## 5G.13

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

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

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

Luyan Zhang - 2D
Posts: 103
Joined: Sat Jul 20, 2019 12:16 am

### 5G.13

(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!

Alexis Webb 2B
Posts: 124
Joined: Thu Jul 11, 2019 12:15 am

### Re: 5G.13

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.

Esha Chawla 2E
Posts: 108
Joined: Thu Jul 25, 2019 12:17 am

### Re: 5G.13

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.

Megan Vu 1J
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

### Re: 5G.13

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.

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

### Re: 5G.13

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.

Sjeffrey_1C
Posts: 108
Joined: Wed Feb 20, 2019 12:17 am

### Re: 5G.13

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

Bryan Chen 1H
Posts: 58
Joined: Mon Jun 17, 2019 7:24 am

### Re: 5G.13

you should use delta G = -RTlnK + RTlnQ.

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

### Re: 5G.13

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

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

### Re: 5G.13

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 $\Delta G$ knot first and then plug it into the other equation?

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