Please help me answer this

[drpduck]

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Ok so this is off Topscore #1, and I cannot figure it out:

Given the standard enthalpy of formations of NO is 90.25 kj/mole calculate the free energy change for the following reaction at 25 degrees C

N2 + O2 -> 2NO

Standard Entropies
------------------
N2 192kj/mol
NO 211kj/mol
O2 205kj/mol


The answer is: 180.5-(298)(25)(1x10/\-3)

Ok now I know you use G=H+T(delta)S, but I still can't figure it out. I get that the Enthalpy is 90.25 x 2=180.5 and the T is 25+273=298, but the rest I have no clue.

 

[fleisch]

Hey there-
Ok, regarding the problem..
You are solving for H, so your equation is: H=G-TdeltaS
G= 2*90.25 = 180.5
T=25+273= 298
S= (S of Products - S of Reactants) = (2*211) - (192 +205) = 25
Then it is all multiplied by (1*10^-3) since the units need to be in kJ (as opposed to J)

Hope that helps!🙂
I am taking my test in two days, so I feel like I should be on top of this!🙂

 

[Literati81]

I'm sure someone will probably answer you before me. But, what the heck...

So, you are correct that you need to use the delta G=deltaH -T (delta S) equation. The remaining numbers are probably delta S and a conversion factor.

For example, if you calculate delta S, you would get 2(211) -(192+205)=25. So there's the reason behind one number. However the (1X 10^-3) is a bit confusing as the majority of units are in kJ. Just to clarify, the negative sign was not a typo? If it was a typo, then the answer would make sense as Topscore wanted you to convert from kJ-->J.

Or...maybe they wanted you to calculate the number in microJoules? Doubtful.

Yeah...this feedback probably was not as beneficial as you may have hoped. Good luck!

 

[drpduck]

grrr, I should have known that, stupid mistakes! Thanks a lot!

 

[Literati81]

So, wait. The 10^-3 was a mistake? Or were you replying to the previous post? Sorry, DAT nerves make me hyper-curious.

 

[drpduck]

Just a conversion of kJ to J

 

[sinned]

wait, the answer of 25 is in KJ, so if you wanted J...wouldn't you have to times it by 10^3, not 10^-3 ???

 

[Literati81]

ditto.

 

[ScottW3]

Sorry, I'm a little confused. Its asking for the free energy change right? Why would you be solving for delta H and not for G? I've been studying too long and my brain is melting I think.

 

[cash]

Okay, this is what I came up with.
The equation you need to use is G=H-T(delta)S. The question gives you enthalpy for one mole of NO. In the reaction, two moles are produced so multiply the enthalpy they give you by 2. You will get H=180.5. In the little table, they give you the entropies. You need to solve for delta S by remembering: (delta)S= sum of products - sum of reactants). When you calculate this you will get 25. The temperature is 298K. Now just plug in your numbers into the equation.

(delta)G = 180.5 - (298)(25).

The other part is just a conversion factor.

Let me know if this helps.

Brandon

 

[ScottW3]

Thanks Cash, that does help a lot. I honestly think I need to get a day of rest in. I'm taking my DAT on Aug. 13 (I know, Friday the 13th 🙁 ) and have been studying since the end of june for at least 6 hours, 5-6 days a week. I think my brain is dying. Anyway, thanks again.

 

[cash]

Good Luck on the DAT. Let us know how it goes. We all get those brain farts once in a while. I am studying for the DAT now and its tough trying to study and get graduate school work done at the same time. Anywhos, good luck to ya.

Brandon