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molly16
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1.0 mol of N2O4 placed in a constant pressure vessel at P = 1bar and T = 298 K. The system is allowed to slowly (reversibly) come to equilibrium. Given gibbs energy of formation, enthalpy of formation and entropy (the values are below) calculate the entropy change to the surroundings.
N2O4:
gibbs energy of formation = 99.8 kJ/mol
Enthalpy of formation =11.1 kJ/mol
Entropy = 304.3 J/mol K
NO2 :
gibbs energy of formation = 51.3 kJ/mol
enthalpy of formation = 33.2 kJ/mol
entropy = 240.1 J/mol K
N2O4 (g) <=> 2NO2 (g)
Attempt at a solution:
So first I found the enthalpy and free energies of the reaction
delta Hrxn = 2(33.2) - 11.1 = 55.3 kJ/mol
delta Grxn = 2(51.3) - 99.8 = 2.8 kJ/mol
since the surroundings are at constant pressure I know:
delta Ssurroundings = qsurroundigs/T = delta H surroundings/ T
but I'm not sure where to go from here. Can anyone help?
N2O4:
gibbs energy of formation = 99.8 kJ/mol
Enthalpy of formation =11.1 kJ/mol
Entropy = 304.3 J/mol K
NO2 :
gibbs energy of formation = 51.3 kJ/mol
enthalpy of formation = 33.2 kJ/mol
entropy = 240.1 J/mol K
N2O4 (g) <=> 2NO2 (g)
Attempt at a solution:
So first I found the enthalpy and free energies of the reaction
delta Hrxn = 2(33.2) - 11.1 = 55.3 kJ/mol
delta Grxn = 2(51.3) - 99.8 = 2.8 kJ/mol
since the surroundings are at constant pressure I know:
delta Ssurroundings = qsurroundigs/T = delta H surroundings/ T
but I'm not sure where to go from here. Can anyone help?