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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: mholds on March 29, 2015, 07:02:56 PM

Title: Kinetics: Temperature Dependence of a First Order Reaction.
Post by: mholds on March 29, 2015, 07:02:56 PM

Hi,

I'm having trouble with the following homework problem:
The first-order rate constant for the gas phase decomposition of dimethyl ether is 3.2 x 10–4 s–1 at 450 °C. The reaction is carried out in a constant volume container. Initially only the reactant, dimethyl ether, is present and the pressure is 0.325 atm. What is the pressure (in atm) of the reactant after 499 seconds? Assume ideal-gas behaviour.

CH₃OCH₃ --> CH₄ + H₂ + CO₂

I'm confused about how to relate pressure with the equation I have in my notes for temperature dependent rate constants k = Ae^(-Ea/RT). Am I supposed to use the ideal gas law somehow?

Attempt:
Pressure at t=0 is 0.325 atm for CH₃OCH₃, 0 atm for products

Pressure at t = 499s is 0.325-x for dimethyl ether and x atm for products

Total pressure at 499s is P_total = 0.325 + 2x

But now I'm stuck because I don't know what to do with this expression for pressure.



Thanks.

Title: Re: Kinetics: Temperature Dependence of a First Order Reaction.
Post by: mjc123 on March 30, 2015, 05:47:34 AM

First, this is not a question about temperature dependence. Everything happens at the same temperature, 450°C.
"Ideal gas behaviour", in this context, means only that the pressure is exactly proportional to the number of moles. You don't need to do any calculations with the gas equation.
What is the integrated rate law for a first-order reaction?
And note that the question (if you have copied it correctly) only asks for the pressure of the reactant.

Title: Re: Kinetics: Temperature Dependence of a First Order Reaction.
Post by: mholds on March 31, 2015, 04:10:43 PM

Thanks for your reply.

The question is copied directly from the homework problem. I'm still having trouble comprehending how I will be able to solve for the pressure of the reactant.

Thank you for pointing out that the question asks for the pressure of the reactant -- that was a misunderstanding on my part. The integrated rate law for the first order reaction is [A] = [A]₀e^-kt. My next attempt will be to express pressure in terms of moles and solve for the moles of the reactant.

Am I on the right track with my thinking?

Title: Re: Kinetics: Temperature Dependence of a First Order Reaction.
Post by: magician4 on April 01, 2015, 05:07:38 AM

Quote

My next attempt will be to express pressure in terms of moles and solve for the moles of the reactant.

Am I on the right track with my thinking?

in my opinion you listed one of the eqtns. you'd need, and correctly at that.

However, I can't see how from this one alone you'll get to the where you wanna be..

hint:
try to include a second eqtn. , the ideal gas law.
p*V = n * R * T  :rarrow: n/V = p/RT

identify what n/V might be, and introduce this into your first eqtn.
... and solve for [A](t=499 secnds)


regards

Ingo

Title: Re: Kinetics: Temperature Dependence of a First Order Reaction.
Post by: mjc123 on April 01, 2015, 06:23:26 AM

It's simpler than that. As the reaction is first order, you can use the equation with any measure of the amount of A - number of moles, concentration, pressure - with the same rate constant. (Ideal gas behaviour means that all these are directly proportional to each other.) So  you can write P_A = P_A0e^-kt. Simples.