## gibbs and temperature, 9.67 6th ed

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

shaunajava2e
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

### gibbs and temperature, 9.67 6th ed

I'm a little confused, for number 9.67 in the 6th edition it says:
Assume that deltaH and deltaS are independent of temp and use data from appendix 2a to calculate deltaG for each of the following reactions at 80C. Over what temperature range will each of the reactions be spontaneous under standard conditions?
b) CaC2 (s) + 2HCl (aq) --> CaCl2 (aq) + C2H2 (g)

In the second half of the question where it asks for the temp range, the solution manual says that since delta H is negative and delta S is positive, using the equation delta G= delta H - T*delta S, "the reaction will be spontaneous at all temperatures."

Are we supposed to assume that temperature can only be positive????? Because if T is less than 0, then the reaction won't be spontaneous anymore...
Thanks!!

Edward Xie 2E
Posts: 63
Joined: Fri Sep 28, 2018 12:23 am

### Re: gibbs and temperature, 9.67 6th ed

If T is negative, that means it goes below 0 K, which is impossible as you cannot go below absolute zero. So yes, temperature (at least in Kelvins) will always be positive.

Ethan Yi 1K
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Joined: Fri Sep 28, 2018 12:28 am

### Re: gibbs and temperature, 9.67 6th ed

if T is 0 it is at absolute zero, which is impossible, and so yes, T has to be greater than 0K.

Max Hayama 4K
Posts: 63
Joined: Fri Sep 28, 2018 12:16 am

### Re: gibbs and temperature, 9.67 6th ed

Because temperature is measured in Kelvin, it is not possible to have a value below 0, since we call the lowest possible temperature absolute 0. Therefore, if deltaH is negative and deltaS is positive, it will not matter what temperature the reaction is at. It will always be spontaneous.

katie_sutton1B
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Joined: Fri Sep 28, 2018 12:17 am

### Re: gibbs and temperature, 9.67 6th ed

In the seventh edition, it says to set delta G = to zero and solve for T to find the range of temperature using deltaG=delatH - TdeltaS. Any number from 0 to the found T value will be a spontaneous reaction.

shaunajava2e
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

thanks!