## Textbook Problem 4J.17

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

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

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

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

### Textbook Problem 4J.17

Assume that Δ H ° and Δ S ° are independent of temperature and use data in Appendix 2A to calculate Δ G ° for each of the following reactions at 80. ° C . Over what temperature range will each reaction be spontaneous under standard conditions? (a) B 2 O 3 (s ) + 6 HF (g ) → 2 BF 3 (g ) + 3 H 2 O (l ) (b) CaC 2 (s ) + 2 HCl (aq ) → CaCl 2 (aq ) + C 2 H 2 (g ) (c) C ( s , graphite ) → C ( s , diamond )

I thought in order to solve this problem, I would have to first use the appendix formation values to calculate the $\Delta$S and $\Delta$H of the equation, and then use the equation $\Delta G$ is equal to the $\Delta H$ - T times $\Delta S$. However, the answer I'm getting is not right. Could someone provide a little guidance on where I could be going wrong?

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

### Re: Textbook Problem 4J.17

a.
deltaH=2deltaH(BF3,g)+3deltaH(H2O,l)-(deltaH(B2O3)+6deltaH(HF,g))
=2(-1137kJ/mol)+3(-285.83kJ/mol)-((-1272.8kJ/mol)+6(-271.1kJ/mol))
=-232.1kJ/mol
deltaS=2Sm(BF3,g)+3Sm(H2O,l)-(Sm(B2O3,s)+6Sm(HF,g)
=2(254.12J/K.mol)+3(69.91J/K.mol)-(53.97J/K.mol+6(173.78J/K.mol))
=-378.68J/K.mol
deltaG=-232.1J/K.mol-(353K)(-378.68J/K.mol)/(1000J/kJ)
=-98.42kJ/mol

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

### Re: Textbook Problem 4J.17

Morgan Carrington 2H wrote:Assume that Δ H ° and Δ S ° are independent of temperature and use data in Appendix 2A to calculate Δ G ° for each of the following reactions at 80. ° C . Over what temperature range will each reaction be spontaneous under standard conditions? (a) B 2 O 3 (s ) + 6 HF (g ) → 2 BF 3 (g ) + 3 H 2 O (l ) (b) CaC 2 (s ) + 2 HCl (aq ) → CaCl 2 (aq ) + C 2 H 2 (g ) (c) C ( s , graphite ) → C ( s , diamond )

I thought in order to solve this problem, I would have to first use the appendix formation values to calculate the $\Delta$S and $\Delta$H of the equation, and then use the equation $\Delta G$ is equal to the $\Delta H$ - T times $\Delta S$. However, the answer I'm getting is not right. Could someone provide a little guidance on where I could be going wrong?

Can someone also explain how you are able to decide at what temperature the reactions are spontaneous? Especially how to understand this when it is spontaneous at all temperatures.

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