Question

3.2 moles of HI(g) were heated in a sealed bulb at 444C till the equilibrium was reached. Its degree of dissociation was found to be 20%.Calculate the number odf hydrogen iodide,hydrogen and iodide present at the equilibrium point and determine the value of equilibrium constant.
Reaction: 2HI-->H2+I2.
Volume of container 1L.

Answer 1

Concentration of HI is 3.2moles/1litrre=3.2

Degree of dissociation=20%

Degree of dissociation=no.of moles of reactant dissociated/no.of moles of reactants present initially 20/100=x3/2

x=0.64moles

According to the reaction

2Hl(g)=H2(g)+l2(g)

Two M of Hl gives one M of H2 & l2 each so 64M of HI would give 32M OF H2 & I2 each hence equilibrium concentration of HI is 2.56M

Kc=(0.32)(0.32)/(2.56)2=0.0156