Question

Part G Calculate AH for the following reaction: CH3OH(1) +O2(g) + HCOH(1) +H2O(1) Express the enthalpy in kilojoules to one decimal place. AC O O ? AH = Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining

Answer 1

part G

Balanced chemical reaction is

CH₃OH_(l) + O_2(g) → HCO₂H_(l)   + H₂O(l)

∆H_f⁰ CH₃OH_(l) = -238.6 KJ/mol

∆H_f⁰ O_2(g) = 0.0 KJ / mol

∆H_f⁰ HCO₂H_(l) = -425.5 KJ/mol

∆H_f⁰ H₂O_(l) = -285.8 KJ/mol

we know the equation

∆H⁰_rxn = ∑ ∆H_f⁰_(product) -   ∑∆H_f⁰_(reactant)

∆H⁰_rxn = [(∆H_f⁰ HCO₂H_(l)) + ( ∆H_f⁰ H₂O) ] - [( ∆H_f⁰ CH₃OH_(l)) + ( ∆H_f⁰ O_2(g))]

substitute the vlaue

∆H⁰_rxn   = [( -425.5 KJ) + (-285.8 KJ) ] - [(-238.6 KJ) + ( 0 )]

∆H⁰_rxn   = [( - 711.3 KJ)] - [-238.6 KJ]

∆H⁰_rxn = [-711.3 KJ] + [238.6 KJ]

∆H⁰_rxn = - 472.7 KJ

Part H

Balanced reaction is

CH₃OH_(l) + O_2(g) → HCO₂H_(l)   + H₂O(l)

∆S⁰ CH₃OH_(l) = 126.8 J/mol.K

∆S⁰ O_2(g) = 205 J/mol.K

∆S⁰​​​​​​​HCO2H(l)  = 129 J/mol.K

∆S⁰​​​​​​​ H₂O_(l) = 69.9 J/mol.K

∆S⁰​​​​​​​_rxn = ∑ ∆S⁰​​​​​​​ (product) -  ∑ ∆S⁰​​​​​​​ (reactant)

∆S⁰​​​​​​​_rxn = [(∆S⁰HCO₂H(l)) + ( ∆S⁰H₂O_(l))] - [(∆S⁰CH₃OH_(l) ) + (∆S⁰O_2(g) )]

Substitute the value

∆S⁰​​​​​​​_rxn = [(129 J/K) + ( 69.9 J/K) ] - [ ( 126.8J/K) + (205 J/K)]

∆S⁰​​​​​​​_rxn = [198.9 J/K] - [ 331.8 J/K]

∆S⁰​​​​​​​_rxn = -132.9 J/K

Part I

we know the equation

∆G⁰ = ∆H⁰ - T∆S⁰

where,

∆H⁰ = - 472.7 KJ/mol

∆S⁰ = -132.9 J/mol.K = -0.1329 KJ/mol.K

(1 J = 0.001 KJ then -132.9 J = -132.9 X 0.001 = -0.1329 KJ)

T = 298.15 K

( 25⁰C = 25⁰C + 273.15 = 298.15 K )

Substitute the value in above formula

∆G⁰ = (-427.7 KJ) - (-0.1329 KJ/mol X 298.15)

∆G⁰ = (-427.7) - (-39.62)

∆G⁰ = (-427.7) + (39.62)

∆G⁰ = -388.08 KJ/mol