Question

At a certain temperature the equilibrium constant, Kc, equals 0.110 for the reaction: 2 ICl (g) ⇌ I2 (g) +Cl2 (g) What is the equilibrium concentration of ICl if 0.750 mol of I2 and 0.750 mol of Cl2 are initially mixed in a 1.00-L flask?

Answer 1

The equilibrium concentration of ICl is 2.26 M

Step-by-step explanation:

Chemical equilibrium is a state in which no changes are observed as time passes, despite the fact that the substances present continue to react. This is because chemical equilibrium is established when the forward and reverse reaction take place simultaneously at the same rate.

For the study of chemical equilibrium, the so-called equilibrium constant Kc is useful. Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is:

`Kc=([C]^(c) *[D]^(d) )/([A]^(a)*[B]^(b) )`

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

For the reaction:

2 ICl (g) ⇌ I₂ (g) +Cl₂ (g)

the constant Kc is:

`Kc=([I_(2) ]*[Cl_(2) ])/([ICl]^(2) )`

Being Kc =0.110 and the concentration being the amount of moles of solute that appear dissolved in each liter of the mixture and being calculated by dividing the moles of the solute by the liters of the solution:

• 
  `[I_(2) ]=(0.750 moles)/(1 L) =0.750 (moles)/(L)`

• 
  `[Cl_(2) ]=(0.750 moles)/(1 L) =0.750 (moles)/(L)`

and replacing in the constant we get:

`0.110=(0.750*0.750)/([ICl]^(2) )`

Solving, you get the ICl concentration at equilibrium:

`[ICl]^(2) =(0.750*0.750)/(0.110 )`

`[ICl] =\sqrt{(0.750*0.750)/(0.110 )}`

[ICl]= 2.26 M

The equilibrium concentration of ICl is 2.26 M