part G
Balanced chemical reaction is
CH3OH(l) + O2(g) → HCO2H(l) + H2O(l)
∆Hf0 CH3OH(l) = -238.6 KJ/mol
∆Hf0 O2(g) = 0.0 KJ / mol
∆Hf0 HCO2H(l) = -425.5 KJ/mol
∆Hf0 H2O(l) = -285.8 KJ/mol
we know the equation
∆H0rxn = ∑ ∆Hf0(product) - ∑∆Hf0(reactant)
∆H0rxn = [(∆Hf0 HCO2H(l)) + ( ∆Hf0 H2O) ] - [( ∆Hf0 CH3OH(l)) + ( ∆Hf0 O2(g))]
substitute the vlaue
∆H0rxn = [( -425.5 KJ) + (-285.8 KJ) ] - [(-238.6 KJ) + ( 0 )]
∆H0rxn = [( - 711.3 KJ)] - [-238.6 KJ]
∆H0rxn = [-711.3 KJ] + [238.6 KJ]
∆H0rxn = - 472.7 KJ
Part H
Balanced reaction is
CH3OH(l) + O2(g) → HCO2H(l) + H2O(l)
∆S0 CH3OH(l) = 126.8 J/mol.K
∆S0 O2(g) = 205 J/mol.K
∆S0HCO2H(l) = 129 J/mol.K
∆S0 H2O(l) = 69.9 J/mol.K
∆S0rxn = ∑ ∆S0 (product) - ∑ ∆S0 (reactant)
∆S0rxn = [(∆S0HCO2H(l)) + ( ∆S0H2O(l))] - [(∆S0CH3OH(l) ) + (∆S0O2(g) )]
Substitute the value
∆S0rxn = [(129 J/K) + ( 69.9 J/K) ] - [ ( 126.8J/K) + (205 J/K)]
∆S0rxn = [198.9 J/K] - [ 331.8 J/K]
∆S0rxn = -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
Part G Calculate AH for the following reaction: CH3OH(1) +O2(g) + HCOH(1) +H2O(1) Express the enthalpy...
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Part A Calculate A H xn for the following reaction: CHA(g) + 4Cl2(g) → CCl4(g) + 4HCl(g) given these reactions and their AH values: | C(s) + C(s) + H2(g) + Express the enthalpy in kilojoules to one decimal place. 2H2 (g) 2Cl2(g) Cl2(g) + CH4(g), + CCL(g), + 2HCl(g), AH = -74.6 kJ AH = -95.7 kJ AH = -184.6 kJ %AM * o aj ? AH,,n= Submit Request Answer
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