Question

Which of the following standards is correct for a reversible process in a state of equilibrium?
(a)- $\Delta {{G}^{\circ }}=-2.303\ \text{RT logK}$
(b)- $\Delta {{G}^{\circ }}=2.303\ \text{RT logK}$
(c)- $\Delta G=-2.303\ \text{RT logK}$
(d)- $\Delta G=2.303\ \text{RT logK}$

Answer 1

Hint: The relation between standard free energy and the equilibrium constant is given by: $\Delta G=\Delta {{G}^{\circ }}+\ \text{RT lnK}$. This formula can be written when the reaction is in equilibrium.

Complete answer:
Gibbs free energy is a thermodynamic quantity that can be used to define the property of the reaction of a system. The Gibbs free energy of the system can be calculated by the difference between the energy factor and the entropy factor. The equation is:
$\Delta G=\Delta H-T\Delta S$
This Gibbs free energy is also equated with the equilibrium constant as the equation given below:
$\Delta G=\Delta {{G}^{\circ }}+\ \text{RT lnK}$
Where, $\Delta G$ is the change in the free energy of the system, $\Delta {{G}^{\circ }}$ is the standard free energy, R is the gas constant, T is the temperature and K is the equilibrium constant.
This equation can be modified if the system is in equilibrium. For an equilibrium system, the change in free energy will be zero. The equation becomes:
$0=\Delta {{G}^{\circ }}+\ \text{RT lnK}$
Now this equation can be written as:
$\Delta {{G}^{\circ }}=-\ \text{RT lnK}$
This equation is in the natural log form. We can convert this equation into a log form by multiplying the equation with 2.303. So, the equation becomes:
$\Delta {{G}^{\circ }}=-2.303\ \text{RT logK}$
So, this is the equation that is used for the reversible process when the system is in equilibrium.

Therefore, the correct answer is an option (a)- $\Delta {{G}^{\circ }}=-2.303\ \text{RT logK}$.

Note:
The equation $\Delta {{G}^{\circ }}=-2.303\ \text{RT logK}$can be written as $K={{e}^{-{{\Delta }_{r}}{{G}^{\circ }}/RT}}$ or we can also write this equation as $K={{10}^{-{{\Delta }_{r}}{{G}^{\circ }}/2.303RT}}$. For an endothermic process the K will be small and for an exothermic reaction the K will be large.