Q and Gibbs Free energy relationship

[Susanna Givan 2B]

What is the conceptual relationship behind how Q influences and changes Gibbs free energy?

[Hannah Alltucker 3L]

I believe it just has something to do with how a reaction proceeds and the spontaneity of said reaction. If we're given the Gibbs free energy and the K value, and then are able to find a Q to compare to said K value, we can identify which way the reaction naturally wants to move. Because this push for equilibrium occurs naturally in these reactions as they find stability, we can relate our Q to our Gibbs free energy.

I found this online as a little summary from https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry/7%3A_Equilibrium_and_Thermodynamics/7.11%3A_Gibbs_Free_Energy_and_Equilibrium#:~:text=%CE%94G%20is%20related%20to%20Q,RTlnQK.&text=If%20%CE%94G%20%3E%200%2C%20then%20K,the%20reaction%20is%20at%20equilibrium.

"If a system is not at equilibrium, ΔG and Q can be used to tell us in which direction the reaction must proceed to reach equilibrium. ΔG is related to Q by the equation ΔG=RTlnQK.

If ΔG < 0, then K > Q, and the reaction must proceed to the right to reach equilibrium.
If ΔG > 0, then K < Q, and the reaction must proceed to the left to reach equilibrium.
If ΔG = 0, then K = Q, and the reaction is at equilibrium.
We can use the measured equilibrium constant K at one temperature, along with ΔH° to estimate the equilibrium constant for a reaction at any other temperature."