A mixture of 3.00 mol of Cl2 and 2.10 mol I2 is initially placed into a rigid 1.00-L container at 350 °C. When the mixture has come to equilibrium, the concentration of ICl is 2.32 M. What is the equilibrium constant for this reaction at 350 °C? Cl2 (g) + I2 (g) ⇌ 2 ICl (g)

The equilibrium constant is 3.11

Explanation:

Step 1: Data given

Number of moles Cl2 = 3.00 moles

Number of moles I2 = 2.10 moles

Volume = 1.00L

Temperature = 350 °C

Concentration of ICl at the equilibrium = 2.32 M

Step 2: The balanced equation

Cl2 + I2 → 2ICl

Step 3: Calculate initial concentrations

[Cl2] = 3.00 moles / 1L = 3.00 M

[I2] = 2.10 moles / 1L = 2.10 M

[ICl] = 0M

Step 4: Calculate concentration at equilibrium

[Cl2] = 3.00 - X M

[I2] = 2.10 - X M

[ICl] = 2X = 2.32 M

X = 2.32 / 2 = 1.16 M

[Cl2] = 3.00 - 1.16 M = 1.84 M

[I2] = 2.10 - 1.16 M = 0.94 M

[ICl] = 2X = 2.32 M

Step 5: Calculate Kc

Kc = [ICl]²/[Cl2][I2]

Kc = 2.32² / (1.84*0.94)

Kc = 3.11

The equilibrium constant is 3.11

Kc = [ICl]² / [Cl₂]. [I₂] →2.32² / 1.84. 0.94 = 3.11

Explanation:

Let's propose the equilibrium reaction:

Cl₂ (g) + I₂ (g) ⇄ 2ICl (g)

Initial 3 m 2.10m -

React x x 2x

X amount has reacted of chlorine and iodine, so by stoichiometry, we made 2X of ICl.

As we have this molar concentration we can determine the x → 2x = 2.32. Then x = 2.32 / 2 = 1.16

Eq (3 m - 1.16) (2.10-1.16) 2.32

Molar concentrations in the equilibrium are: [Cl₂] = 1.84M, [I₂] = 0.94M

Let's make the expression for Kc:

Kc = [ICl]² / [Cl₂]. [I₂] →2.32² / 1.84. 0.94 = 3.11