## 5G. 13

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

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

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

ayushibanerjee06
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### 5G. 13

Can someone find where my error is?
The question is: (a) Calculate the reaction Gibbs free energy of I2(g) -> 2I(g) at 1200. K (K 5 6.8) when the partial pressures of I2 and I are 0.13 bar and 0.98 bar, respectively. (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

I used the Delta G = G +RTlnQ equation with G being 19.33 kJ, R is 8.314, T is 1200K and Q is 7.3877. I keep getting Delta G = 39281.76.

Selena Yu 1H
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### Re: 5G. 13

For G, I got -19.124 kJ instead of 19.33 kJ. You would use the equation G = -RTlnK to find what G is.

ayushibanerjee06
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### Re: 5G. 13

Selena Yu 1H wrote:For G, I got -19.124 kJ instead of 19.33 kJ. You would use the equation G = -RTlnK to find what G is.

So, you use G=-RTlnK to find the standard GFE and not the ones in the back of the book?

Labiba Sardar 2A
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### Re: 5G. 13

I'm confused on what value of K you're supposed to use for delta G = -RTlnK. The problem gives you the partial pressures, so are you supposed to find Kp? But at the same time, it says that K = 6.8, so which one are you supposed to use?

Luyan Zhang - 2D
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Joined: Sat Jul 20, 2019 12:16 am

### Re: 5G. 13

ayushibanerjee06 wrote:Can someone find where my error is?
The question is: (a) Calculate the reaction Gibbs free energy of I2(g) -> 2I(g) at 1200. K (K 5 6.8) when the partial pressures of I2 and I are 0.13 bar and 0.98 bar, respectively. (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

I used the Delta G = G +RTlnQ equation with G being 19.33 kJ, R is 8.314, T is 1200K and Q is 7.3877. I keep getting Delta G = 39281.76.

Your approach is right. Check the signs.

Selena Yu 1H
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### Re: 5G. 13

ayushibanerjee06 wrote:
Selena Yu 1H wrote:For G, I got -19.124 kJ instead of 19.33 kJ. You would use the equation G = -RTlnK to find what G is.

So, you use G=-RTlnK to find the standard GFE and not the ones in the back of the book?

Yup! You would use the ones in the back of the book if they didn't give you equilibrium and values to find Q.

Pegah Nasseri 1K
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Joined: Wed Feb 27, 2019 12:15 am

### Re: 5G. 13

You would use the equation $\Delta G = \Delta G + RTlnQ = -RTlnK + RTlnQ$. Substitute the 6.8 for the K value in the equation and use all the partial pressures given to you to calculate Q (partial pressure of the products over the partial pressure of the reactants). You use the Delta G=-RTlnK part because the K is given to you in the problem.