
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
Answer
220.5k+ views
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.
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.
Recently Updated Pages
Difference Between Alcohol and Phenol: Structure, Tests & Uses

Class 12 Chemistry Mock Test Series for JEE Main – Free Online Practice

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

The D and F Block Elements Class 12 Chemistry Chapter 4 CBSE Notes - 2025-26

NCERT Solutions for Class 12 Chemistry Chapter Chapter 7 Alcohol Phenol and Ether

NCERT Solutions ForClass 12 Chemistry Chapter Chapter 8 Aldehydes Ketones And Carboxylic Acids

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

