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Question

32 moles of HI(g) were heated in a sealed bulb at 444oC till the equilibrium was reached. Its 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 the equilibrium constant for the reaction 2HI(g)H2(g)+I2(g). Considering the volume of the container 1L.

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Solution

Equilibrium process is:
2HIH2+I2
Initial concentration (mol/L):
3200
At equilibrium:
322xxx
Since degree of dissociation, α=0.2 (20%) = fraction of reactant dissociated
α=2x/32
Given 0.2=2x/32
or, x=3.2
Therefore equilibrium concentrations (in 1L volume) are:
[HI]=322×3.2=25.6 M, [I2]=x=3.2 M=[H2]
Using the relation for equilibrium constant as:
KC=[H2][I2][HI]2
Upon substitution we get:
KC=[3.2][3.2](25.6)2
KC=0.0156 and number of moles of reactant and product present at equilibrium are:
[HI]=322×3.2=25.6 mol, [I2]=x=3.2 mol=[H2]

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