At 80 degrees celsius, Kc = 1.87 x 10^-3 for the reaction, PH3BCl3 (s) <-----> P

Question

At 80 degrees celsius, Kc = 1.87 x 10^-3 for the reaction,

PH3BCl3 (s) <-----> PH3 (g) + BCl3 (g).

When a solid sample of phosphine-borontrichloride complex is placed in a closed vessel and decomposes until equilibrium, the initial concentrations of PH3 and BCl3 are 0.0432 M each.

When a flask has a volume of 0.500 L, the minimum mass of PH3BCl3 that must be added to the flask to reach equilibrium is 3.265g.

QUESTION: After the system reached equilibirum, an addition of 0.5g PH3BCl3 is added to the equilibrium mixture. Calculate the equilibirum concentrations of PH3 once equilibirum is again reached.

Steps would be helpful!

Explanation / Answer

(a)
Kp = Kc(RT)^(?n) where ?n = (moles of gaseous products) - (moles of gaseous reactants)

0.052 = Kc[(0.0821 atm-L/mol-K)(60 + 273K)]^(2 - 0)
Kc = 6.96e-5

(b)
Kc = [PH3][BCl3]

Equilibrium starting with the solid PH3BCl3:
Kc = x^2
6.96e-5 = x^2
x = [PH3] = [BCl3] = 8.34e-3 M

Adding the extra BCl3:
initial: [PH3] = 8.34e-3 M, [BCl3] = (8.34e-3 + 0.0135/0.5) M
change: [PH3] = -y, [BCl3] = -y
final: [PH3] = 8.34e-3 - y M, [BCl3] = 8.34e-3 + 0.0135/0.5 - y M

6.96e-5 = (8.34e-3 - y)(8.34e-3 + 0.0135/0.5 - y)
6.96e-5 = (8.34e-3 - y)(3.534e-2 - y)
0 = y^2 - 4.368e-2 y + 2.251e-4

By quadratic equation:
y = 3.77e-2 M (not possible since this is greater than 8.34e-3 M of PH3 we started with)
or
y = 5.97e-3 M

[PH3] = 8.34e-3 - 5.97e-3 = 2.37e-3 M

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