Question

3.2 moles of HI (g) were heated in a sealed bulb at ${\text{44}}{{\text{4}}^{\text{o}}}{\text{C}}$ 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 equilibrium constant for the reaction ${\text{2H}}{{\text{I}}_{\left( {\text{g}} \right)}} \rightleftharpoons {{\text{H}}_{\text{2}}}_{\left( {\text{g}} \right)}{\text{ + }}{{\text{I}}_{\text{2}}}_{\left( {\text{g}} \right)}$.Considering the volume of the container 1L.

Answer 1

Hint: The number of moles of all the species in the above reaction can be known from the degree of dissociation ($\alpha$) and from the initial moles of HI. Degree of dissociation is the extent of which the given molecules of the substance are dissociated or separated out.

Complete Solution :
In the question it is given that, initially 3.2 moles of HI was heated in a sealed tube to about a temperature of ${\text{44}}{{\text{4}}^{\text{o}}}{\text{C}}$, till it reaches the equilibrium and the degree of dissociation is given as 20% or it can be also taken as 0.2 as per the percentage calculation. And we have to find the equilibrium constant and number of moles of HI, H and I at equilibrium.

The given reaction is :${\text{2H}}{{\text{I}}_{\left( {\text{g}} \right)}} \rightleftharpoons {{\text{H}}_{\text{2}}}_{\left( {\text{g}} \right)}{\text{ + }}{{\text{I}}_{\text{2}}}_{\left( {\text{g}} \right)}$
At time, t=o	3.2 moles	0 moles	0 moles
After some time,t=t sec	3.2-(2x) moles	x moles	x moles

Moles dissociated is = $\dfrac{3.2\times 20}{100}$
Moles dissociated is = 0.64 moles = (2x)
So to find out the concentrations of all the species of the above reaction let us substitute the value of 2x at time t.

And the number of moles of HI, H and I at equilibrium is,
The number of moles of HI is, 3.2-(2x) = 3.2-0.64 = 2.56 moles
The number of moles of hydrogen is x, 0.64/2 = 0.32moles
The number of moles of iodine is x, 0.64/2 = 0.32moles.

And now we have to find the equilibrium constant of the reaction (${{K}_{c}}$), which can be calculated by the formulae,
${{K}_{c}}=\dfrac{\left[ {{H}_{2}} \right]\left[ {{I}_{2}} \right]}{{{\left[ HI \right]}^{2}}}$${{K}_{c}}$
\[{{K}_{c}}=\dfrac{\left[ 0.32 \right]\left[ 0.32 \right]}{{{\left[ 2.56 \right]}^{2}}}\]
${{K}_{c}}$ = 0.0156

Note: Since the degree of dissociation given is 20%, it means that 20% of 3.2 moles of HI has been dissociated and the final concentration remaining at equilibrium would be the 0.2 times that of 3.2 moles.