Part A
Using the following data, calculate Delta G at 298 K. Delta G (NO2(g))=51.84 kJ/mol. Delta G (NO2O4(g))=98.28 kJ/mol.

Part B
Calculate Delta G at 298K if the partial pressures of NO2 and N2O4 are 0.36atm and 1.63atm , respectively.

Please answer both questions Delta G in kJ format. Thank you :)

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4)  = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84  kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG =  ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

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