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# For which reaction from the following, $\Delta S$ will be maximum?(A) $Ca(s)+\dfrac{1}{2}{{O}_{2}}(g)\to CaO(s)$ (B) $CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)$ (C) $C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)$ (D) ${{N}_{2}}(g)+{{O}_{2}}(g)\to 2NO(g)$

Last updated date: 11th Aug 2024
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Hint: $\Delta S$ tells the change in the randomness of a system. Randomness increases as the number of moles increases. So we have to calculate the reaction in which the number of moles has increased.

Complete step by step solution:
$\Delta S$ is known as a change in entropy. The change in entropy is equal to the change in the randomness of the system. The randomness of the system is due to the molecules present in the system. As the number of molecules or the number of moles increases, the randomness increases. As a result, the change in entropy $\Delta S$ increases.
So, we have to calculate the $\Delta n$ (change in a number of moles) of the reaction. Change in the number of moles can be calculated by subtracting the number of moles of reactant from the number of moles product. Or we can say that,
$\Delta n={{n}_{P}}-{{n}_{R}}$
(a)- $Ca(s)+\dfrac{1}{2}{{O}_{2}}(g)\to CaO(s)$
In this reaction, the number of moles in the product is 0 and the number of moles in the reactant is 1/2.
$\Delta n={{n}_{P}}-{{n}_{R}}=0-\dfrac{1}{2}=-\dfrac{1}{2}$
So, $\Delta n$for this reaction is -1.
(b)- $CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)$
In this reaction, the number of moles in the product is 1 and the number of moles in the reactant is 0.
$\Delta n={{n}_{P}}-{{n}_{R}}=1-0=1$
So, $\Delta n$for this reaction is 1.
(c)- $C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)$
In this reaction, the number of moles in the product is 1 and the number of moles in the reactant is 1.
$\Delta n={{n}_{P}}-{{n}_{R}}=1-1=0$
So, $\Delta n$for this reaction is 0.
(d)- ${{N}_{2}}(g)+{{O}_{2}}(g)\to 2NO(g)$
In this reaction, the number of moles in the product is 1 and the number of moles in the reactant is 2.
$\Delta n={{n}_{P}}-{{n}_{R}}=1-2=-1$
So, $\Delta n$for this reaction is -1.
So, the $\Delta n$is maximum for reaction, $CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)$, so this reaction will have maximum $\Delta S$.

The correct answer is an option (b)- $CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)$.

Note: For calculating the change in the number of moles, only the moles of gases are considered. The moles of solids and liquids are not considered because an increase in randomness has a negligible effect.