5.39 HW

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rachelle1K
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

5.39 HW

Postby rachelle1K » Mon Jan 20, 2020 8:20 pm

5.39 states,
In an experiment, 0.020 mol NO2 was introduced into a flask of volume 1.00 L and the reaction 2 NO2(g) ∆ N2O4(g) was allowed to come to equilibrium at 298 K. (a) Using information in Table 5E.2, calculate the equilibrium concentrations of the two gases. (b) The volume of the flask is reduced to half its original value. Calculate the new equilibrium concentrations of the gases.

I was wondering how to calculate the equilibrium concentration? I took the inverse of Kc given in the table but my answers are not adding up. Any help would be appreciated, thanks!

DarrenKim_1H
Posts: 123
Joined: Fri Sep 20, 2019 12:17 am
Been upvoted: 1 time

Re: 5.39 HW

Postby DarrenKim_1H » Mon Jan 20, 2020 10:32 pm

It is an ice box question, so you need to find x. From the table, the information you need to find is Kc to calculate X

Skyllar Kuppinger 1F
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am

Re: 5.39 HW

Postby Skyllar Kuppinger 1F » Tue Jan 21, 2020 11:47 pm

I'm having the same problem. In the solution manual it says that kc= 1/6.1x10^-3. But according to the table, it should definitely be 1/6.1x10^23. But it seems that we are doing the process correctly. There must be an error somewhere, perhaps the value in the table is a typo or the value in the solution manual is a typo. either way, I would not worry about it.


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