For the equilibrium

2IBr(g) ⇌ I₂(g) + Br₂(g) Kc = 8.50 × 10⁻³ at 150°C.

If 0.0600 mol of IBr is placed in a 1.0-L container, what is the partial pressure of I₂(g) in atm after equilibrium is reached?

If 0.0600 mol of IBr is placed in a 1.0-L container, the partial pressure of I₂(g) after equilibrium is reached is 0.176 atm.

0.0600 mol of IBr is placed in a 1.0-L container. The initial molar concentration is:

We will make an ICE chart for the following reaction.

        2 IBr(g)  ⇌  I₂(g)  +  Br₂(g)

I         0.060         0          0

C         -2x           +x         +x

E      0.060-2x      x           x

The concentration equilibrium constant (Kc) is:

The concentration (C) of I₂ at equilibrium is x = 0.00507 M. We can calculate its partial pressure (P) using the following expression.

where,

If 0.0600 mol of IBr is placed in a 1.0-L container, the partial pressure of I₂(g) after equilibrium is reached is 0.176 atm.

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Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

   Kc = 8.50 × 10⁻³

n(IBr) = 0.0600 mol

     V = 1.0 L

1. Calculate [IBr]

2. Set up an ICE table.

3. Calculate [I₂]

4. Convert the temperature to kelvins

T = (150 + 273.15) K = 423.15 K

5. Calculate p(I₂)

2b2t

explanation:

2b2t