Question

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?

Answer 1

The percentage of I2 converted = 78.6 %

Step-by-step 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 %