## ΔGionization

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

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

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

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Abigail Menchaca_1H
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### ΔGionization

Is there a different way to calculate ΔG ionization?

Rhea Shah 2F
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Joined: Thu Jul 25, 2019 12:17 am

### Re: ΔGionization

I don't think so! It should be the same as the regular way of calculating standard gibbs free energy.

Ashley Nguyen 2L
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### Re: ΔGionization

I believe that deltaG ionization is calculated the same as normal standard gibbs free energy.

Diana A 2L
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### Re: ΔGionization

Would delta G ionization be the same for other temperatures besides 25 degrees Celsius? Let's say for example you're taking the delta G of ionization for a reaction at 30 degrees Celsius, would you use just the standard equations for delta G? I hope that question makes sense, please someone help I want to know.

Tauhid Islam- 1H
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### Re: ΔGionization

I'm pretty sure all standard state quantities are dependent on time. Gibb's free energy change is a function of temperature so at different temperatures, you would have different energies.

Diana A 2L
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### Re: ΔGionization

Tauhid Islam- 1H wrote:I'm pretty sure all standard state quantities are dependent on time. Gibb's free energy change is a function of temperature so at different temperatures, you would have different energies.

In that case, how would you calculate Gibbs Free Energy at non-standard temperature?

Ying Yan 1F
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Joined: Fri Aug 02, 2019 12:16 am

### Re: ΔGionization

I don't think there is another way, calculating for delta Gionization is the same as calculating for delta Go.

Ellen Amico 2L
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Joined: Thu Sep 19, 2019 12:16 am

### Re: ΔGionization

Nope! you can use any of the equations for calculating deltaG. I think it's just a way to label it relating to the reaction.

Diana A 2L
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### Re: ΔGionization

Ellen Amico 2L wrote:Nope! you can use any of the equations for calculating deltaG. I think it's just a way to label it relating to the reaction.

Thank you! I understand now:)

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