Question

The incorrect match in the following is:
(A)- \[\Delta G{}^\circ <0;K<1\]
(B)- $\Delta G{}^\circ <0;K=1$
(C)- $\Delta G{}^\circ >0;K<1$
(D)- $\Delta G{}^\circ <0;K>1$

Answer 1

Hint:. A quantity which is used to measure the maximum amount of work done in a thermodynamic system keeping the temperature and pressure constant, is known as Gibbs free energy and is represented by the symbol ‘G’.

Complete step by step answer:
-Gibbs free energy is equal to the enthalpy of the system subtracted from the product of the temperature and entropy, that is-
G = H – TS
Where G = Gibbs free energy
H = Enthalpy
T = Temperature
S = Entropy

-Gibbs free energy does not depend on the path, hence it is a state function. Therefore, the formula of Gibbs free energy can now be written as-
$\Delta G=\Delta H-\Delta (TS)$
If the temperature is kept constant during the chemical reaction,
$\Delta G=\Delta H-T\Delta S$

The above equation is known as the Gibbs Helmholtz equation.
When $\Delta G>0$ then the reaction is nonspontaneous and endergonic.
When $\Delta G<0$ then the reaction is spontaneous and exergonic.
When $\Delta G=0$ then the reaction is at equilibrium.

-The free energy change of the reaction in any state $(\Delta G)$ is related to the standard free energy change of the reaction $(G{}^\circ )$ according to the equation-
$\Delta G=\Delta G{}^\circ +RT\ln Q$
Where Q = reaction quotient
When $\Delta G = 0$ and here q becomes equal to the equilibrium constant. Hence the equation now becomes,
$\Delta G{}^\circ = -RT\ln K(eq)$
$\Delta G{}^\circ = -2.303\text{ RT log K(eq)}$
Where $R=8.314\text{ J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}$
T = temperature on the Kelvin scale

-The above equation at equilibrium can be written as-
\[\begin{align}
  & \Delta G{}^\circ =-RT\text{ ln K} \\
 & \Delta G{}^\circ =0;K=1 \\
 & \Delta G{}^\circ >0;K>1 \\
 & \Delta G{}^\circ <0;K<1 \\
\end{align}\]

So, the correct answer is “Option A”.

Note: $\Delta G$ determining the direction and extent of chemical change during a chemical reaction. It is meaningful only for the reactions in which the temperature and pressure remain constant. $\Delta G$ serves as the single master variable that determines whether a given chemical change is thermodynamically possible or not. If free energy of the reactants is greater than that of the products, the entropy of the chemical reaction will increase and so the reaction will tend to take spontaneously. When $\Delta G$ is negative, the process will take place spontaneously and is referred to as exergonic.