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For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2NO(g)+O2(g)↽−−⇀2NO2(g) 2NO(g)+O2(g)↽−−⇀2NO2(g) the standard change in Gibbs free energy is ΔG°=−72.6 kJ/molΔG°=−72.6 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are PNO=0.350 atm NO=0.350 atm , PO2=0.300 atm PO2=0.300 atm , and PNO2=0.950 atm PCO2=0.950 atm

Respuesta :

Answer:

ΔG = -64671.8 J/mol = -64.7 kJ/mol

Explanation:

Step 1: Data given

Temperature = 298 K

Pressure = 1 atm

ΔG°=−72.6 kJ/mol

PNO=0.350 atm

PO2=0.300 atm

PNO2=0.950 atm

Step 2: The balanced equation

2NO(g) + O2(g) ⇆ 2NO2(g)

Step 3: Calculate Q

Q = (PNO2)² / (PO2)*(PNO)²

Q = 0.950² / (0.300 * 0.350²)

Q = 24.56

Step 4: Calculate ΔG

ΔG = ΔG° + RT ln Q

⇒with ΔG = TO BE DETERMINED

⇒with ΔG° = -72.6 kJ/mol = -72600 J/mol

⇒with R = 8.314 J/mol*K

⇒with T = 298 K

⇒ ln Q = ln 24.56 = 3.20

ΔG = -72600 J/mol   + (8.314 J/mol * K * 298) * 3.20

ΔG = -64671.8 J/mol = -64.7 kJ/mol

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