Question

The Kc for the following reaction is 56.2 at 305°C.

A2(g) + B2(g) 2 C(g)

If a reaction is initiated with 2.4 M of each substance, what is the equilibrium concentration of C?

a. 0.76 M

b. 5.7 M

c. 4.3 M

d. 1.2 M

e. 2.4 M

please show work

Answer 1

A₂(g) + B₂(g) 2 C(g)

IC: 2.4 2.4 0
C: -x -x 2x
EC: 2.4 - x 2.4 - x 2x

So,

Kc = [C]² / {[A₂] [B₂]}

56.2 = (2x)² / [ (2.4 - x) (2.4 - x) ]

56.2 = 4x² / (5.76 - 4.8x + x²)

56.2 (5.76 - 4.8x + x²) = 4x²

323.71 - 269.76x + 56.2x² = 4x²

323.71 - 269.76x + 52.2x² = 0

52.2x² - 269.76x + 323.71 = 0

Solving the quadratic equation, we get

x = 3.27 and x = 1.89

Since, x value can't be greater than 2.4, so x = 3.27 is discarded. Hence

x = 1.89

Now, equilibirium [C] = 2x = 2 x 1.89 M = 3.8 M