Chemistry Gibbs Energy Change and Equilibrium

Topics Covered :

● Gibbs Energy Change and Equilibrium

Gibbs Energy Change and Equilibrium :

`=>` 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.
Q 3007634588

At `60°C`, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.


`N_2O_4 (g) ⇌ 2NO_2 (g)`

If `N_2O_4` is `50%` dissociated, the mole fraction of both the substances is given by

`χ_(N_2O_4) = (1-0.5)/(1+0.5) ; χ_(NO_2) = (2xx0.5)/(1+0.5)`

`p_(N_2O_4) = (0.5)/(1.5) xx 1 atm = p_(NO_2) = 1/(1.5) xx 1 atm`

The equilibrium constant `K_p` is given by

`K_p = (p_(NO_2))^2/(p_(N_2O_4)) = (1.5)/((1.5)^2 (0.5))`

` = 1.33 atm`

since `Delta_rG^(⊖) =- RT ln K_p`

`Delta_rG^(⊖) = (-8.314 J K^(-1) mol^(-1) ) xx (333 K) xx (2.303) xx (0.1239)`

` = -763.8 kJ mol^(-1)`
Q 3037534482

Find out the value of equilibrium constant for the following reaction at 298 K.

`2NH_3 (g) + CO_2 (g) ⇌ NH_2 CONH_2(aq) +H_2O(l)`

Standard Gibbs energy change, `Delta_rG^(⊖)` at the given temperature is `–13.6 kJ mol^(–1)`.


We know, `log K = ( - Delta_r G^(⊖))/(2.303 RT)`

` = (-13.6xx10^3 J mol^(-1))/(2.303(8.314J K^(-1) mol^(-1) ) (298 K))`

`= 2.38`

Hence `K = antilog 2.38 = 2.4 × 10^2`.
Q 3027434381

Calculate `Delta_rG^(⊖) ` for conversion of oxygen to ozone, `3/2O_2 (g) → O_3 (g)` at `298 K`. if `K_p` for this conversion is `2.47xx10^(-29)`


We know `Delta_rG^(⊖) = -2.303 RT log K_p` and `R = 8.314 J K^(-1) mol^(-1)`

Therefore `Delta_r G^(⊖ ) = -2.303 (8.314J K^(-1) mol^(-1) ) xx (298 K) (log 2.47 xx 10^(-20))`

` = 163000 J mol^(-1)`

` = 163 kJ mol^(-1)`