Question

One mole of ${{\text{N}}_2}{{\text{O}}_4}$ at $300{\text{K}}$ is kept in a closed container under $1{\text{atm}}$ pressure. It is heated to $600{\text{K}}$ when $20\%$ by mass of ${{\text{N}}_2}{{\text{O}}_4}$ decomposes to ${\text{N}}{{\text{O}}_2}\left( {\text{g}} \right)$ . The resultant pressure is:
A.$1.2{\text{atm}}$
B.$2.4{\text{atm}}$
C.$2{\text{atm}}$
D.$1{\text{atm}}$

Answer 1

To solve this question, first try to find out the final number of moles present in the container and use the ideal gas equation to compare initial and final conditions. Always remember that a closed container corresponding to that volume remains constant.

Complete step by step answer:
The balanced decomposition reaction of ${{\text{N}}_2}{{\text{O}}_4}$ to produce ${\text{N}}{{\text{O}}_2}\left( {\text{g}} \right)$ can be given as:
${{\text{N}}_2}{{\text{O}}_4} \to 2{\text{N}}{{\text{O}}_2}$ .
As we can see 1 mole of ${{\text{N}}_2}{{\text{O}}_4}$ decomposes to produce 2 mole of ${\text{N}}{{\text{O}}_2}\left( {\text{g}} \right)$ . But as it is given that only $20\%$ by mass of ${{\text{N}}_2}{{\text{O}}_4}$ decomposes, as mass of 1 mol of ${{\text{N}}_2}{{\text{O}}_4}$ is $92{\text{g}}$ and $20\% {\text{ of }}92{\text{g}} = 18.4{\text{g}}$ and in terms of moles we can say ${\text{moles}} = \dfrac{{18.4}}{{92}} = 0.2{\text{mol}}$ of ${{\text{N}}_2}{{\text{O}}_4}$ decomposes to produce $2 \times 0.2 = 0.4{\text{mol}}$ of ${\text{N}}{{\text{O}}_2}\left( {\text{g}} \right)$ . This can be given as:

Time	Moles of ${{\text{N}}_2}{{\text{O}}_4}$	Moles of ${\text{N}}{{\text{O}}_{2\left( {\text{g}} \right)}}$
Initially	1	0
final	$1 - 0.2 = 0.8$	$2 \times 0.2 = 0.4$

As it is given in question that the reaction is performing in a closed container, it corresponds to that volume remaining constant. Now the given data is initial temperature is $300{\text{K}}$ and final temperature is $600{\text{K}}$ , initial number of moles is 1 and final number of moles is $0.8 + 0.4 = 1.2{\text{mol}}$ , initial pressure is $1{\text{atm}}$ and final pressure we have to find out. This can be do so by using ideal gas equation:${\text{PV}} = {\text{nRT}}$ ,
Now as volume remain constant we can say ${{\text{V}}_1} = {{\text{V}}_2}$ , thereby it can be given as:
$\dfrac{{{{\text{n}}_1}{\text{R}}{{\text{T}}_1}}}{{{{\text{P}}_1}}} = \dfrac{{{{\text{n}}_2}{\text{R}}{{\text{T}}_2}}}{{{{\text{P}}_2}}}$ .
Putting the values in the equation we get: $\dfrac{{{\text{1}} \times {\text{R}} \times 300}}{1} = \dfrac{{{\text{1}}{\text{.2}} \times {\text{R}} \times 600}}{{{{\text{P}}_2}}}$
On solving the final pressure comes out to be $2.4{\text{atm}}$ .

Thus, the correct option is B.

Note:
The final pressure comes out to be $2.4{\text{atm}}$ which is the total pressure of both the gases ${{\text{N}}_2}{{\text{O}}_4}$ and ${\text{N}}{{\text{O}}_2}\left( {\text{g}} \right)$ . We can also find pressure of each gas known as partial pressure using total pressure of all the gases in a container and mole fraction of the gas.