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The equilibrium constant Kp for the reac...

The equilibrium constant Kp for the reaction `N_(2)O_(4(g)) hArr 2NO_(2(g))` is 4.5. What would be the average molar naSs (in g/mol) of an eyuilibriunm mixture of `N_(2)O_(4) and NO_(2)` formed by the dissociation of pure `V_(2) O_(4)` at a total pressure of 2 atm ?

A

69

B

57.5

C

80.5

D

85.5

Text Solution

Verified by Experts

The correct Answer is:
B
`underset(1-alpha)underset(1)(N_(2)O_(4(g))) harr underset(2alpha)underset(0)(NO_(2(g)))`
Total no. of moles at equlibrium = `1+alpha`
`P_(N_(2)O_(4))=(1-alpha)/(1+alpha).P, P_(NO_(2))=(2alpha)/(1+alpha).P`
Hence, `Kp=(P_(NO_(2)))/(P_(N_(2)O_(4))) implies 4.5=(4alpha^(2))/((1-alpha^(2))) xx 2`
`implies alpha=0.6`
`D//d=1+(n-1)alpha=1.6 implies d=(92)/(1.6)=57.5`
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