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2IC1 rarr I(2) + C1(2) K(C) = 0.14 In...

`2IC1 rarr I_(2) + C1_(2)` `K_(C) = 0.14`
Intitial concentration of IC1 is `0.6` M then equilibrium concentration of `I_(2)` is:

A

`0.37` M

B

`0.128` M

C

`0.224` M

D

`0.748` M

Text Solution

Verified by Experts

The correct Answer is:
B
`overset (2IC1) underset (0.6)` = `I_(2)` + `C1_(2)`
`06-2x` x x
`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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