Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: integral0 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?
---------------------------------------
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?
-
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.
-
thanks for the help.
Actually, I found out that my original equation was right...I just was calculating it wrong. Thanks anyways!