    Question

# Consider the following equilibrium in a closed container N2O4(g)⇌2NO2(g) At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant Kp and degree of dissociation (α)?

A
neither Kp nor α changes
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
both Kp and α change
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Kp changes, but α does not changes
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Kp does not change, but α changes
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D Kp does not change, but α changes At constnnt temperature Kp or Kc remains constant. For the equilibria: N2O4(g)⇌2NO2(g) Kp=Kc×(RT)△n Kp=Kc because here △n=1 Since temperature is constant so Kp or Kc will remain constant. Further since volume is halved, the pressure will be doubled so α will decrease so as to maintain the constancy of Kp or Kc. N2O4⇌2NO2initial 10at equilibrium(1−α)2α ∴ Total mole=1−α+2α=1+α Let total pressure = P ∴pN2O4=1−α1+α.P pNO2=2α1−αP Kp=p2NO2pN2O4=4α2×P(1−α)(1+α)=4α2P1−α2 Since Kp= constant, so α∝1√P So when volume is halved, pressure gets doubled and thus α will decrease.  Suggest Corrections  0      Similar questions  Explore more