
Which of the following process best describes atomization of
\[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\]?
(A) Exothermic
(B) Endothermic
(C) Non-spontaneous
(D) Spontaneous
(E) Both Endothermic and Non-spontaneous
Answer
563.4k+ views
Hint: As we know that when any molecule is formed and deformed there is either energy supplied or absorbed. The solid particles are attracted with great force of attraction, if we want to change its phase, we will have to supply extra energy.
Complete Step by step answer: As we also know that chemical reactions involve the breaking and making of chemical bonds. When a bond is formed between the atoms, energy is released which is also known as an exothermic process. Obviously, the same amount of energy will be needed to break the bond enthalpy which is known as endothermic process.
Now, let’s discuss what is atomisation?
Atomisation is the process when a molecule breaks into its atoms, the energy which is used to break the molecule into its atoms is known as enthalpy of atomisation. The enthalpy of atomisation is also known as bond dissociation enthalpy because in this process breaking of bonds occurs.
Now come to our given question,
Atomization of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\]occurs as
\[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\, \to \,{\rm{C(g)}}\,{\rm{ + }}\,{\rm{4}}\,{\rm{H(g)}}\,\,\,\,\,\,\,\Delta {\rm{H}}\,{\rm{ = }}\,{\rm{ + ve, }} \Delta {\rm{S}}\,{\rm{ = }}\,{\rm{ + ve}}\]
As we have seen in above explanation, that atomization of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\]occurs by the endothermic process.
It is also a non- spontaneous because Gibbs free energy is positive by the formula
\[{\rm{\Delta G = }}\,{\rm{\Delta H}}\,{\rm{ - }}\,{\rm{T\Delta S}}\,\]
Therefore, the correct option is option (E)
Note: There are two different terms used in thermodynamics for expressing enthalpy change associated with the chemical bonds. These are:
(a) Bond dissociation enthalpy- The amount of energy to break one mole of molecule into its gaseous atoms.
(b) Mean or average bond enthalpy- The average bond energy to break molecules into gaseous atoms.
Complete Step by step answer: As we also know that chemical reactions involve the breaking and making of chemical bonds. When a bond is formed between the atoms, energy is released which is also known as an exothermic process. Obviously, the same amount of energy will be needed to break the bond enthalpy which is known as endothermic process.
Now, let’s discuss what is atomisation?
Atomisation is the process when a molecule breaks into its atoms, the energy which is used to break the molecule into its atoms is known as enthalpy of atomisation. The enthalpy of atomisation is also known as bond dissociation enthalpy because in this process breaking of bonds occurs.
Now come to our given question,
Atomization of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\]occurs as
\[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\, \to \,{\rm{C(g)}}\,{\rm{ + }}\,{\rm{4}}\,{\rm{H(g)}}\,\,\,\,\,\,\,\Delta {\rm{H}}\,{\rm{ = }}\,{\rm{ + ve, }} \Delta {\rm{S}}\,{\rm{ = }}\,{\rm{ + ve}}\]
As we have seen in above explanation, that atomization of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\]occurs by the endothermic process.
It is also a non- spontaneous because Gibbs free energy is positive by the formula
\[{\rm{\Delta G = }}\,{\rm{\Delta H}}\,{\rm{ - }}\,{\rm{T\Delta S}}\,\]
Therefore, the correct option is option (E)
Note: There are two different terms used in thermodynamics for expressing enthalpy change associated with the chemical bonds. These are:
(a) Bond dissociation enthalpy- The amount of energy to break one mole of molecule into its gaseous atoms.
(b) Mean or average bond enthalpy- The average bond energy to break molecules into gaseous atoms.
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