Given the information calculate ΔG° at 298 K for the reaction A + B --> 2C

Question

A + B --> 2D. Delta H deegree= -666.3 kJ Delta S deegree= 301.0 J/K

C--> D Delta H deegree= 509.0 kJ Delta S deegree= -163.0 J/K

Answer 1

Hess's Law

A + B ==> 2C .. TARGET EQUATION

Given:

eq.1: A + B ==> 2D .. ∆Hº = -666.3 kJ/mol and ∆Sº = 301.0 J/molK

eq.2: C ==> D .. ∆Hº = 590.0 kJ/mol and ∆Sº = -163.0 J/molK

Procedure:

copy eq.1: A + B ==> 2D .. ∆Hº = -666.3 kJ/mol .. ∆Sº = 301.0 J/Kmol

reverse eq.2 and x 2: 2D ==> 2C .. ∆Hº = -1180 kJ/mol .. ∆Sº = + 326 J/Kmol

Add these together and combine/cancel like terms:

A + B + 2D ==> 2D + 2C

A + B ==> 2C .. TARGET EQUATION

For this reaction, we have...

∆Hº = -666.3 kJ/mol + -1180 kJ/mol = -1846 kJ/mol

∆Sº = 301 J/Kmol + 326 J/Kmol = 627 J/Kmol = 0.627 kJ/Kmol

To find ∆Gº, use

∆Gº = ∆Hº - T∆Sº and be sure to change units of ∆Sº to kJ/Kmol to agree with those of ∆Hº

∆Gº = -1846 -(298)(0.627)

∆Gº = -1846 - 187

∆Gº = -2033 kJ/mol

(be sure to check all of the math)