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Initially 0.4 mole of N(2) and 0.9 mole ...

Initially 0.4 mole of `N_(2)` and 0.9 mole of `O_(2)` were mixed then moles of `N_(2),O_(2)` and NO at equilibrium are (approximate):

A

`0.4, 0.9, 6 xx 10^(-21)`

B

`4 xx 10^(-21) , 6 xx 10^(-21), 6 xx 10^(-21), 6 xx 10^(-21)`

C

0.4, 0.9, 0.9

D

`6 xx 10^(-21), 0.4, 0.9`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)(g) + O_(2)(g) ltimplies 2NO(g)`
`t=0, 0.4, 0.9`
`K_(c ) = ([NO]^(2))/([N_(2)][O_(2)])`
`[NO]^(2) = 0.4 xx 0.9 xx 10^(-40)`
[NO] = `6 xx 10^(-21)`
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