%3Cp%3E%3Cspan%20class%3D%22c1%22%3EFor%20the%20equilibrium%20%3C%2Fspan%3E%3Cbr
ID: 826676 • Letter: #
Question
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/ Answer
Kc = [PH3][BCl3]
Equilibrium starting with the solid PH3BCl3:
Kc = x^2
6.96*10^-5 = x^2
x = [PH3] = [BCl3] = 8.34*10^-3 M
Adding the extra BCl3:
initial: [PH3] = 8.34*10^-3 M, [BCl3] = (8.34*10^-3 + 0.05/(1.5*117.17)) M = 8.624*10^-3 M
change: [PH3] = -y, [BCl3] = -y
final: [PH3] = 8.34*10^-3 - y M, [BCl3] = 8.624*10^-3 - y M
6.96*10^-5 = ( 8.34*10^-3 - y)(8.624*10^-3 - y)
0 = y^2 - 1.6964*10^-2 y + 2.324*10^-6
By quadratic equation:
y = 0.0168 M (not possible since this is greater than 8.34*10^-3 M of PH3 we started with)
or
y = 1.38*10^-4 M
[PH3] = 8.34*10^-3-1.38*10^-4 = 8.202*10^-3 M
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