A mixture containing 0.392 M A(g) and 0.452 M B(g) is allowed to come to equilibrium at 300 K. The reaction 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs. At equilibrium, [C] = 0.00128 M. What is the value of Kc?

Question

Quizplus · Accepted Answer

First, let's set up the equilibrium expression:
Kc = ([C][D])/([A]^3[B]^2)

We are given [C] = 0.00128 M at equilibrium. However, we don't know [D], [A], or [B].

We can use an ICE chart to figure out the equilibrium concentrations of A, B, C, and D.

Initial:    0.392 M      0.452 M        0            0 
Change:    -3x             -2x            +x         +x 
Equilibrium: 0.392-3x  0.452-2x     x           x

We can plug these equilibrium concentrations into the Kc expression:
Kc = ([C][D])/([A]^3[B]^2)
Kc = (0.00128)(x)/(0.392-3x)^3(0.452-2x)^2

To solve for x, we can use the quadratic formula since there is no simple way to solve for x in this equation.

After doing the math, we get x = 0.00000416 M.

Now we can plug in all the equilibrium concentrations into the Kc expression and solve:
Kc = ([C][D])/([A]^3[B]^2)
Kc = (0.00128)(0.00000416)/(0.392-3(0.00000416))^3(0.452-2(0.00000416))^2

Kc = 1.39 × 10^-4

Therefore, the answer is B.

A Mixture Containing 0