Answer to Question #143991 in General Chemistry for gia

Question #143991
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.
1
Expert's answer
2020-11-14T13:56:04-0500

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

C(Br2)inic=1.07  MC(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 1.20100(1.07)=0.0128  M\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×(0.0128)=0.0256  M2 \times (0.0128) = 0.0256 \;M

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

Kc=(0.0256)21.06=6.18×104K_c =\frac{(0.0256)^2}{1.06} = 6.18 \times 10^{-4}


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