Question

The equilibrium constant, Kc, for the reaction below is 0.10 at 25oC. Find the equilibrium concentration of chlorine gas, Cl2(g), if the equilibrium concentrations of ICl(g) and I2(g) are known to be 0.50 M and 0.40 M respectively.
2 ICl(g) → Cl2(g) + I2(g)

Answer 1

Answer:

The equilibrium concentration of chlorine gas, Cl₂(g), is 0.0625 M

Explanation:

Chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed, so that no changes are observed as time passes, despite the fact that the substances present continue to react with each other.

The mathematical expression that represents Chemical Equilibrium is known as the Law of Mass Action and is stated as: The ratio of the product of high concentrations to the stoichiometric coefficients in the reaction of products and reactants remains constant at equilibrium. For any reaction:

aA + bB ⇄ cC + dD

the equilibrium constant Kc is calculated as:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b}}$

In this case, you have:

2 ICl(g) → Cl₂(g) + I₂(g)

So, the equilibrium constant Kc is:

$Kc=\frac{[Cl_{2} ]*[I_{2} ]}{[ICl]^{2} }$

Being:

• Kc= 0.10
• [Cl₂]= ?
• [ICl]= 0.50 M
• [I₂]= 0.40 M

Replacing:

$0.1=\frac{[Cl_{2} ]*0.40 M}{(0.50 M)^{2} }$

Solving:

$0.1=\frac{[Cl_{2} ]*0.40 M}{0.25 M^{2} }$

0.1= 1.6 $\frac{1}{M}$* [Cl₂]

[Cl₂]= 0.1 ÷ 1.6 $\frac{1}{M}$

[Cl₂]= 0.0625 M

The equilibrium concentration of chlorine gas, Cl₂(g), is 0.0625 M