The equilibrium constant for the reaction H2 (g) + I2 (g) 2 HI (g)

Kp is 54.4. What percent of I2 (g) will be converted to HI (g) is 0.200 moled each of H2 (g) and I2 (g) are mixed and allowed to come to equlibrium in a 1.00 liter container?

The percentage of I2 converted = 78.6 %

Explanation:

Write down the given values

Kp = 54.4

The percentage of I2 will be converted to HI is 0.200 moled each of H2 and I2.

It should come to an equilibrium in 1.00 lit container.

Change in I2 (iodine) and H2 (hydrogen) = x in each

Change in HI = 2x

total ni (nickel)

number of moles = 0.2 - x + 0.2 - x + 2x

=0.4 moles

Mole fractions:

I2 = 0.2-x / 0.4 H2

=0.2-x / 0.4 HI

= 2x / 0.4

Kp = HI ^2 / H2 * I2

= (2x) ^2 / (0.2-x) ^2 = 54.4

by taking square root:

2x / 0.2-x = 7.375

= x=0.157

percentage of I2 converted = 78.6 %