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How is delta G related to the equilibrium constant?

Answer
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Hint: Thermodynamics is a branch of physical chemistry that deals with many forms of energy and heat. It tells us about how thermal energy is converted into other forms of energy. Gibbs free energy is used for the measurement of the maximum amount of work done in a thermodynamic system.

Complete step-by-step answer:Gibbs free energy is denoted with symbol G and its value is expressed in Joules.
Now let us discuss the relationship between the change in Gibbs free energy and equilibrium constant-
The change in Gibbs free energy is related to the change in standard Gibbs free energy that is represented by $\Delta {{G}^{{}^\circ }}$ .
Now let us see the equation
$\Delta G=\Delta {{G}^{{}^\circ }}+RT\ln Q$
Where, $\Delta G$ is change in Gibbs free energy
$\Delta {{G}^{{}^\circ }}$ change in standard Gibbs free energy
$R$ is the gas constant $\left( R=0.08314 kJ mol{{l}^{-1}}{{K}^{-1}} \right)$
$T$ is the temperature on Kelvin
$Q$ is the reaction quotient
At equilibrium,
$\Delta G=0$ and $Q=K$
Here, $K$ is the equilibrium constant
Therefore, on substituting the value in the above formula we get,
$\Delta {{G}^{{}^\circ }}=-RT\ln K$
As $\ln =2.303\log $
On further substituting we get,
$\Delta {{G}^{{}^\circ }}=-2.303RT\log K$
If the value of equilibrium constant is large then it will result in negative value of change in Gibbs free energy. It indicates reaction spontaneity. It is an irreversible process that can only be reversed by external agents.

Note: Change in Gibbs free energy helps in determining the extent of chemical change and direction.
If the change in Gibbs free energy is negative then it is a spontaneous reaction.
Spontaneity depends upon the temperature of the system.
Spontaneous process releases free energy and moves to a more stable energy state.