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The degree of dissociation of N(2) O(2) ...

The degree of dissociation of `N_(2) O_(2)` into `NO_(2)` at one atmosphere and `40^(@)`C is 0.310. For this :

A

`K_(P)` = 0.42 atm

B

`K_(c)` = 0.003 mole `lit^(-)`

C

Degree of dissociation at 10 atm pressure at same temperature `alpha` = 0.1025

D

Degree of dissociation (`alpha`) decreases with increase pressure a given temperature

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`underset(underset(0.7)(1-0.3))underset(1)(N_(2)O_(4(g))) harr underset(underset(0.6)(0.6))underset(0)(2NO_(2(g)))`
Total moles = 0.7 +0.6=1.3
`Kp=(P^(2)NO_(2))/(PN_(2)O_(4))=(((0.6)/(1.3))^(2))/((0.7)/(1.3))=(0.6 xx 0.6)/(1.3 xx 0.7)`
`=0.4 atm implies Kp=Kc(RT)^(Delta n_((g)))`
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