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Topic: Kinetics  (Read 6552 times)

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integral0

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Kinetics
« on: August 01, 2005, 01:37:40 PM »
The decomposition of dimethyl ether, (CH3)2O, at 510 deg C is a first-order process with a rate constant of 6.8 x 10^-4 s^-1.

(CH3)2O (g) ---> CH4(g) + H2(g) + CO(g)

If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420s?

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I think you have to use

ln[(CH3)2O]t = -kt + ln[(CH3)2O]0

I tried plugging in t, k, and the initial pressure but i'm not getting 51 torr as the final answer even after I take e^x.  

What am I doing wrong?

Offline sdekivit

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Re:Kinetics
« Reply #1 on: August 01, 2005, 01:55:00 PM »
state -d[(CH3)2O]/dt = k * [(CH3)2O]

then solve this differential equation. You can then use for the partial pressure  p(i) = c(i)RT and substitute c.

You then get an equation for the partial pressure in the time t and when you this properly , the answer will be 51,4 torr. If you still don't get it, I'll help you further.
« Last Edit: August 01, 2005, 02:14:35 PM by sdekivit »

integral0

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Re:Kinetics
« Reply #2 on: August 01, 2005, 02:16:50 PM »
thanks for the help.

Actually, I found out that my original equation was right...I just was calculating it wrong.  Thanks anyways!

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