What Temperature Makes the Reaction N2O4 --> 2NO2 Spontaneous?

[CasanovaFrankenstein]

Alright,

This is a followup to my previous post: thanks for the advice on the two questions. I got the HH one. That was quite easy.

However, I'm not getting this one correctly:

N2O4 --> 2NO2
Delta G is 2.8kJ (under standard conditions). At what temp will it become spontaneous?

So. I know Gibb's is: Delta G = Delta H - T(delta S)

The delta H for N2O4 is 9.66 and NO2 is 33.85
The delta S for N2O4 is 304.3 and NO2 is 240.46
This is per my book

I plug those into the Delta H and Delta S equations:
2(33.85) - 9.66 = 58
2(240.46) - 304.3 = 176.9

T = Delta H / Delta S
This comes out to 0.328 K

That can't be right! What did I do wrong?

 

[GCT]

You should be able to find a derivation which includes the $$\Delta G^{0},~and~ \Delta G$$ The reaction is spontaneous when $$\Delta G$$ is zero. Try to find the simplest equation.

 

[ravenprp]

Welcome back! It's great to see you back with a new question. It looks like you're working on a thermodynamics problem involving the reaction N2O4 --> 2NO2. To determine at what temperature the reaction will become spontaneous, you need to use the equation Delta G = Delta H - T(delta S), just as you did. However, it seems like you may have made a mistake in your calculations. When plugging in the values for delta H and delta S, you need to use the values for the products (2NO2) minus the values for the reactant (N2O4). This will give you a delta G value of -58 kJ/mol. To solve for T, you would divide this value by the delta S value of 176.9 J/mol*K, giving you a temperature of approximately 328 K. This makes more sense because at higher temperatures, the reaction becomes more spontaneous. I hope this helps clarify things for you. Keep up the good work!