Question

Consider the reaction

CO(g) + Cl₂(g)---COCl₂(g)

Use the standard thermodynamic data in the tables linked above. Calculate delta G for this reaction at 298.15K if the pressure of COCl₂(g) is reduced to 20.15 mm Hg, while the pressures of CO(g) and Cl₂(g) remain at 1 atm.

ANSWER: kJ/mol

Consider the reaction

2SO₂(g) + O₂(g)----2SO₃(g)

Using the standard thermodynamic data in the tables linked above, calculate delta G_rxn for this reaction at 298.15K if the pressure of each gas is 25.11 mm Hg.

ANSWER: kJ/mol

Answer 1

Sol 1.

Reaction :

CO(g) + Cl2(g) ----> COCl2(g)

Standard free energy of reaction = deltaG°  

= deltaGf°(COCl2 , g) - deltaGf°(CO , g) - deltaGf°(Cl2 , g )

= - 204.6 - ( - 137.2 ) - 0  

= - 67.4 KJ / mol

Now , PCOCl2 = 20.15 mm Hg = 20.15 / 760 atm = 0.0265 atm

PCO = PCl2 = 1 atm

So , Reaction Quotient = Q = PCOCl2 / ( PCO × PCl2 )

= 0.0265 / ( 1 × 1 ) = 0.0265  

Also , Gas constant = R = 0.008314 KJ / K mol

Temperature = T = 298.15 K

So , Free energy of reaction = deltaG

= deltaG° + RTln(Q)

= - 67.4 + 0.008314 × 298.15 × ln(0.0265)

= - 76.400   KJ / mol   

Sol 2.

Reaction :

2SO2(g) + O2(g) ----> 2SO3(g)

Standard free energy of reaction = deltaG°  

= 2 × deltaGf°(SO3 , g) - deltaGf°(O2 , g) - 2 × deltaGf°(SO2 , g )

= 2 × ( - 371.1 ) - 0 - 2 × ( - 300.2 )

= - 141.8 KJ / mol

Now , PSO2 = PO2 = PSO3 = 25.11 mm Hg = 25.11 / 760 atm = 0.033 atm

So , Reaction Quotient = Q

= (PSO3)² / ( PO2 × (PSO2)² )

= (0.033)² / ( 0.033 × (0.033)² )  

= 30.303

Also , Gas constant = R = 0.008314 KJ / K mol

Temperature = T = 298.15 K

So , Free energy of reaction = deltaG

= deltaG° + RTln(Q)

= - 141.8 + 0.008314 × 298.15 × ln(30.303)

= - 133.344 KJ / mol