10:53 AM 64% o AT&T; LTE Session masteringchemistry.con C CHEM 1308-003 Fall 201
ID: 916661 • Letter: 1
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10:53 AM 64% o AT&T; LTE Session masteringchemistry.con C CHEM 1308-003 Fall 2015 Dr. Patrick McLaurin Help Close 7.70 a previous l 8 of 12 l next Exercise 17.70 Use AGE values from Appendix llB to calculate the equilibrium constants at 25 Cfor each of the following reactions. Part A 2NO2 (g) N2O4 (g) Express your answer using two significant figures. K 2.08.10 25 Submit My Answers Give Up incorrect; Try Again Part B H2 (g) P4 (g) PH3 (g) Express your answer using two significant figures. Submit My Answers Give Up Provide Feedback ContinueExplanation / Answer
1) Given reaction
2NO2 (g) <------------> N2O4 (g)
Gfo [NO2(g)] = +51.9 kJ/mol
Gfo [N2O4(g)] = + 98.3 kJ/mol
Go = Gfo(products) - Gfo( reactants)
= Gfo [N2O4(g)] - (2) Gfo [NO2(g)]
= 98.3 kJ/mol - (2) 51.9 kJ/mol
= -5.5 kJ/mol
= -5500 J/mol
Therefore, Go = -5500 J/mol
Calculation of Equilibrium constant K :
Go = -RT ln K where R = 8.314 J/mol/K
T = 25oC = 298 K
K = e -(Go/RT)
= e-[ -5500/ (8.314 x 298)]
= 9.21
K = 9.21
Therefore, Equilibrium constant, K = 9.21
2) 3/2 H2 (g) + 1/4 P4 (g) <------------> PH3 (g)
Gfo [H2(g)] = 0 kJ/mol
Gfo [P4(g)] = 0 kJ/mol
Gfo [PH3(g)] = + 18.4 kJ/mol
Go = Gfo(products) - Gfo( reactants)
= Gfo [PH3(g)] - { (3/2) Gfo [H2(g)] + (1/4) Gfo [P4(g)]}
= 18.4 kJ/mol - { 0 + 0}
= 18.4 kJ/mol
= 18400 J/mol
Therefore, Go = 18400 J/mol
Calculation of Equilibrium constant K :
Go = -RT ln K where R = 8.314 J/mol/K
T = 25oC = 298 K
K = e -( Go/RT)
= e -[18400/ (8.314 x 298)]
= 5.95 x10-4
K = 5.95 x10-4
Therefore, Equilibrium constant, K = 5.95 x10-4
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