Question

Find Delta G^o for the following reaction, using delta Hf^o and S^o values.
MnO2(s) + 2CO(g) -> Mn(s) + 2CO2(g)

Answer 1

Sol:-

Values of standard enthalpy of formation( delta H⁰_f ) :-

delta H⁰_f of CO(g) = -110.53 KJ/mol

delta H⁰_f of CO₂(g) = -393.51 KJ/mol

delta H⁰_f of MnO₂ (s) = - 520.0 KJ/mol

delta H⁰_f of Mn (s) = 0 KJ/mol

Values of standard entropy ( S⁰ ) :-

S⁰ of CO(g) = 197.6 J/Kmol

S⁰ of CO₂(g) = 213.6 J/Kmol

S⁰ of MnO₂ (s) = 53.0 J/Kmol

S⁰ of Mn (s) = 32.0 J/Kmol

Given reaction is :

MnO₂ (s) + 2 CO (g) ---------------> Mn (s) + 2 CO₂ (g)

first determine the values of standard enthalpy of the reaction ( i.e delta H⁰_r ) and change entropy of the reaction ( i.e delta S⁰_r ) .

delta H⁰_r = [sum of strandard enthalpy of formation of products] - [sum of standard enthalpy of formation of reactants]

delta H⁰_r = [delta H⁰_f of Mn (s) +2 x delta H⁰_f of CO₂(g) ] -[delta H⁰_f of MnO₂ (s) + 2 x delta H⁰_f of CO(g) ]

delta H⁰_r = [ 0 KJ /mol + 2 x - 393.5 KJ/mol] - [ -520.0 KJ/mol + 2 x - 110.53 KJ/mol ]

delta H⁰_r = [ - 787.02 KJ/mol ] - [ - 520.0 - 221.06 ]

delta H⁰_r = - 787.02 KJ/mol + 741.06 KJ/mol

delta H⁰_r = - 45.96 KJ/mol

delta H⁰_r = - 45960 J/mol

also delta S⁰_r = [sum of strandard entropy of products] - [sum of standard entropy of reactants]

delta S⁰_r = [ S⁰ of Mn (s) + 2 x S⁰ of CO₂(g) ] - [ S⁰ of MnO₂ (s) + 2 x S⁰ CO (g) ]

delta S⁰_r = [ 32.0 J/Kmol + 2 x 213.6 J/Kmol ] - [ 53.0 J/Kmol + 2 x 197.6 J/Kmol ]

delta S⁰_r = [ 32.0 J/Kmol + 427.2 J/Kmol ] - [ 53.0 J/Kmol + 395.2 J/Kmol ]

delta S⁰_r = 459.2 J/Kmol - 448.2 J/Kmol

delta S⁰_r = 11 J/Kmol

we know delta G⁰_r = delta H⁰_r - T delta S⁰_r  

delta G⁰_r = - 45960 J/mol - 298 K x 11 KJ/Kmol

delta G⁰_r = - 45960 J/mol - 3278 J/mol

delta G⁰_r = - 49238 J/mol

delta G⁰_r = - 49.238 J/mol