Answer to Question #143991 in General Chemistry for gia

Question

Consider the following reaction at 1600 degree C.
Br2(g) <=> 2Br(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

We can calculate molar concentration of Br₂. No Br is present initially.

$C(Br_2)_{inic} = 1.07 \;M$

It is given that 1.20 % of the Br₂ undergoes dissociation. That is, the amount of Br₂ after dissociation is $\frac{1.20}{100}(1.07) = 0.0128 \;M$ . The equilibrium concentration of Br2 is (1.07 – 0.0128) = 1.06 M. The concentration change in Br is $2 \times (0.0128) = 0.0256 \;M$

$K_c =\frac{[Br]^2}{[Br2]}$

$K_c =\frac{(0.0256)^2}{1.06} = 6.18 \times 10^{-4}$