An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 670 K in a % litre flask contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. Calculate equilibrium constant.

Reaction will be given as:
H2 (g) + I2 (g) = 2 HI (g)
Therefore, the equilibrium constant K will be given as:
K = [HI] 2 / [H2] [I2]
The molar concentrations of the species at equilibrium are:
[H2] = 0.4 / 5 = 0.08 mol L-1
[I2] = 0.4 / 5 = 0.08 mol L-1
[HI] = 2.4 / 5 = 0.48 mol L-1
Substituting these values in the expression of K, we will get:
K = (0.48) 2 / (0.08) (0.08)
K = 36.0
Hence the equilibrium constant is equal to 36.0

Category: Free Energy and Chemical Equilibria

More Questions

1 Response to " An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 670 K in a % litre flask contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. Calculate equilibrium constant. "

  1. hamza says:

    how is 0.4(the no. of moles of hydrogen and iodine) are being divided by 5. i mean from where 5 has come?

Leave a Reply

Copyright © All rights reserved. TheBigger.com | Privacy Policy | Contact Us | Copyright Policy | Useful Resources