Quiz 1 Prep, #8

[Coco Hailey 2E]

For the reaction 2C (s) + 2H2 (g) --> C2H4(g)

∆Hr is 52.3kJ/mol and ∆S = -53.07J/Kmol at 298K
At what temperature will this reaction be spontaneous?

I set the ∆G to zero, and got T + ∆H/∆S which = 52300/-53.07, the calculated temperature I got was -985K. So for the reaction to be spontaneous does T > -985 K or T< -985 K???

[Diana Catapusan 1N]

This is actually kind of a trick question! It is actually NOT spontaneous at any temperature. When ∆H is positive and ∆S is negative, ∆G is never spontaneous. Hope this helps!

[Ariana de Souza 4C]

So I get that from the table. But from the math, are we supposed to assume that since we get a negative K value, it's impossible? And from that should we know that it would never be spontaneous?

And additionally, when we get values for T, is it always that when T is greater than that value, it's spontaneous for the forward reaction, and then when T is less than that value, it's spontaneous for the reverse?

[Bronson_Barretto_2C]

So I understand that since ∆H is positive and ∆S is negative= non-spontaneous ∆G, but for this question in particular do we show all this work and then say it's non-spontaneous or just write down non-spontaneous as an answer right off the bat?

[Thomas Hui 2J]

In addition to seeing that it is a trick question, you can actually derive the Temperature yourself, to prove mathematically that it is completely impossible. When I did the equation myself, I got -936K as the temperature that it needs to be, so that G will equal 0. As we know, this is impossible, as the lowest temperature possible is 0K.