Question

Suppose a 500.mL flask is filled with 0.30mol of I2 and 0.60mol of HI . The following reaction becomes possible: +H2gI2g 2HIg The equilibrium constant K for this reaction is 0.247 at the temperature of the flask. Calculate the equilibrium molarity of H2 . Round your answer to one decimal place.

Answer 1

[I₂] = 0.3 mol / 0.5 L = 0.6 M

[HI] = 0.6 mol / 0.5 L = 1.2 M

H₂(g) + I₂(g) 2HI(g)
IC: 0 0.6 1.2
C: +x +x -2x
EC: x 0.6 + x 1.2 - 2x

Now,

K = [HI]² / [H₂] [I₂]

0.247 = (1.2 - 2x)² / [ x (0.6 + x)]

0.247 = (1.44 - 4.8x + 4x²) / (0.6x + x²)

0.247 (0.6x + x²) = (1.44 - 4.8x + 4x²)

0.247x² + 0.1482x = 1.44 - 4.8x + 4x²

3.753x² - 4.9482x + 1.44 = 0

Solving the quadratic equation, we get;

x = 0.88 and x = 0.43

Since for x = 0.88, 2x = 1.76, which is greater than 1.2 (initial concentration of HI).

So, x = 0.43 is accepted.

Hence, [H₂] = x = 0.43 M = 0.4 M