## HW question 5G.15

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

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

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

kennedyp
Posts: 56
Joined: Tue Nov 13, 2018 12:18 am

### HW question 5G.15

I keep getting the wrong answer for this homework problem and I don't know what I am doing wrong. ( 5G.15 (a) Calculate the reaction Gibbs free energy of N2(g) 1 3 H2(g) S 2 NH3(g) when the partial pressures of N2, H2, and NH3 are 4.2 bar, 1.8 bar, and 21 bar, respectively, and the tempera- ture is 400. K. For this reaction, K 5 41 at 400. K) I keep getting positive 7 kJ/mol and the answer at the back of the book is -27 kJ/mol. I googled and the book made a mistake (the actual answer is -2.7 kJ/mol) but I'm still not even getting that! Pls help!!!

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

### Re: HW question 5G.15

I also don't know how to do this? Like how would we incorporate the pressures given into this? Do we use delta G = -RTlnK?

AniP_2D
Posts: 95
Joined: Sat Aug 17, 2019 12:17 am

### Re: HW question 5G.15

You would first use deltaG(naught)=-RTlnk and then you would use deltaG(reaction)=deltaG(naught)+RTlnQ. Since the question is asking for the deltaG(reaction), you would have to use these 2 equations.

705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

### Re: HW question 5G.15

Since you are given the k value and the partial pressures to find Q, you would use deltaG=-RTlnk + RTlnQ and this should give you the answers in joules

Uisa_Manumaleuna_3E
Posts: 60
Joined: Wed Sep 21, 2016 2:56 pm

### Re: HW question 5G.15

So it took me a while to work this one out but you basically do it in 3 steps
1. use the partial pressures to calculate the reaction quotient, Q
2. use the equation delta G naught = - RTlnK and substitute it for delta g naught in the equation delta g = delta g naught + RTlnQ, so you get:
delta G = (-RTlnK) + RTlnkQ
3. plug in your Q from step 1, the given K and temp into the formula from step 2
and i got the answer that way!