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Standard Gibbs free energy change $\Delta {{G}^{\circ }}$ for a reaction is zero. The value of equilibrium constant of the reaction will be:
[A] 0
[B] 1
[C] 2
[D] 3

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Last updated date: 14th Apr 2024
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MVSAT 2024
Answer
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Hint: Standard Gibbs free energy is the change in free energy at S.T.P i.e. standard temperature and pressure. At equilibrium, the equilibrium constant K and standard free energy change is related as-
\[\Delta {{G}^{\circ }}=-RT\ln K\]
We can put the values in the above equation to solve this question.

Complete step-by-step answer:
In thermodynamics, we use Gibbs free energy to calculate maximum reversible work done on a system at a constant temperature and pressure. It is basically the net work done on the spontaneous.
If we carry out a spontaneous process in such a way that no net work is involved which is, $d{{w}_{net}}=0$ where, $d{{w}_{net}}$ is the total work done then, the Gibbs energy at constant temperature and pressure is less than zero and we can write this as-
${{(dG)}_{T,P}} < 0$
For any spontaneous system, there occurs a decrease in free energy and this decrease is continued till the system reaches equilibrium. Therefore, at equilibrium, change in free energy at constant temperature and pressure becomes zero. We can write this as-
${{(dG)}_{T,P}}=0$
We denote the standard Gibbs free energy change as $\Delta {{G}^{\circ }}$ which is the change in free energy under standard condition i.e. 1atm pressure and 25 degrees Celsius temperature.
There exists an equation which gives us the relation between the equilibrium constant of a chemical reaction and the Gibbs free energy and the equation is-
\[\Delta G=\Delta {{G}^{\circ }}+RT\ln K\]
Where, $\Delta G$ is the change in free energy
$\Delta {{G}^{\circ }}$ is the standard change in free energy
T is the temperature
K is the equilibrium constant
R is the universal gas constant and its value is fixed for every gas.
In the question, it is given to us that the standard free energy change is zero. So, putting this in the above equation we will get-
\[\Delta G=0+RT\ln K\]
As we have discussed above, change in free energy becomes zero at equilibrium, therefore we can write that-
\[\begin{align}
  & RT\ln K=0 \\
 & \Rightarrow \ln K=0 \\
\end{align}\]
\[or,\ln K={{e}^{0}}=1\]
Therefore, the value of equilibrium constant is 1.
Therefore, the correct answer is option [B] 1.

Note: When a reaction takes place at standard state, the Gibbs free energy change is equal to standard free energy change.
If the products and the reactant are at standard state and if $\Delta {{G}^{\circ }} < 0$ then, K is greater than 1 and the formation or product is more favourable over the reactant and this is called a spontaneous reaction and when $\Delta {{G}^{\circ }} > 0$ then k is less than 1 and the reactant is favoured thus, the reaction is nonspontaneous and when $\Delta {{G}^{\circ }}=0$, K is 1 and the reaction is at equilibrium.