From the given expression, ∆G∘=+4.73KJ∕mol&RT=+2.5KJ∕mol. This gives T=300K Initially, PNO, →0 hence
PNO32
PN=0.
→0&ln[
PNO22
PN,04
]→−∞ Hence∆G expression is (-)ve & reaction moves in forward direction spontaneously. Mole fraction of N2O4 is much larger than that of NO2. Thus PN,o4>PNO, But for
PNO2
PN,O4
we can't say, hence comparison of Kp value =+ antilog [
−∆G0
RT
]=1.909 with Qp=
Pp2
PN,0,O4
is inclusive. Hence Qp>Kp or Qp<Kp or even Qp=Kp can be possible