Relation between Equilibrium Constant, Reaction Quotient and The Gibbs Energy

RELATIONSHIP BETWEEN EQUILIBRIUM CONSTANT K, REACTION QUOTIENT Q AND GIBBS ENERGY G

The value of K_c for a reaction does not depend on the rate of the reaction. However, as you have studied in Unit 6, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy,\(\Delta\)G If,

1. \(\Delta\)G is negative, then the reaction is spontaneous and proceeds in the forward direction.

2. \(\Delta\)G is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative \(\Delta\)G, the products of the forward reaction shall be converted to the reactants.

3. \(\Delta\)G is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:

\(\Delta\)G=+\(\Delta\) +RT lnQ          (7.21)

where, is standard Gibbs energy.

At equilibrium, when \(\Delta\)G = 0 and Q = K_c, the equation (7.21) becomes,

\(\Delta\)G=+\(\Delta\) +RT ln K=0 (7.22)

\(\Delta\) = – RT lnK

lnK =-\(\Delta\) /RT

Taking antilog of both sides, we get,

K=e^{-\(\Delta\) /RT}      (7.23)

Hence, using the equation (7.23), the reaction spontaneity can be interpreted in terms of the value of \(\Delta\)  . 

If \(\Delta\)  < 0, then - \(\Delta\)  /RT is positive and , making K >1, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.

If  \(\Delta\)  <0 then - \(\Delta\)  /RT is negative, and ,that is , K < 1, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

PROBLEM 10

The value of for  \(\Delta\) the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of K_c at 298 K.

SOLUTION

 \(\Delta\)  = 13.8 kJ/mol = 13.8 × 10³J/mol

Also  \(\Delta\)  = – RT lnK_c

Hence, ln K_c = –13.8 × 10³J/mol

(8.314 J mol K_c × 298 K)

ln K_c = – 5.569

K_c = e^–5.569

K_c = 3.81 × 10^–3

PROBLEM 11

Hydrolysis of sucrose gives,Sucrose + H₂OGlucose + Fructose Equilibrium constant K_c for the reaction is 2 ×10¹³ at 300K. Calculate  at 300K.

SOLUTION

 \(\Delta\) = – RT lnK_c

  \(\Delta\) =

= – 8.314J mol^–1K^–1×

300K × ln(2×10¹³⁾

 \(\Delta\) = – 7.64 ×10⁴ J mol^–1