Answer:-
Given:-
partial pressure of NO gas (PNO) = 0.350 atm
partial pressure of O2 gas (PO2) = 0.500 atm
partial pressure of NO2 gas (PNO2) = 0.900 atm
temperature (T) = 298 K
standard free energy change of reaction (G0) = - 72.6 kJ.mol-1 = - 72600 J.mol-1
free energy change of reaction (G) = ?
As we know that
gas constant (R) = 8.314 J.K-1.mol-1
Also we know that
2NO(g) + O2(g) 2NO2(g)
Equilibrium constant (Kp) = P2NO2 / P2NO PO2
Equilibrium constant (Kp) = (0.900 )2 / (0.350)2 0.500
Equilibrium constant (Kp) = 0.81 / 0.1225 0.500
Equilibrium constant (Kp) = 0.81 / 0.06125
Equilibrium constant (Kp) = 13.22
According to formula
G = G0 + 2.303RTlog(Kp)
then
free energy change of reaction (G) = standard free energy change of reaction (G0) + 2.303RTlog(Kp)
free energy change of reaction (G) = (- 72600 J.mol-1 ) + 2.303 8.314 J.K-1.mol-1 298 K log(13.22)
free energy change of reaction (G) = (- 72600 J.mol-1 ) + 2.303 8.314 J.K-1.mol-1 298 K 1.12
free energy change of reaction (G) = (- 72600 J.mol-1 ) + 6390.55 J.mol-1
free energy change of reaction (G) = - 72600 J.mol-1 + 6390.55 J.mol-1
free energy change of reaction (G) = - 66209.45 J.mol-1
free energy change of reaction (G) = - 66.21 kJ.mol-1 (i.e the answer)
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