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

Submitted by caseyj1155 on Tue, 07/07/2009 - 02:34

The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8 ? 10–4s–1:

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

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

how do i solve this??


You would use the integrated rate law:

ln [A] = - kt +  ln [A]0

where [A] is the concentration of the reactant at time= t, k is the rate constant, and [A]0 is the initial concentration of A (t=0), and t = time in seconds.

In your case you should be able to use the pressure in place of concentration since P is directly related to concentration (n/V) at constant Temp ( P = (n/V) RT  )