For a dimerization reaction \[2A(g)\xrightarrow{{}}{A_2}\] at \[298\,K\] , \[\Delta U = - 20\,kJ\,mo{l^{ - 1}}\] , \[\Delta S = 30\,kJ\,mo{l^{ - 1}}\] , then the \[\Delta G\] will be

Hint: We know that the term "dimerized" refers to the combination of two molecules of the same chemical to form a dimer. We must determine enthalpy to determine Gibbs free energy.

Formula Used:
The change in enthalpy is equal to the energy supplied as heat at constant pressure. The equation is as follows:
\[\Delta G = \Delta H - T\Delta S\]
Here, \[\Delta G\] is a change in Gibbs free energy, \[\Delta H\] is a change in enthalpy, \[T\] is temperature, and \[\Delta S\] is a change in entropy.
The equation to calculate the change in enthalpy is as follows:
\[\Delta H = \Delta U + \Delta {n_g}RT\]
Here, \[\Delta H\] is change in enthalpy, \[\Delta U\] is a change in internal energy, \[\Delta {n_g}\] is a change in the mol of gases, \[R\] is gas constant, and \[T\] is temperature.

Complete Step by Step Solution:
The Gibbs free energy sometimes referred to as the Gibbs function, Gibbs energy, or free enthalpy, is a thermodynamic potential that is employed to calculate the greatest amount of work that may be accomplished in a particular thermodynamic system when the system's temperature and pressure are held constant. Since it is also energy, its value is typically stated in Joules or Kilojoules. The most work that may be accomplished from a closed thermodynamic system is referred to as Gibbs free energy.

The amount of heat in a system is measured in enthalpy. This heat is used to cause a process to occur. Hence, enthalpy is any system involving heat that is referred to as a thermodynamic system.
The given value of change in internal energy is as follows:
\[ \Delta U = - 20\,kJ\,mo{l^{ - 1}} \\
   \Rightarrow \Delta U = - 20\,J\,mo{l^{ - 1}} \\ \]

Let us calculate the change in gaseous mole as follows:
\[ \Delta {n_g} = 1 - 2 \\
   \Rightarrow \Delta {n_g} = - 1 \\ \]

Let us calculate the value of change in enthalpy as follows:
\[ \Delta H = \Delta U + \Delta {n_g}RT \\
   \Rightarrow \Delta H = - 20000 + \left( { - 1} \right) \times 8.314 \times 298 \\
   \Rightarrow \Delta H = - 22477.52\,J \\ \]

Let us calculate the change in free energy as follows:
\[ \Delta G = \Delta H - T\Delta S \\
   \Rightarrow \Delta G = - 22477.52 - \left( {298} \right) \times \left( { - 30} \right) \\
   \Rightarrow \Delta G = - 13537.5\,J \\ \]

Converting the value in kilojoules,
\[ \Delta G = - 13537.5\,J \\
   \Rightarrow \Delta G = - 13537.5 \times 1000\,kJ \\
   \Rightarrow \Delta G = - 13.5\,kJ \\ \]
Therefore, the value of \[\Delta G\] is \[ - 13.5\,kJ\].

Note: Since the Gibbs free energy is a state function, the path has no bearing on it (i.e., its path-independent entity). As a result, the difference between the change in enthalpy and the system's change in temperature and entropy will equal the change in Gibbs free energy.