`=>` We have seen how a knowledge of the sign and magnitude of the free energy change of a chemical reaction allows :
(i) Prediction of the spontaneity of the chemical reaction.
(ii) Prediction of the useful work that could be extracted from it.
`=>` So far we have considered free energy changes in irreversible reactions.
`=>` Let us now examine the free energy changes in reversible reactions.
`=>` 'Reversible' under strict thermodynamic sense is a special way of carrying out a process such that system is at all times in perfect equilibrium with its surroundings.
● When applied to a chemical reaction, the term ‘reversible’ indicates that a given reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.
● This means that the reactions in both the directions should proceed with a decrease in free energy, which seems impossible.
● It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would spontaneously change to configuration of lower free energy.
● So, the criterion for equilibrium
`color{purple}(A + B ⇌ C +D `; is `color{purple}(Delta_r G = 0)`
● Gibbs energy for a reaction in which all reactants and products are in standard state, `color{purple}(Delta_rG^(⊖))` is related to the equilibrium constant of the reaction as follows :
`color{purple}(0 = Delta_r G^(⊖)+RT ln K)`
or `color{purple}(Delta_rG^(⊖) = -RT ln K)`
or `color{purple}(Delta_rG^(⊖) = -2.303 RT log K)` .......(6.23)
We also know that
`color{purple}(Delta_r G^(⊖) = Delta_r H^(⊖) - T Delta_r S^(⊖) = -RT ln K)` .....(6.24)
`=>` For strongly endothermic reactions, the value of `color{purple}(Delta_rH^(⊖))` may be large and positive. In such a case, value of `color{purple}(K)` will be much smaller than `color{purple}(1)` and the reaction is unlikely to form much product.
`=>` In case of exothermic reactions, `color{purple}(Delta_r H^(⊖))` is large and negative, and `color{purple}(Delta_r G^(⊖))` is likely to be large and negative too. In such cases, `color{purple}(K)` will be much larger than `1`.
● We may expect strongly exothermic reactions to have a large `color{purple}(K)`, and hence can go to near completion.
`=>` `color{purple}(Delta_rG ^(⊖))` also depends upon `color{purple}(Delta_r S^(⊖))`, if the changes in the entropy of reaction is also taken into account, the value of `color{purple}(K)` or extent of chemical reaction will also be affected, depending upon whether `color{purple}(Delta_rS^(⊖))` is positive or negative.
● Using equation (6.24),
(i) It is possible to obtain an estimate of `color{purple}(DeltaG^(⊖))` from the measurement of `color{purple}(DeltaH^(⊖))` and `color{purple}(DeltaS^(⊖))`, and then calculate `color{purple}(K)` at any temperature for economic yields of the products.
(ii) If `color{purple}(K)` is measured directly in the laboratory, value of `color{purple}(DeltaG^(⊖))` at any other temperature can be calculated.
`=>` We have seen how a knowledge of the sign and magnitude of the free energy change of a chemical reaction allows :
(i) Prediction of the spontaneity of the chemical reaction.
(ii) Prediction of the useful work that could be extracted from it.
`=>` So far we have considered free energy changes in irreversible reactions.
`=>` Let us now examine the free energy changes in reversible reactions.
`=>` 'Reversible' under strict thermodynamic sense is a special way of carrying out a process such that system is at all times in perfect equilibrium with its surroundings.
● When applied to a chemical reaction, the term ‘reversible’ indicates that a given reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.
● This means that the reactions in both the directions should proceed with a decrease in free energy, which seems impossible.
● It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would spontaneously change to configuration of lower free energy.
● So, the criterion for equilibrium
`color{purple}(A + B ⇌ C +D `; is `color{purple}(Delta_r G = 0)`
● Gibbs energy for a reaction in which all reactants and products are in standard state, `color{purple}(Delta_rG^(⊖))` is related to the equilibrium constant of the reaction as follows :
`color{purple}(0 = Delta_r G^(⊖)+RT ln K)`
or `color{purple}(Delta_rG^(⊖) = -RT ln K)`
or `color{purple}(Delta_rG^(⊖) = -2.303 RT log K)` .......(6.23)
We also know that
`color{purple}(Delta_r G^(⊖) = Delta_r H^(⊖) - T Delta_r S^(⊖) = -RT ln K)` .....(6.24)
`=>` For strongly endothermic reactions, the value of `color{purple}(Delta_rH^(⊖))` may be large and positive. In such a case, value of `color{purple}(K)` will be much smaller than `color{purple}(1)` and the reaction is unlikely to form much product.
`=>` In case of exothermic reactions, `color{purple}(Delta_r H^(⊖))` is large and negative, and `color{purple}(Delta_r G^(⊖))` is likely to be large and negative too. In such cases, `color{purple}(K)` will be much larger than `1`.
● We may expect strongly exothermic reactions to have a large `color{purple}(K)`, and hence can go to near completion.
`=>` `color{purple}(Delta_rG ^(⊖))` also depends upon `color{purple}(Delta_r S^(⊖))`, if the changes in the entropy of reaction is also taken into account, the value of `color{purple}(K)` or extent of chemical reaction will also be affected, depending upon whether `color{purple}(Delta_rS^(⊖))` is positive or negative.
● Using equation (6.24),
(i) It is possible to obtain an estimate of `color{purple}(DeltaG^(⊖))` from the measurement of `color{purple}(DeltaH^(⊖))` and `color{purple}(DeltaS^(⊖))`, and then calculate `color{purple}(K)` at any temperature for economic yields of the products.
(ii) If `color{purple}(K)` is measured directly in the laboratory, value of `color{purple}(DeltaG^(⊖))` at any other temperature can be calculated.
At `60°C`, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
