Solveeit Logo
Login

Question

Question: Which of the following standards is correct for a reversible process in a state of equilibrium? (a...

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

Explanation

Solution

The relation between standard free energy and the equilibrium constant is given by: ΔG=ΔG∘+ RT lnK\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:
ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta S
This Gibbs free energy is also equated with the equilibrium constant as the equation given below:
ΔG=ΔG∘+ RT lnK\Delta G=\Delta {{G}^{\circ }}+\ \text{RT lnK}
Where, ΔG\Delta G is the change in the free energy of the system, ΔG∘\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=ΔG∘+ RT lnK0=\Delta {{G}^{\circ }}+\ \text{RT lnK}
Now this equation can be written as:
ΔG∘=− RT lnK\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:
ΔG∘=−2.303 RT logK\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)- ΔG∘=−2.303 RT logK\Delta {{G}^{\circ }}=-2.303\ \text{RT logK}.

Note:
The equation ΔG∘=−2.303 RT logK\Delta {{G}^{\circ }}=-2.303\ \text{RT logK}can be written as K=e−ΔrG∘/RTK={{e}^{-{{\Delta }_{r}}{{G}^{\circ }}/RT}} or we can also write this equation as K=10−ΔrG∘/2.303RTK={{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.