## 5G.15

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

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

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

ayushibanerjee06
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### 5G.15

For 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 temperature is 400. K. For this reaction, K 5 41 at 400. K. (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium, I keep getting my answer to be -23.29 kJ even though the answer is -27 kJ/mol. My standard GFE is -32.9 KJ/mol (times 3), R is 8.314, T is 400 K, and Q is 18.004.

Selena Yu 1H
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### Re: 5G.15

For 5G.15, for the standard GFE I got -12.35 kJ and for my final answer, I actually got -2.7 kJ/mol instead of -27 kJ/mol

ayushibanerjee06
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### Re: 5G.15

Selena Yu 1H wrote:For 5G.15, for the standard GFE I got -12.35 kJ and for my final answer, I actually got -2.7 kJ/mol instead of -27 kJ/mol

How did you get -12.35 kJ? I used the one for NH3 in the back of the book that was -32.9 kJ/mol since the standard GFE for the reactants are zero.

Hiba Alnajjar_2C
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Joined: Fri Aug 09, 2019 12:17 am

### Re: 5G.15

Selena Yu 1H wrote:For 5G.15, for the standard GFE I got -12.35 kJ and for my final answer, I actually got -2.7 kJ/mol instead of -27 kJ/mol

I also got -2.7 kJ/mol, did you happen to figure out what went wrong? Thank you!!

Selena Yu 1H
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### Re: 5G.15

Hiba Alnajjar_2C wrote:
Selena Yu 1H wrote:For 5G.15, for the standard GFE I got -12.35 kJ and for my final answer, I actually got -2.7 kJ/mol instead of -27 kJ/mol

I also got -2.7 kJ/mol, did you happen to figure out what went wrong? Thank you!!

I'm not sure I think the book just made an error but I checked with other people and they also got -2.7 kJ/mol. Hope that helps!

Selena Yu 1H
Posts: 108
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### Re: 5G.15

ayushibanerjee06 wrote:
Selena Yu 1H wrote:For 5G.15, for the standard GFE I got -12.35 kJ and for my final answer, I actually got -2.7 kJ/mol instead of -27 kJ/mol

How did you get -12.35 kJ? I used the one for NH3 in the back of the book that was -32.9 kJ/mol since the standard GFE for the reactants are zero.

For this question, you also use the equation deltaG = -RTlnK to find the standard GFE because you were given K and values to find Q.

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