Question

12. Considering the following electrochemical cell carried out at 25°C and using only the information in the table below: MnO2 (s) Cu (s)4 H (a) Mn2 (a) Cu2 (aq)2 H2O (0) Substance AH (kJ/mol) 0.00 219.4 0.00 285.8 64.9 520.0 AS (J/mol Cu (s) 33.2 78.8 0.00 70.0 98.0 53.1 H2O Cu (aq) MnO2 (s) a) If this system were at equilibrium, in what direction would the reaction go to re-attain its state of equilibrium if the temperature were incre ased? b) Determine the equilibrium constant for this reaction.

Answer 1

Solution :-

MnO2(s) + Cu(s) + 4H^+(aq) ---- > Mn^2+(aq) + Cu^2+(aq) + 2H2O(l)

Part a) using the given thermochemical data we can find the enthalpy change of the reaction and determine the shift in the equilibrium when temperature is increased.

Delta H rxn= sum of delta H product – sum of delta H reactant

=[(Mn^2+*1)+(Cu^2+*1)+(H2O*2)]-[(MnO2*1)+(Cu*1)+(H^+*4)]

=[(-219.4*1)+(64.9*1)+(-285.8*2)]-[(-520*1)+(0*1)+(0*4)]

= -206.1 kJ/mol

Since the reaction is having negative enthalpy change therefore reaction is exothermic

For the exothermic reaction when the temperature is increased then reaction will shift to left (reactant ) side.

Part b)

Lets calculate the entropy change for the reaction

Delta S rxn= sum of delta S product – sum of delta S reactant

=[(Mn^2+*1)+(Cu^2+*1)+(H2O*2)]-[(MnO2*1)+(Cu*1)+(H^+*4)]

=[(-78.8*1)+(-98.0*1)+(70.0*2)]-[(53.1*1)+(33.2*1)+(0.00*4)]

= -122.4 J/K

Now using the delta H and delta S we can find the delta G

Delta G= delta H – (T*delta S)

= -206.1 kJ/mol – (298K * (-0.1224 kJ/K))

= -169.6 kJ

-169.6 kJ * 1000 J/1kJ= -169600 J

Now using the delta G value we can find the equilibrium constant

Delta G= - RT ln K

-169600 J = - 8.314 J per mol K 298 K ln K

-169600 J / ( - 8.314 J per mol K * 298 K) = ln K

68.45 = ln K

Anti ln (68.45 ) = K

5.34*10^29 = K

Therefore the equilibrium constant of the reaction is 5.34*10^29