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.

Solution:

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).

Solution:

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)

Solution:

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)