#### Question

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) ⇌ I_{2}(g) + Cl_{2}(g) ; K_{C} = 0.14

#### Solution

The given reaction is: 2 ICl(g) ⇌ I_{2}(g) + Cl_{2}(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 - 0.748x`

`=> 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

Is there an error in this question or solution?

Advertisement

Advertisement

What is the Equilibrium Concentration of Each of the Substances in the Equilibrium When the Initial Concentration of Icl Was 0.78 M? Concept: Law of Chemical Equilibrium and Equilibrium Constant.

Advertisement