Comments on: Chemical Equilibria

By: Shweta Sharma

Sat, 01 May 2021 06:51:42 +0000

I think it should be standard Gibbs free energy in the expression rather than just gibbs free energy ….. is standard gibbs free energy the diff of gibbs free energy of products and reactants

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By: Shweta Sharma

Sat, 01 May 2021 06:46:01 +0000

In the expression relating gibbs free energy to equilibrium constant what exactly is del G. Is it for the products or reactants …it cant be for the system since in equilibrium del G is 0

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By: Alton

Thu, 18 Jun 2020 19:12:41 +0000

This clarified that for me. Thank you!

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By: James Ashenhurst

Wed, 03 Jun 2020 14:43:10 +0000

In reply to Katelyn.

That is a *fantastic* question.

The difference is between delta G of the *system* and delta G° (the energy of formation) of the individual components.

For a *system* at equilibrium, delta G is zero. There is no net change of energy.

If the equilibrium constant was 1 (concentration of the products = concentration of reactants) then we could *also* say that the delta G° (delta G of formation) of the products is equivalent to that of the reactants.

But most of the time, the equilibrium constant is greater (or lower) than 1 (depending on which direction you draw the equation). That means one component is more *stable* than the other, and we can relate their relative stabilities to the equilibrium by using the Gibbs free energy equation, delta G° = RT ln K. If we know the equilibrium constant at a given temperature we can rearrange the equation to solve for the difference in energies.

Does that make sense?

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By: Katelyn

Sat, 30 May 2020 22:01:55 +0000

Why do you calculate delta G for an equilibirum reaction if delta G is zero at equilibrium? What is there to calculate?

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