a. For a reaction with ?Ho = -12 kcal/mol and ?So =

Solution 24P

Step 1 of 3:

A.

Here, we are asked to calculate the ΔGo and the equilibrium constant at given temperatures.

Given: ΔHo = -12 kcal/mol

ΔSo = 0.01kcal mol-1 K-1.

The standard free energy is calculated using the formula:

         ΔGo =  ΔHo - T ΔSo

Where, ΔHo = change in enthalpy

 ΔSo = change in entropy

(1) at 30 oC:

T = 30 oC = 303 K        

First, let’s calculate for ΔGo :

 ΔGo =  ΔHo - T ΔSo

         =  -12 kcal/mol - 303K0.01kcal/mol/K

 ΔGo = -15.03 kcal/mol.

Now, let’s calculate for Keq:

        Go = _ RT ln Keq

        ln Keq = _ 

                   = _

                         = 25.05

        Keq = e25.05

        Keq  = 7.21010

(2) 150 oC:

T = 150 oC = 423 K        

First, let’s calculate for ΔGo :

 ΔGo =  ΔHo - T ΔSo

         =  -12 kcal/mol - 423K0.01kcal/mol/K

 ΔGo = -16.23 kcal/mol.

Now, let’s calculate for Keq:

        Go = _ RT ln Keq

        ln Keq = _ 

                   = _

                         = 19.34

        Keq = e19.34

        Keq  = 2.4108