Question

Under constant temperature conditions, the equilibrium constant (Kc) value was measured to be 0.85 for the following chemical reaction: 2 ICl (g) ⇋ I2 (g) + Cl2 (g). Please determine the equilibrium concentration of ICl if 0.57 mol of I2 and 0.57 mol of Cl2 are mixed in a 3.0-L flask with no ICl present initially.(A) 0.067 M(B) 0.40 M(C) 0.20 M(D) 0.10 M(E) 0.13 M

Answer 1

Answer:

E) 0,13 M.

Explanation:

For the equilibrium reaction:

2 ICl (g) ⇋ I₂ (g) + Cl₂(g)

The equilibrium constant, kc, is:

$K_{c} = \frac{[I_{2}][Cl_{2}]}{[ICl]^2}$ (1)

The equilibrium moles are:

[I₂] = 0,57 mol - x

[Cl₂] = 0,57 mol -x

[ICl] = 2x

-Where x are the moles that react-

Replacing in (1):

$0,85 = \frac{[0,57-x][0,57-x]}{[2x]^2}$

2,4x² + 1,14x - 0,3249 = 0

Solving:

X =-0,675 -No chemical sense. There are not negative concentrations-

X = 0,200 -Real answer-

Thus, moles of ICl are 2×0,200 = 0,400 moles

As the volume is 3 L

The concentration of ICl in equilibrium is 0,400mol/3L = 0,13 M

I hope it helps!

Answer 2

Answer:

E) [ICI] ≅ 0.13 M

Explanation:

• 2 ICI(g) ↔ I2(g)    +    Cl2(g)

  ni:      0           0.57          0.57

  nf:      X       0.57 - X     0.57 - X    

∴ Kc = 0.85

⇒ Kc = [I2] * [Cl2] / [ICI]² = 0.85

⇒ 0.85 = (((0.57-X)/3.0)*((0.57-X)/3.0)) / X²

⇒ 0.85 = ((0.57-X)/3.0)² / X²

⇒ 0.922 = ((0.57-X)/3.0)) / X

⇒ 0.922X = (0.57 - X) / 3.0

⇒ 3.0*(0.922X) = 0.57 - X

⇒ 2.766X = 0.57 - X

⇒ 2.766X + X = 0.57

⇒ 3.766X = 0.57

⇒ X = 0.57/3.766

⇒ X = 0.151 M

⇒ [ICI] = 0.15 M ≅ 0.13 M