In order to account for changes in both entropy and enthalpy J.Willard Gibbs defined a new function, which is called Gibbs free energy (Gibbs potential) G. The maximum useful work is the amount of energy produced, is given by the decrease in another thermodynamic property known as the Gibbs free energy.

**Gibbs Free Energy Formula** is independent of pressure and temperature and can be stated as

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Solved problems based on Gibbs free energy are given below. ### Solved Examples

**Question 1: **What is the standard free energy change $\Delta$G for the following reaction at 25^{o}C?

N_{2} + 3H_{2} $\rightarrow$ 2NH_{3}

The values for $\Delta$H ang S is -91.8KJ and -198.0J/K

** Solution: **

We can use the equation

Substitute the values in the above equation

$\Delta$G = -91.8KJ - (298K)(-0.1980KJ/K)

$\Delta$G = -32.8KJ

**Question 2: **Calculate the standard Gibbs free energy change for the formation of methane from carbon and hydrogen at 298K.

Given the $\Delta$H value is -74.9KJ/mol $\Delta$S = -80.7J/K.mol

** Solution: **

Given

$\Delta$H = -74.9KJ/mole

$\Delta$S = -80.7J/K.mol

Now use the equation

$\Delta$G = $\Delta$H - T$\Delta$S

Substitute the values in the above equation

$\Delta$G = -74.9KJ/mole - (298K)(-80.7J/K.mol)(1KJ/1000J)

$\Delta$G = -50.9KJ/mole

N

The values for $\Delta$H ang S is -91.8KJ and -198.0J/K

We can use the equation

Substitute the values in the above equation

$\Delta$G = -91.8KJ - (298K)(-0.1980KJ/K)

$\Delta$G = -32.8KJ

Given the $\Delta$H value is -74.9KJ/mol $\Delta$S = -80.7J/K.mol

Given

$\Delta$H = -74.9KJ/mole

$\Delta$S = -80.7J/K.mol

Now use the equation

$\Delta$G = $\Delta$H - T$\Delta$S

Substitute the values in the above equation

$\Delta$G = -74.9KJ/mole - (298K)(-80.7J/K.mol)(1KJ/1000J)

$\Delta$G = -50.9KJ/mole

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