Question

25 mL of ${ H }_{ 2 }$ and 18 mL of ${ I }_{ 2 }$ vapors were heated in a sealed glass tube at ${ 465 }^{ \circ }{ C }$ and at equilibrium ${ 30.8mL }$ of HI is formed. Calculate the percentage degree of dissociation of HI at ${ 465 }^{ \circ }{ C }$.

Answer 1

Hint: In a chemical equilibrium, the degree of dissociation is the fraction of molecules that have dissociated. To answer this question, you can consider the relation-
${ Degree\quad of\quad dissociation= }\dfrac { number\quad of\quad molecules\quad dissociated }{ total\quad number\quad of\quad molecules }$

Complete step by step answer:
As we know, that according to the Avogadro’s law-
The number of moles of a gas is directly proportional to its volume. The volumes in mL of gases can be used as the number of moles in the equilibrium constant expression.
In the question, the initial volume of hydrogen and iodine is given to us. So, let us consider that at equilibrium, x mL of hydrogen reacted with x mL of iodine to form 2x mL of HI. So, the volume at equilibrium will be the difference between initial volume and the volume reacted. We can write that –
\[{{H}_{2}}+{{I}_{2}}\to 2HI\]

	${{H}_{2}}$	${{I}_{2}}$	HI
Initial volume (mL)	25	18	-
Volume at equilibrium(mL)	25 - x	18 - x	2x

But in the question, the volume of HI at equilibrium is given to us as 30.8 mL.
So, we can write that –
 $\begin{align}
  & 2x=30.8 \\
 & x\text{ }=\frac{30.8}{2}=15.4mL \\
\end{align}$

Therefore, we can write that volume of ${ H }_{ 2 }$ at equilibrium = ${ (25-x) = (25-15.4) = 9.6mL }$
Volume of ${ I }_{ 2 }$ at equilibrium = (${ 18-x) = (18- 15.4) = 2.6mL }$

Now, let us consider V litre to be the total volume-
We know ${{K}_{c}}$ is the concentration of product divided by concentration of the reactants. So, we can write that -
${ K }_{ c }{ = }\dfrac { \left[ HI \right] ^{ 2 } }{ \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] }$ ……(1)
Now, if we put the values in equation (1), we will get
${ K }_{ c }{ = }\dfrac { \left( 30.8 \right) ^{ 2 } }{ \left( 18-15.4 \right) \left( 25-15.4 \right) }$
By solving this, the value of ${ K }_{ c }$ will be –
${ K }_{ c }$ = ${ 38.0 }$

Now, consider the reaction of decomposition of HI-
${ 2HI(g)\rightarrow H }_{ 2 }{ (g)+I }_{ 2 }{ (g) }$
${{K}_{c}}^{'}=\dfrac{1}{{{K}_{c}}}$
After putting the value of ${ K }_{ c }$ in the above formula, we get
${{K}_{c}}^{'}=\dfrac{1}{38.0}$
${{K}_{c}}^{'}=0.0263$

Let ‘y’ be the degree of dissociation,
If we start with 2 moles of HI, 0 moles of hydrogen and 0 moles of iodine, then at equilibrium, we will have 2(1-y) moles of HI, y moles of hydrogen and y moles of iodine will be present.
So,
${{K}_{c}}^{'}=\dfrac{\left[ {{H}_{2}} \right]\left[ {{I}_{2}} \right]}{{{\left[ HI \right]}^{2}}}$
${{K}_{c}}^{'}=\dfrac{\left( y/V \right)\left( y/V \right)}{\left( 2(1-y)/{{V}^{2}} \right)}$
${ 0.0263=\dfrac { { y }^{ 2 } }{ 4(1-y)^{ 2 } } }$
${ 0.105=\dfrac { { y }^{ 2 } }{ (1-y)^{ 2 } } }$
Taking square root on both the sides, we get
${ 0.3244=\dfrac { { y } }{ (1-y) } }$
${ 1-y=\dfrac { { y } }{ 0.3244 } }$
${ 1-y = 3.08y }$
${ 1 = 4.08y }$
${ y= }\dfrac { 1 }{ 4.08 } { \times 100 }$
y = ${ 24.5 }$%
Hence, the percentage degree of dissociation of HI at ${ 465 }^{ \circ }{ C }$ is ${ 24.5 }$% and this is the required answer.

Note: The solubility product can be calculated by using the concentration of the ions of the dissociated salt. It is dependent on other surrounding factors like temperature, pressure and nature of the electrolyte. These factors affect the concentration of the ions and thus affect the solubility product too.