Question

Suppose a 250.mL flask is filled with 1.3mol of H2 and 0.30mol of I2 . The following reaction becomes possible: H2(g) + I2(g) <--> 2HI(g) The equilibrium constant K for this reaction is 3.99 at the temperature of the flask. Calculate the equilibrium molarity of H2 . Round your answer to two decimal places.

Answer 1

H2(g)    +    I2(g) <----------> 2HI(g)

5.2          1.2 M             0

5.2-x          1.2-x                     2x

K = (2x)^2 / (5.2-x)(1.2-x)

3.99 = 4x^2 / x^2 - 6.4x + 6.24

3.99 x^2 - 25.536 x + 24.90 = 4x^2

0.01 x^2 + 25.536 x - 24.90 = 0

x = 0.975

Equilibrium Molarity of H2 = 4.22 M