Question

1) A(g) + 2B(g) → 2C(g)

If K_p for the above reaction is 4.50 x 10^-2, what is the value of Q if the pressures of A, B, and C are 0.250 atm, 0.124 atm, and 0.450atm respectively?

2) A(aq) + B(aq) → 2C(aq)

If the value of K_c for the reaction is 471, what is the concentration of C at equilibrium if initial concentrations of A and B are both 0.500 M. (Hint: Everthing is squared after you set-up the equilibrium expression with the values given.

Answer 1

Q1.

Calculate Kp

Kp = P-C / (P-A)(P-B)^2

Qp is the "actual" value

so:

Qp = (0.45) /((0.25)(0.124^2)) = 117.065

Since Qp > Kp

then expect equilibrium to shift toward products

Q2.

if

Kc = 471

find concentration of C if

[A] = 0.5

[B] = 0.5

[C] = 0

in equilibrium

[A] = 0.5 - x

[B] = 0.5 - x

[C] = 0 + 2x

Kc = [C]^2/ ([A][B])

471 = (2x)^2 / (0.5-x)(0.5-x)

sqrt(471) = 2x / (0.5-x)

sqrt(471) *  (0.5-x) = 2 x

21.702*0.5 - 21.702x = 2x

10.851 = 23.7x

x = 10.851 /23.7 = 0.457848

[C] = 2 x = 2*0.457848= 0.915696 M