HW question 5G.15

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HW question 5G.15

Postby kennedyp » Tue Feb 25, 2020 4:09 pm

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!!!

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Re: HW question 5G.15

Postby AKatukota » Tue Feb 25, 2020 6:10 pm

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?

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Re: HW question 5G.15

Postby AniP_2D » Thu Feb 27, 2020 9:41 pm

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.

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Re: HW question 5G.15

Postby 705121606 » Fri Feb 28, 2020 2:24 pm

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

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Re: HW question 5G.15

Postby Uisa_Manumaleuna_3E » Fri Feb 28, 2020 3:18 pm

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!

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