2011 Midterm Q1B


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Amanda Nguyen 2E
Posts: 19
Joined: Tue Nov 25, 2014 3:00 am

2011 Midterm Q1B

Postby Amanda Nguyen 2E » Mon Mar 07, 2016 12:25 pm

"1.00 mol of methane is combusted irreversibly at constant room temp. Calculate the work associated with this process."

Why is the change in moles -2.00? Why don't we use 1.00 mol for n?

704628249
Posts: 43
Joined: Fri Sep 25, 2015 3:00 am

Re: 2011 Midterm Q1B

Postby 704628249 » Mon Mar 07, 2016 1:46 pm

We'll work for irreversible processes is P x delta V. P delta V is also equal to delta nRT. You can calculate it from there.

Mary Anastasi
Posts: 25
Joined: Fri Sep 25, 2015 3:00 am

Re: 2011 Midterm Q1B

Postby Mary Anastasi » Mon Mar 07, 2016 1:46 pm

I think you look at the reaction itself and see that on the reactants side there are 3 moles of gas and on the products side there is 1 mole of gas (seeing that the H2O is liquid), so delta n is -2 moles.


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