What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2 ICl(g) ⇌ I₂(g) + Cl₂(g) ; K_C = 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

Hence, at equilibrium,
$[ICl] = [I_{2}] = 0.167 M [ICl] = (0.78 - 2 \times 0.167) M = 0.446 M$

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Similar questions

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2 I C l (g) ⇌ I_{2} (g) + C l_{2} (g); K_{C} = 0.25

I_{C l} \to I_{2} + C l_{2}

2 I C l \to I_{2} + C l_{2}; K_{c} = 0.14

I C L

0.6 M

I_{2}

K c

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M, [S O_{3}] = 1.90 M 2 S O_{2} + O_{2} ⇌ 2 S O_{3}

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