Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K.4A +5B → 1C + 3D + 4E ΔHf0(kJ/mol)S0(J/mol/K)A-7.6917.43B5.2616.74C-0.756.75D-8.5711.50E10.706.30Remember, pK = -log(K)

The given reaction is as follows. 4A+5B →1C+3D+4E We calculate the value of standard Gibb's free energy change at 25 0C from the following equations. ∆G0=∆H0-T∆S0∆H0=∑n∆Hf0products-∑n∆Hf0reactants∆S0=∑n∆S0products-∑n∆S0reactants∆G0=standard Gibb's free energy change∆S0=change in entropy of the reaction∆H0=change in enthalpy of the reactionn=no of molesNow we can calculate the enthalpy change and entropy change of the given reaction. ∆H0=∑n∆Hf0products-∑n∆Hf0reactants=4∆Hf0E+3∆Hf0D+∆Hf0C-4∆Hf0A+5∆Hf0B=4×10.70+3×-8.57+-0.75-4×-7.69+5×5.26=42.8-25.71-0.75--30.76+26.3=16.34--4.46=20.8 kJ/mol ∆H0=∑n∆S0products-∑n∆S0reactants= 4∆S0E+3∆S0D+∆S0C-4∆S0A+5∆S0B=4×6.30+3×11.50+6.75-4×17.43+5×16.74 =25.2+34.5+6.75-69.72+83.7 =66.45-153.42 =-86.97 J/mol/K =-0.08697 KJ/mol/K Now we can calculate the standard Gibb's free energy change at 25 0C. ∆H0=20.8 KJ/molT=25+273=298 K∆S0=-0.08697 KJ/mol/K∆G0=∆H0-T∆S0=20.8 KJ/mol-298 K×-0.08697 KJ/mol/K=20.8 KJ/mol--25.92 KJ/mol=46.72 KJ/mol The standard Gibb's free energy change and equilibrium constant K have the following relationship. ∆G0=-RTlnK From the above equation we can find the value of pK. ∆G0=-RTlnK-lnK=∆G0RT=46720 J/mol8.314 Jmol.K×298 K-2.303×lnK=2.303×18.857 J/mol -log K=43.43 J/mol= pK Hence the value of pK is 43.43 J/mol