Question

A certain gas A polymerizes to a small extent at a given temperature and pressure, $nA \rightleftarrows {A_n}$. Show that the gas obeys the approx. equation $\dfrac{PV}{RT}=[1-\dfrac{(n-1){{K}_{c}}}{{{V}^{n-1}}}]$ . Whereas \[{{K}_{c}}={ }\dfrac{[{{A}_{n}}]}{{{[A]}^{n}}}\] and \[V\] is the volume of the container. Assume that initially one mole of A was taken in the container.

Answer 1

Hint: Polymerization is the process of production of polymers that is, repeating units of monomers by undergoing certain specific reactions.
A reversible reaction is a type of reaction in which the product which is formed from the reactant, can turn back to the reactant.

Complete answer:
In the given question, we can see that the reaction given which is given is $nA \rightleftarrows {A_n}$ which is a reversible reaction. Where $n$ is the number of moles of gas A, and ${{A}_{n}}$ is the polymerized product which is formed from the gas.
At time $t=0$ as our assumption we should take $nA$ as one mole and the mole of ${{A}_{n}}$ is zero
So now, accordingly at time $t={{t}_{equilibrium}}$ we get moles for $nA$ is $1-nx$ and ${{A}_{n}}$ has $x$ moles
$nA \rightleftarrows {A_n}$

At $t=0$	$1$	$0$
$t={{t}_{equilibrium}}$	$1-nx$	$x$

So total number of moles ${{n}_{T}}=1-nx+x=1-x(n-1)$
Formula for the concentration is moles divided by volume, we can mathematically express this equation as,
\[Concentration=\dfrac{mol}{volume}\]
Now, putting all the symbols, we get,
$[{{A}_{n}}]=\dfrac{x}{V}$
Where $x$ is the number of moles and $V$ is the volume of the gas.
${{[A]}^{n}}=[\dfrac{1-nx}{V}]$
We know that the value of equilibrium constant was provided to us in the question, so substituting in equilibrium constant \[{{K}_{c}}={ }\dfrac{[{{A}_{n}}]}{{{[A]}^{n}}}\]
Now we will put the values of numerator and denominator as calculated above, and we get,
\[{{K}_{c}}={ }\dfrac{[{{A}_{n}}]}{{{[A]}^{n}}}=\dfrac{[\dfrac{x}{V}]}{[\dfrac{1-nx}{V}]}\]
As per question we know it exerts very low amount so, we can neglect its value,
$1-nx\approx 1$
Then we \[{{K}_{c}}=x\times {{V}^{n-1}}\]
And now we take the volume term on the left side of the equation in order to get the value in terms of mole,
\[\dfrac{{{K}_{c}}}{{{V}^{n-1}}}=x\]
We know that the ideal gas equation is $PV=nRT$ , where $R$ is the gas constant whose value is known to us, and the $T$ denotes the value of temperature. Rest of the symbols have their former significance. Now if we rearrange the ideal gas equation,
$\dfrac{PV}{RT}={{n}_{total}}$
We know that the value of number of moles is,
${{n}_{T}}=1-x(n-1)$
Now we will substitute this value in the above rearranged equation, to get,
$\dfrac{PV}{RT}=1-x(n-1)\to (1)$
We got \[\dfrac{{{K}_{c}}}{{{V}^{n-1}}}=x\] substitute in equation$(1)$
We get $\dfrac{PV}{RT}=[1-\dfrac{(n-1){{K}_{c}}}{{{V}^{n-1}}}]$
Since this expression was given in the question and we justified it, hence this is the final required answer.

Additional information: An equilibrium constant is stated to be equal to the ratio of the forward and backward reaction rate constants has the dimension of concentration, but the equilibrium constant \[K\] is always dimensionless. \[{{K}_{p}}\] depends on partial pressure .
\[{{K}_{c}}\]= Equilibrium constant measured in moles per liter.
\[{{K}_{p}}\]= Equilibrium constant calculated from the partial pressures
\[\vartriangle G{ }<{ }0\] and \[{{Q}_{c}}<{{K}_{c}}\] or \[{{K}_{p}}\] at the start of the reaction: The reaction will proceed to form products.
\[\vartriangle G{ }={ }0\] and \[{{Q}_{c}}={{K}_{c}}\]or \[{{K}_{p}}\] at equilibrium and no changes in the concentration of the mixture.
\[\vartriangle G{ }>{ }0\] and \[{{Q}_{c}}>{{K}_{c}}\] or \[{{K}_{p}}\] after equilibrium: The reaction will proceed in the direction to form reactants.

Note:
-Concentration of a solution or a solvent can be expressed in many ways, one of which is molarity which is the number of moles of solute present per liter of solution.
-The ideal gas equation is applicable for every gas, but under certain specific conditions, in other words, the gases tend to deviate from their ideal behavior beyond certain specific conditions of temperature and pressure.