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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nernst on August 03, 2009, 05:20:37 PM

Title: General Chem II - Are my answers correct?
Post by: nernst on August 03, 2009, 05:20:37 PM
1. Calculate the  :delta:S for decomposition of ozone from oxygen:
2O3(g)  :rarrow: 3O2(g)

So = 205 J/molK for O2 and 239 J/molK for O3 at 25oC.

I calculated that  :delta:S =  :delta:S = 137 J/molK because I calculated it this way:

3(205J/molK) - 2(239J/molK)


2. In the reaction, 2NO(g) + O2(g)  :rarrow: 2NO2(g)  :delta:Go = -77.4 kJ and  :delta:So= -146.5 at 251 K and 1 atm.

This reaction is product(?) favored under standard condtions at 251 K.

The standard enthalpy change  :delta:Ho for the reaction of 1.55 moles of NO(g) at this temperature would be ___?

I calculated that  :delta:H = -96.68 kJ because I calculated that:

1.55 mol NO * ((-77.4kJ)/(2mol NO)) = -59.98 =  :delta:Go

And since  :delta:G =  :delta:H -  :delta:S(T):

-59.98kJ =  :delta:H - (-.1465kJ)(251K)
Hence, :delta:H = -96.68 kJ
Title: Re: General Chem II - Are my answers correct?
Post by: plankk on August 04, 2009, 01:58:14 AM
First is correct.

About second task: remember that enthropy is inversely proportional to the number of moles.
Title: Re: General Chem II - Are my answers correct?
Post by: nernst on August 04, 2009, 07:16:06 AM
Quote from: plankk on August 04, 2009, 01:58:14 AM
First is correct.

About second task: remember that enthropy is inversely proportional to the number of moles.

does that mean i just divide 1/1.55 mol?
Title: Re: General Chem II - Are my answers correct?
Post by: MrTeo on August 04, 2009, 08:17:11 AM
Quote from: nernst on August 03, 2009, 05:20:37 PM
-59.98kJ =  :delta:H - (-.1465kJ)(251K)
Hence, :delta:H = -96.68 kJ

First of all a (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+S%5E0+%7D&hash=e47c9deab1fd4b507000d63522a0f65852aa8e3f) of -146.5 kJ/K⋅mol sounds very strange to me...
Personally I think that it's -146.5 J/K⋅mol... anyway I would suggest you to find the enthalpy variation for 2 mol and then calculate the (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+H%5E0+%7D&hash=a5c171fb37c4cd34c2888a88c45c176ee7e44f82) for 1.55 mol, but if you want you can also apply what plankk said, I just think doing this way is more straightforward...  ;) But you can't use the absolute temperature of 251 K as they ask you (notice the 0 superscript) to calculate enthalpy at 298.15 K and 1 atm (normal conditions).

Moreover I don't really understand what did you mean with that question mark after product... if you hadn't clear what it meant consider simply that a reaction with an equilibrium moved to the right at about -22º C must be esothermic ((https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+H%5E0%26gt%3B0%7D&hash=2a075a35a2bcbbcd51c4e62d8e28d26c193ae922))...

Here is the answer if you don't understand my explanation or if you simply want to check your answers:

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+G%5E0%3D%5CDelta+H%5E0-T%5CDelta+S%5E0+%7D&hash=cc272e0e84e5f4aea9eeaf1b7a74beeda0d10ffc)
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+H%5E0%3D%5CDelta+G%5E0%2BT%5CDelta+S%5E0%3D-121.08%5C%3BkJ+%5Ccdot+mol%5E%7B+-+1%7D+%7D&hash=ec3d0d5b80ef9d5277819d649032e790627a231d)
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5Cfrac%7B%7B%5CDelta+H%5E0+_%7B2mol%7D+%7D%7D%7B2%7D%3D%5Cfrac%7B%7B%5CDelta+H%5E0+_%7B1.55mol%7D+%7D%7D%7B%7B1.55%7D%7D+%7D&hash=6b3b3665d0194f9c1a9c795e7f310f618fc2cb6c)
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+H%5E0+_%7B1.55mol%7D%3D%5Cfrac%7B%7B1.55%7D%7D%7B2%7D%5CDelta+H%5E0+_%7B2mol%7D%3D93.83%5C%3BkJ+%5Ccdot+mol%5E%7B+-+1%7D+%7D&hash=ae4fd690d5799fd3492a7957257bd1776676371c)