Question

Which of the following processes show a decrease in entropy of the system$?$
(i) $2NO\,\left( g \right)\, + \,{O_2}\,\left( g \right)\, \to \,2N{O_2}\,\left( g \right)$
(ii) $COC{l_2}\,\left( g \right)\, \to \,CO\,\left( g \right)\, + \,C{l_2}\,\left( g \right)$
(iii) $C{H_3}OH\,\left( l \right)\, \to \,CO\,\left( g \right)\, + \,2{H_{2\,}}\,\left( g \right)$
(iv) $NaCl{O_{3\,}}\,\left( s \right)\, \to \,N{a^ + }\,\left( {aq} \right)\, + \,Cl{O_3}^ - \,\left( {aq} \right)$

Answer 1

Hint: Entropy is a measurable physical property that is associated with a state of disorder and randomness. The state of disorder is maximum for gases, followed by liquids than gases. Consider the reactants and products in each process and check in which process the randomness is decreasing on the product side.

Complete step-by-step answer:
Entropy is the measure of degree of disorder or randomness of a system and the entropy of a system increases whenever its particles have more freedom of motion. Thus, the entropy increases when there are more moles of gaseous products on the product side than on the reactant side. Also if there are more product particles in solution than the reactant particles then the entropy increases. In the case of the opposite situation entropy will decrease.

$2NO\,\left( g \right)\, + \,{O_2}\,\left( g \right)\, \to \,2N{O_2}\,\left( g \right)$
On the reactant side there are $2\,moles$of $NO$ and $1\,mole$ of ${O_2}$ in the gaseous phase. Hence the total number of moles in the gaseous phase on the reactant side $ = \,3$
While on the product side there are $2\,moles$of $N{O_2}$in the gaseous phase. Hence the total number of moles in the gaseous phase on the product side $ = \,2$. Hence the entropy is decreasing.

$COC{l_2}\,\left( g \right)\, \to \,CO\,\left( g \right)\, + \,C{l_2}\,\left( g \right)$
On the reactant side there is $1\,mole$of $COC{l_2}$ in the gaseous phase. Hence the total number of moles in the gaseous phase on the reactant side $ = \,1$
While on the product side there are $1\,mole$ of $CO$and $1\,mole$ of $C{l_2}$ in the gaseous phase. Hence the total number of moles in the gaseous phase on the product side $ = \,2$. Hence the entropy is increasing.

$C{H_3}OH\,\left( l \right)\, \to \,CO\,\left( g \right)\, + \,2{H_{2\,}}\,\left( g \right)$
On the reactant side there are no moles present in the gaseous phase. Hence the total number of moles in the gaseous phase on the reactant side $ = \,0$ While on the product side there are $1\,mole$ of $CO$and $1\,mole$ of ${H_2}$ in the gaseous phase. Hence the total number of moles in the gaseous phase on the product side $ = \,2$.Hence the entropy is increasing.

$NaCl{O_{3\,}}\,\left( s \right)\, \to \,N{a^ + }\,\left( {aq} \right)\, + \,Cl{O_3}^ - \,\left( {aq} \right)$
$NaCl{O_3}$ is undergoing dissociation. So on the product side there are more particles present in the solution compared to the reactant side.Hence the entropy is increasing.

So the correct answer is (i) $2NO\,\left( g \right)\, + \,{O_2}\,\left( g \right)\, \to \,2N{O_2}\,\left( g \right)$.

Note: The most important step in these types of questions is to find the number of moles of reactants and products in the different phases. If the calculation of the number of moles goes wrong there are high chances you will make errors in interpreting the result and hence marking the wrong option.