Question

Consider the following reaction at 1600°C.Br2(g) => 2 Br(g)When 1.05 moles of Br2 are put in a 0.980 L flask, 1.20 percent of the Br2 undergoes dissociation. Calculate the equilibrium constant Kc for the reaction.

Answer 1

Answer:

Kc = 6.18 × 10⁻⁴

Explanation:

Let's consider the following reaction.

Br₂(g) ⇄ 2 Br(g)

The initial concentration of Br₂ is:

$\frac{1.05mol}{0.980L} =1.07M$

To reach the equilibrium, 1.20% of 1.07 M reacts, that is 0.0128 M.

We can represent these changes through an ICE Chart. There are 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in the concentration.

            Br₂(g) ⇄ 2 Br(g)

I             1.07          0

C      - 0.0128     +2 × 0.0128

E           1.06        0.0256

The equlibrium constant (Kc) is:

$Kc=\frac{[Br]^{2} }{[Br_{2}]} =\frac{(0.0256)^{2} }{1.06} =6.18 \times 10^{-4}$