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Write the relationship between Gibbs energy, equilibrium constant and change in enthalpy.

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Last updated date: 13th Jun 2024
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Answer
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Hint: There is no single formula relating the three terms, but two different formulae for the relation. Gibbs energy is represented as $\vartriangle \text{G}$, equilibrium constant as ${{\text{K}}_{\text{eq}}}$ and change in enthalpy as $\vartriangle \text{H}$. One is the formula of thermodynamics and spontaneity and another is telling the reaction is spontaneous if the reaction with equilibrium constants.

Complete step by step answer:
Let us first write the relation between Gibbs energy and change in enthalpy.
The Gibbs free energy of a system is defined as the enthalpy of a system (amount of heat absorbed or evolved) minus the product of temperature multiplied to entropy of a system (randomness). $\text{G= H-TS}$; but this is with enthalpy not with change in enthalpy.
Gibbs free energy is a state function, so the change of Gibbs free energy is defined as change of enthalpy minus the change in temperature multiplied with entropy of a system.
It is written as $\vartriangle \text{G= }\vartriangle \text{H-}\left( \vartriangle \text{TS} \right)$. If we take temperature to be constant, the equation becomes
$\vartriangle \text{G= }\vartriangle \text{H- T}\vartriangle \text{S}$.
This is the relation we generally use while solving the questions or checking the spontaneity of reactions.
Now, move to the relation between Gibbs free energy and equilibrium constant.
The relation between Gibbs free energy $\left( \vartriangle \text{G} \right)$, standard Gibbs free energy which is at ${{25}^{\text{o}}}\text{C}$ is $\left( \vartriangle {{\text{G}}^{\text{o}}} \right)$ and reaction quotient (Q) at any moment of time is $\vartriangle \text{G = }\vartriangle {{\text{G}}^{\text{o}}}+\text{RTlnQ}$; where R is ideal gas constant with value $8.314\text{ J}\text{.mo}{{\text{l}}^{-1}}{{\text{K}}^{-1}}$, T is the temperature in Kelvin. Q is written like equilibrium constant but is defined at any moment other than equilibrium.
When, the driving force of a chemical reaction is zero, then $\vartriangle \text{G}$ is zero and Q becomes ${{\text{K}}_{\text{eq}}}$ is $0=\vartriangle {{\text{G}}^{\text{o}}}+\text{RTln}{{\text{K}}_{\text{eq}}}$.
This expression is moulded to $\vartriangle {{\text{G}}^{\text{o}}}=-\text{RTln}{{\text{K}}_{\text{eq}}}$. The smaller the value of $\vartriangle {{\text{G}}^{\text{o}}}$, the closer the standard state is to equilibrium.
The relation between Gibbs energy and equilibrium constant is $\vartriangle {{\text{G}}^{\text{o}}}=-\text{RTln}{{\text{K}}_{\text{eq}}}$.
The relation between Gibbs energy and change in enthalpy is $\vartriangle \text{G= }\vartriangle \text{H- T}\vartriangle \text{S}$.

Note:
The beauty of $\vartriangle \text{G= }\vartriangle \text{H- T}\vartriangle \text{S}$ is the ability to determine the relative importance of the enthalpy and entropy as driving forces behind a reaction. $\vartriangle \text{G}$ measures the balance between the driving forces to determine whether a chemical reaction is spontaneous. If its value is less than zero, the reaction will be spontaneous . For positive values, the reaction is non spontaneous.