11.19 and 11.83

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

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

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

Dahriel Aron 3A
Posts: 60
Joined: Fri Sep 25, 2015 3:00 am

11.19 and 11.83

Hello,
both 11.19 and 11.83 ask to calculate the equilibrium constant k for chemical equations, however 19 just simply asks to calculat k at 25 degrees celsius and 83 asks to calculate k at 25 and at 150 degrees celsius. However, in 83 to solve this problem it uses delta G= delta h - T delta S, and for 83 and then plugs that value in for delta G in the equation delta G = -RTln k, while 19 only uses the equation delta G = -rtlnk. I was wondering why these 2 siimilar problems use different methods to solve for k, and if its just because 83 asks for k at 2 different temps while 19 only asks for k at one temp?

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 820 times

Re: 11.19 and 11.83

Since you are no longer at standard conditions, you do not know the delta G value in 83. Since delta H and delta S are relatively constant over temperature ranges, you can use those to solve for delta G at 150 degress. Then you can use the delta G= -RTlnK equation to solve for K.

haleygamboa2E
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

Re: 11.19 and 11.83

Do you still need to use deltaH and deltaS values to find the equilibrium constant if the initial temperature is 298K?

Helen Shi 1J
Posts: 78
Joined: Sat Jul 22, 2017 3:00 am

Re: 11.19 and 11.83

How do you solve for 11.19?