Question

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of \(ICl\) was \(0.78 \ M\)?
\(2ICl (g) ⇋ I_2 (g) + Cl_2 (g); \ K_c = 0.14\)

Answer 1

Given:

Initial concentration of ICl = 0.78 M
Reaction:
2 ICl(g) ⇌ I₂(g) + Cl₂(g)
Equilibrium constant, K_c = 0.14

---

Step 1: Assume change in concentration

Let the equilibrium concentration of I₂ formed be x M.

Then, according to stoichiometry:
Decrease in ICl = 2x

Equilibrium concentrations:
[ICl] = (0.78 − 2x) M
[I₂] = x M
[Cl₂] = x M

---

Step 2: Write the equilibrium constant expression

K_c = ([I₂][Cl₂]) / [ICl]²

0.14 = x² / (0.78 − 2x)²

---

Step 3: Solve for x

x² = 0.14 (0.78 − 2x)²

Solving the equation,

x = 0.168 M

---

Step 4: Calculate equilibrium concentrations

[ICl] = 0.78 − 2(0.168) = 0.444 M

[I₂] = 0.168 M
[Cl₂] = 0.168 M

---

Final Answer:

Equilibrium concentration of ICl = 0.444 M
Equilibrium concentration of I₂ = 0.168 M
Equilibrium concentration of Cl₂ = 0.168 M