Question

The correct relation between equilibrium constant (K), standard free energy $\Delta {{G}^{\circ }}$ and temperature (T) is……
(A) $\Delta {{G}^{\circ }}=RT\ln K$
(B) $K={{e}^{\Delta {{G}^{\circ }}/2.303RT}}$
(C) $\Delta {{G}^{\circ }}=-RT\log K$
(D) $K={{10}^{-\Delta {{G}^{\circ }}/2.303RT}}$

Answer 1

. Value of $\Delta {{G}^{\circ }}$ is directly proportional to Temperature and value of logK. $\Delta {{G}^{\circ }}$ is Gibbs free energy, so its unit is the same as energy. The reversible work done by electrochemical cells is equal to decrease in Gibbs energy.

Complete step by step answer:
We will need to relate the quantity free energy, temperature and equilibrium constant here. To relate them, we must find how they are related to each other. Let’s find their relationship.
As we know that the value of $\Delta {{G}^{\circ }}$ is directly proportional to Temperature and value of logK , we can write that $\Delta G\propto T$ means as value of T increases, value of free energy will also increase. Same way, $\Delta G\propto \log K$ means as the value of logK increases, the value of free energy will increase and vice-versa. So, we can write those both relations combined and the formula $\Delta {{G}^{\circ }} = RT\ln K$ is obtained, where R is a constant and known as universal gas constant. We have also studied about this formula.
But it is not given in the option. We can clearly see that option (A) and (C) are not true as the value of free energy is said to be somewhat else. But option (B) and (D) is written in a different way. So, let’s find the value of constant K in that formula. We will need to find the value of K in order to check those options.
We know that, $\Delta {{G}^{\circ }}=RT\ln K$
So, we can write this as
$\ln K=\dfrac{-\Delta {{G}^{0}}}{RT}$
- As $\operatorname{lnK}=2.303\operatorname{logK}$ ,
$\log K=\dfrac{-\Delta {{G}^{0}}}{2.303RT}$………………..(1)

Here we are using the logarithm on the base of 10 only. So, we can write that
${{\log }_{10}}K = \dfrac{-\Delta {{G}^{0}}}{2.303RT}$
It is a fundamental logarithmic equation that if${{\log }_{10}}K = x$ , then $K = {{10}^{x}}$ . Putting this formula in our equation (1), we get
$K = {{e}^{\Delta {{G}^{\circ }}/2.303RT}}$
So, the correct answer is “Option B”.

- In a broader sense, the formula to convert logarithmic equations when base is mentioned, is as under.
If ${{\log }_{a}}b = k$ then ${{a}^{k}} = b$.
- Also remember the following outcomes.
$\ln (e) = 1$
$\ln (1) = 0$
${{e}^{0}} = 1$

Note: - Don’t get confused when the direct formula which is given in text does not match with the options in the question. This just requires a simple mathematical simplification or conversion to get to the required answer.
- Always remember the minus sign if it is present in the formula because it is often forgotten when in a hurry.