At 273K and 1atm $\alpha $ L of \[{{N}_{2}}{{O}_{4}}(g)\to 2N{{O}_{2}}(g)\].To what extent has the decomposition proceeded when the original volume is 25% less than that of existing volume ?
(A) $\alpha $ = 40%
(B) $\alpha $ = 10%
(C) $\alpha $ = 33%
(D) $\alpha $ = 67%

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Hint: The initial concentration of ${{N}_{2}}{{O}_{4}}$ could be taken as 100 to proceed with the solution. The conversion of ${{N}_{2}}{{O}_{4}}$ to $N{{O}_{2}}$ is exothermic, because ${{N}_{2}}{{O}_{4}}$ is more stable than $N{{O}_{2}}$, therefore it requires energy for conversion.

Complete step by step answer:
Let the initial volume of ${{N}_{2}}{{O}_{4}}$ be x , concentration of $N{{O}_{2}}$will be zero,
${{N}_{2}}{{O}_{4}}\to 2N{{O}_{2}}$
After time t at equilibrium, the concentration of ${{N}_{2}}{{O}_{4}}$ will be , $x(1-\alpha )$ and concentration of $N{{O}_{2}}$ will become, $2\alpha x$, where $\alpha $ is the de
Total initial volume = x + 0 = x
Total equilibrium volume = $x(1-\alpha )$+$2\alpha x$=$x(1+\alpha )$
According to the given in the question, the initial volume is 25% less than the final volume,
Let us assume that the initial concentration is given 100, so when it is 25% less than the final volume, it will become 75.
\[\begin{align} & x=75\times (1+\alpha ) \\ & \Rightarrow 100=75\times (1+\alpha ) \\
 & \Rightarrow (1+\alpha )=1.33 \\
 & \Rightarrow \alpha =0.33 \\
Therefore, the obtained value of degree of dissociation will be 0.33 and in terms of percentage it will be 33%.
Hence, the correct answer is (C) option.
Note: Decomposition of ${{N}_{2}}{{O}_{4}}$ and hydrogen peroxide is a first order reaction. The conversion of $N{{O}_{2}}$ to ${{N}_{2}}{{O}_{4}}$ is endothermic. Nitrogen dioxide is a toxic gas with pungent odour, it is reddish brown in color. When cooled, the liquid freezes to form a colorless crystalline solid that consists almost entirely of the dimeric form, ${{N}_{2}}{{O}_{4}}$. Nitrogen tetroxide is used as an oxidizer in one of the most important rocket propellants because it can be stored as a liquid at room temperature.