Find ΔrG for reaction (in kJ mol-1) at 298 K?
2 NO2 (g) ⇌ N2O4 (g)
Given Conditions:
Pressure: NO2 = 1.25 bar
Pressure: N2O4 = 0.65 bar
N2O4(g) --
Kp = PN2O4/PNO22
= 0.65/1.252
= 0.416
Now, rG = -RT ln(Kp)
i.e. rG = -(8.314/1000) kJ/mol.K * 298 K * ln(0.416)
Therefore, rG = 2.173 kJ/mol
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