Question

Given,0.1 mole of ${{N}_{2}}{{O}_{4}}$(s) was sealed in a tube under one atmosphere condition at 25 degrees. Calculate the number of moles of $N{{O}_{2}}$ (g) present, if the equilibrium ${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)[{{K}_{sp}}=0.14]$ is reached after some time :

Answer 1

Hint: Consider $\alpha $to be the degree of dissociation, and then analyse the reaction a time t=0 and the time at equilibrium. Calculate the partial pressures of the compounds and from that, find the number of moles.

Complete answer:
In order to answer our question, we need to learn about the equilibrium of a chemical reaction. In this case, the reaction given to us is conversion of ${{N}_{2}}{{O}_{4}}$ to $N{{O}_{2}}$. Now, we will find out the degree of dissociation first, with the data set given to us in the question. Now, degree of dissociation is the constant that helps us to determine the extent by which the reactant gets dissociated. More the degree of dissociation, more is the tendency of the reactant to dissociate more. Let us assume that the degree of dissociation of the following reaction is $\alpha $. When reaction is started, then the time is taken as 0. At that time, we have 0.1 moles of ${{N}_{2}}{{O}_{4}}$ present. As no $N{{O}_{2}}$ is present, its concentration is taken to be 0. After sometime, when the time to reach equilibrium is reached, then $0.1(1-\alpha )$ of ${{N}_{2}}{{O}_{4}}$ is left, and ($0.1\times 2\alpha $) moles of $N{{O}_{2}}$ is left. In this case, the total number of moles of the gas is calculated as:
\[\begin{align}
  & 0.1(1-\alpha )+0.1\times 2\alpha \\
 & =0.1(1+\alpha ) \\
\end{align}\]
Now, we are familiar with the equation $PV=nRT$, so, as the number of moles of the gas increase, the pressure also increases and vice versa. So, pressure will become $1+\alpha $, and now we will calculate the partial pressure of ${{N}_{2}}{{O}_{4}}$ and $N{{O}_{2}}$ which can be written as:
\[\begin{align}
  & {{p}_{{{N}_{2}}{{O}_{4}}}}=\dfrac{0.1(1-\alpha )}{0.1(1+\alpha )}\times (1+\alpha ) \\
 & {{p}_{N{{O}_{2}}}}=\dfrac{0.1\times 2\alpha }{0.1(1+\alpha )}\times (1+\alpha ) \\
\end{align}\]
As we know, that ${{K}_{p}}$ is the ratio of partial pressures of the product and the reactant, so we can say that:
     \[{{K}_{p}}=\dfrac{{{p}_{N{{O}_{2}}}}}{{{p}_{{{N}_{2}}{{O}_{4}}}}}\]
So, the number of moles will be $0.1\times 0.37$, which comes out to be 0.037 mole, which is the required answer for our question.

Note:
It is to be noted that ${{K}_{p}}$ is always constant for a reaction, it does not get changed if temperature is varied, because partial pressures do not depend on temperature.