Question

$\Delta S^o$ will be highest for which of the following reaction?

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

Answer 1

Answer: (B)

${CaCO}_3\left(s\right)\to {CaO\left(s\right)+CO}_2\left(g\right)$

Solution:- (B) ${CaC{O}_3}_{\left(s\right)}⟶{CaO}_{\left(s\right)}+{C{O}_2}_{\left(g\right)}$

The entropy of a system increases whenever its particles have more freedom of motion.

Thus, the entropy increases whenever you have more moles of gaseous products than of reactants and whenever you have more product particles in solution than you have of reactant particles.

Hence among the given reactions, the increase in entropy is associated with-

${CaC{O}_3}_{\left(s\right)}⟶{CaO}_{\left(s\right)}+{C{O}_2}_{\left(g\right)}$

Since the reactant side has no gaseous particle while the product side has $1$ gaseous particle.

Hence the entropy increases for the reaction and will be highest.