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2ICl rarr I(2) + C1(2) K(C) = 0.14 In...

`2ICl rarr I_(2) + C1_(2)` `K_(C) = 0.14`
Initial concentration of ICl is `0.6` M then equilibrium concentration of `I_(2)` is:

A

0.37M

B

0.128M

C

0.224M

D

0.748M

Text Solution

Verified by Experts

The correct Answer is:
B
`underset(0.6)(2lCl)=I_(2)+Cl_(2)`
`0.6-2x" "x" "x`
`therefore K_(c)=0.14=(x^(2))/((0.6-2x)^2`
`0.37=(x)/(0.6-2x)`
`0.224-0.748x`
`" =x"`
`1.748x=0.224`
`x=0.128`
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