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What is the Equilibrium Concentration of Each of the Substances in the Equilibrium When the Initial Concentration of Icl Was 0.78 M? - Chemistry

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) ⇌  I2(g) + Cl2(g) ; KC = 0.14

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 - 0.748x`

`=> 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?
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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.
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