Initial concentration of OCl2 = mol of OCl2 / volume in L
= 0.90 mol / 0.500 L
= 1.8 M
Initial concentration of BrOCl = mol of BrOCl / volume in L
= 0.20 mol / 0.500 L
= 0.40 M
Initial concentration of BrCl = mol of BrCl / volume in L
= 1.4 mol / 0.500 L
= 2.8 M
ICE Table:
Equilibrium constant expression is
Kc = [BrOCl]*[BrCl]/[OCl2]*[Br2]
0.798 = (0.4-1*x)(2.8-1*x)/((1.8 + 1*x)(1*x))
0.798 = (1.12-3.2*x + 1*x^2)/(1.8*x + 1*x^2)
1.436*x + 0.798*x^2 = 1.12-3.2*x + 1*x^2
-1.12 + 4.636*x-0.202*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -0.202
b = 4.636
c = -1.12
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 20.59
roots are :
x = 0.2442 and x = 22.71
x can't be 22.71 as this will make the concentration negative.so,
x = 0.2442
At equilibrium:
[Br2] = +1x = +1*0.2442 = 0.244 M
Answer: 0.24 M
Suppose a 500. mL flask is filled with 0.90 mol of OC1,, 0.20 mol of BroCl...
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