Question

$\Delta S$ will be the highest for which of the following reactions?

• A. $Ca\left(s\right)+\frac{1}{2}{O}_2\to CaO\left(s\right)$
• B. $CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$
• C. $N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)$
• D. All have same entropy

Answer 1

The correct option is B $CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$

Entropy change will be positive in the cases where the number of gaseous moles of the product is more than the number of gaseous moles of the reactants.
Which means, $△n_g>0$
where, $△n_g=n_p-n_R$
$n_P=$ total number of gaseous moles of products
$n_R=$ total number of gaseous moles of reactants

a) A solid $\left(Ca\right)$ and a gas $\left(O_2\right)$ combine to give the solid $CaO$. The value of $\Delta n_g<0$. Hence, the entropy will decrease during the reaction.

b)A solid $\left(CaC{O}_3\right)$ decomposes to give the solid $CaO$ and $C{O}_2$ gas. The value of $\Delta n_g>0$. Hence, the entropy will increase during the reaction.

c) $N_2$ and $O_2$ combine to give $NO$ gas. The value of $\Delta n_g=0$. Hence, the entropy will not change during the reaction.

Therefore, the entropy of $CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$ is maximum.