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### 5G.13

Posted: **Tue Feb 18, 2020 7:48 pm**

by **Luyan Zhang - 2D**

(a) Calculate the reaction Gibbs free energy of I2(g) -->2I(g) at 1200. K (K = 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.

Can someone go through this problem? Thanks!

### Re: 5G.13

Posted: **Tue Feb 18, 2020 8:44 pm**

by **Alexis Webb 2B**

The equation for delta G of the reaction equals -RTlnK (standard delta G for rxn) + RTlnQ. The problem gives you K and T, and the pressures it gives you can be used to solve for Q, which equals the concentration of products over concentration of reactants at a given time. Plug in all the values you have for the variables and solve. If your answer is positive, then the reaction is spontaneous to produce your reactant.

### Re: 5G.13

Posted: **Tue Feb 18, 2020 8:57 pm**

by **Esha Chawla 2E**

Luyan Zhang - 2D wrote:(a) Calculate the reaction Gibbs free energy of I2(g) -->2I(g) at 1200. K (K = 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.

Can someone go through this problem? Thanks!

Given the partial pressure, you can calculate the Q value. Then, use this value and plug into the equation delta G = delta G naught + RTlin(Q). You can calculate delta G naught using -RTln(K), since you are given the K values. Plug everything into the equation and solve for delta G.

If delta G is positive, it has a tendency to form reactants. If delta G is negative, then it is likely to form products. If delta G is 0, it is at equilibrium.

### Re: 5G.13

Posted: **Wed Feb 19, 2020 12:52 pm**

by **Megan Vu 1J**

You are able to first calculate the Q value of the equation.

Then, once getting this, you can use the delta G = delta G knot + RTlnQ, but since we do not know the delta G knot, we can replace it with -RTlnK.

Thus, the whole equation would be delta G = -RTlnK + RTlnQ.

You are able to insert all of the information in order to find the delta G of the whole chemical equation.

### Re: 5G.13

Posted: **Wed Feb 19, 2020 11:38 pm**

by **DesireBrown1J**

This is what I plugged into the equation G = -RTlnK + RTlnQ

delta G=(-(8.314 J*K^{-1}*mol^{-1})(1200K)(ln(6.8))-((8.314J*K^{-1}*mol^{-1})(1200K)(ln(7.38)))

However, my answer was completely off and was not close to the solutions manual.

### Re: 5G.13

Posted: **Thu Feb 27, 2020 5:52 pm**

by **Sjeffrey_1C**

The problem is that you are subtracting. Add delta G naught and RTlnQ and you will get the correct answer!

### Re: 5G.13

Posted: **Thu Feb 27, 2020 11:07 pm**

by **Bryan Chen 1H**

you should use delta G = -RTlnK + RTlnQ.

### Re: 5G.13

Posted: **Sun Mar 01, 2020 11:02 pm**

by **DesireBrown1J**

Thank you! I didn't realize I made that small mistake!

### Re: 5G.13

Posted: **Wed Mar 04, 2020 11:20 pm**

by **Morgan Carrington 2H**

Megan Vu 1J wrote:You are able to first calculate the Q value of the equation.

Then, once getting this, you can use the delta G = delta G knot + RTlnQ, but since we do not know the delta G knot, we can replace it with -RTlnK.

Thus, the whole equation would be delta G = -RTlnK + RTlnQ.

You are able to insert all of the information in order to find the delta G of the whole chemical equation.

Why do you have to first find the

knot first and then plug it into the other equation?