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Question
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
\[\ce{2 ICl(g) ⇌ I2(g) + Cl2(g)}\]; KC = 0.14
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Solution
The given reaction is:
| 2 ICl(g) | ⇌ | I2(g) | + | Cl2(g) | |
| Initial conc. | 0.78 M | 0 | 0 | ||
| At equilibrium | (0.78 - 2x) M | x M | x M |
Now we can write, `(["I"_2]["Cl"_2])/["ICl"]^2 = "K"_"C"`
`=> (x xx x)/(0.78 - 2x)^2 = 0.14`
`=> x^2/(0.78 - 2x)^2` = 0.14
`=> x/(0.78 - 2x) = 0.374`
`=> x= 0.292 - 0.748 x`
`=> 1.748 x = 0.292`
⇒ x = 0.167
Hence, at equilibrium,
[ICl] = [I2] = 0.167 M[ICl]
= (0.78 - 2 × 0.167)M
= 0.446 M
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