5.29 calculating EQ concentration

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705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

5.29 calculating EQ concentration

Postby 705121606 » Sun Jan 19, 2020 12:27 am

"In an experiment, 0.020 mol No2 was introduced into a flask with volume 1.00L and the reaction 2NO2(g) <---> N2O4(g) was allowed to come to equilibrium at 298k
A) using the information in table 5E.2, calculate the EQ concentrations of the two gases"

from the table 5E.2, the k value for the reverse reaction is 6.1x10^23 so when we set up the ice box and set it equal to k it should be 1/(6.1x10^23) right? The solutions manual is saying 1/(6.1x10^-3) I am confused as where they got that value.

Jielena_Bragasin2G
Posts: 104
Joined: Sat Aug 24, 2019 12:18 am
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Re: 5.29 calculating EQ concentration

Postby Jielena_Bragasin2G » Sun Jan 19, 2020 11:02 am

The solution manual sometimes has flaws. You should ask a TA about this if you are still concerned about the real answer.


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