Answer
$$K_c = 6.18 \times 10^{-4}$$
Work Step by Step
1. Calculate the initial concentration of $Br_2$:
$$[Br_2]_{init} = \frac{1.05 \space moles}{0.980-L} = 1.07 \space M$$
2. At equilibrium, these are the concentrations:
$[Br_2] = 1.07 - x$
$[Br] = 0 + 2x$
3. According to the problem: $1.20\% = \frac{x}{1.07} \times 100\%$. Solving for x:
$$x = \frac{1.20 \times 1.07}{100} = 0.0128$$
4. Substituting:
$[Br_2] = 1.06 \space M$
$[Br] = 0.0256 \space M$
5. Write the expression for $K_c$ and evaluate it:
$$K_c = \frac{[Br]^2}{[Br_2]} = \frac{(0.0256)^2}{1.06} = 6.18 \times 10^{-4}$$
