Question

3.2 moles of HI(g) is heated in a sealed bulb at 444°C till the equilibrium was reached. It's degree of dissociation was found to be 20%.
calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of equilibrium constant for the reaction 2HI(g) H2(g)+I2(g). Considering the volume of the container 1L

Answer 1

⭕️Concentration of HI is 3.2 moles/1litre 
=3.2M 

⭕️degree of dissociation= 20% 

⭕️degree of dissociation = no of moles of reactants dissociated/no. of moles of reactants present initially 
20/100=x/3.2 

x=0.64 moles 

2HI(g)=H2(g)+I2(g) 

⭕️two M of HI gives one M of H2 and I2 each so....

64 M of HI would give 32 M of H2 and I2 each


⭕️hence equilibrium concentration of HI is 2.56M 

⭕️2.56 is obtained by 3.2 - 0.64 which is number of moles left at equilibrium.


Kc=[0.32][0.32]/[2.56]2 
=0.0156

Answer 2

Heya....

===== Answer ======

Concentration of HI = 3.2 Moles /litre or 3.2 M

Degree of dissociate = 20%

= no of Moles of reactant dissociated / no of Moles of reactant initially

= 20/100 = X/3.2

So...

2 HI (g) = H2(g) + 12 g

2 Moles of HI gives 1 mole of H2 & 12 each so...

64 Moles of HI give 32 Moles of H2

Equilibrium concentration is = 2.56

KC = [0.32][0.32]/ [2.56]2 = 0.0156

Thank you