Homework Answers
since volume is 2 L, concentration will be half of the number of mol
I2 + Br2 <------> 2IBr
0.200 0.200 1.05 (initial)
0.200-x 0.200-x 1.05+2x (at equilibiurm)
Qc= [IBr]^2 / [I2]/[Br2]
= (1.05)^2 / (0.200*0.200)
= 27.6
it is less than Kc. SO equilibrium will move to right
Kc = (1.05+x)^2 / (0.200-x)^2
110.25 = (1.05+2x)^2 / (0.200-x)^2
sqrt (110.25) = (1.05+2x) / (0.200-x)
10.5 = (1.05+2x) / (0.200-x)
2.10 - 10.5*x = 1.05 + 2x
12.5*x = 1.05
x = 0.084
so,
[Br2] = 0.200 -x = 0.200 - 0.084 = 0.116 M
[IBr] = 1.05+2x = 1.05 + 2*0.084 = 1.22 M
Answer: a
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