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For the reaction of XO with O(2) to form...

For the reaction of XO with `O_(2)` to form `XO_(2)` the equilibrium constant at `398 K` is `1.0 xx 10^(-4) mol^(-1)`. If 1.0 mol of XO and 2,0 mol of `O_(2)` are placed in a 1.0 L flask and allowed to come to equilibrium, the equilibrium concentration of `XO_(2)` will be

A

`1.4 xx 10^(-2) mol L^(-1)`

B

`2.8 xx10^(-2) mol L^(-1)`

C

`2.8 xx 10^(-3) mol L^(-1)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
Units of K indicate that the reaction involved is ,
`{:(,2XO,+,O_(2),hArr ,2XO_(2)),("Initial conc. " (mol L^(-1)),1.0,,2.0,,-),("At equilibrium " (mol L^(-1)),(1.0-2x),,(2.0-x),,2x):}`
`K=([XO_(2)]^(2))/([XO]^(2)[O_(2)])=1.0xx10^(-4)mol^(-1)`
As value of K is very small , x is also very small
`1.0 -2x ~~ 1.0` and `2.0-x =2.0`
`1.0 xx10^(-4)=((2x)^(2))/((1.0)^(2)(2.0))impliesx=7.1xx10^(-3)`
`:. [XO_(2)]=2x=2xx7.1xx10^(-3)`
`=1.4xx10^(-2) mol L^(-1)`
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