure is thus

 

[math]P(t) = P_{DTBP}(t)+P_{decomp}(t) = P_0(3-2e^{-kt})[/math]

 

You want the inverse function. Solving for t given P and P₀,

 

[math]t= -\,\ln\left((3-P/P_0)/2\right)/k[/math]

 

Using P₀=975 mmHg, P=1225 mmHg, and k=8.66e-3/min yields t=15.84 minutes.

[hermanntrude]

my math is a bit rusty. This is kind of the route i was taking, but i was screwing it up.

 

It'll take me a while but i think I can present that to my students now. it;s way beyond anything they'd get in an exam but interesting nevertheless. Thanks. Rep on the way.