Question

Predict the sign of the entropy change, Δ S°, for each of the reaction displayed. Drag the appropriate items to their respective bins.

\(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{AgBr}(\mathrm{s})\)

\(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

\(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)

\(2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})\)

\(\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{~g})+6 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

\(\mathrm{I}_{2}(\mathrm{~s}) \rightarrow \mathrm{I}_{2}(\mathrm{~g})\)

Answer 1

Concepts and reason

The change of entropy $\left( {\Delta {{\rm{S}}^{\rm{o}}}} \right)$ is the difference in entropy of the products and the entropy of reactants.

$\Delta {{\rm{S}}^{\rm{o}}} = \Delta {{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right) - \Delta {{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)$

Its sign is positive if the number of gaseous particles on the product side is greater than the number of gaseous particles on the reactant side.

Fundamentals

The entropy is the amount of disorderness. In solid state, the particles are closely packed; hence, the disorderness is negligible. Liquids have more disorderness when compared to solids, but it is negligible when compared to gaseous particles.

Consider the following reaction:

$\,{\rm{A}}{{\rm{g}}^ + }\left( {aq} \right) + {\rm{B}}{{\rm{r}}^ - }\left( {aq} \right) \to {\rm{AgBr}}\left( s \right)$

Here, products have less entropy than reactants.

Hence, for this reaction, the sign $\Delta {{\rm{S}}^{\rm{o}}}$ is negative.

Consider the following reaction:

${\rm{Ca}}{\left( {{\rm{OH}}} \right)_2}\left( s \right) \to {\rm{CaO}}\left( s \right) + \,{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)$

Here, products have more entropy than reactants.

Hence, for this reaction, the sign $\Delta {{\rm{S}}^{\rm{o}}}$ is positive.

Consider the following reaction:

${\rm{2}}{{\rm{N}}_{\rm{2}}}{\rm{O}}\left( g \right) \to 2{{\rm{N}}_2}\left( g \right) + \,{{\rm{O}}_2}\left( g \right)$

Here, products have more entropy than reactants.

Hence, for this reaction, the sign $\Delta {{\rm{S}}^{\rm{o}}}$ is positive.

Consider the following reaction:

$\,2{\rm{Na}}\left( s \right) + \,{\rm{C}}{{\rm{l}}_2}\left( g \right) \to 2{\rm{NaCl}}\left( s \right)$

Here, products have less entropy than reactants.

Hence, for this reaction, the sign $\Delta {{\rm{S}}^{\rm{o}}}$ is negative.

Consider the following reaction:

$\,{{\rm{C}}_4}{{\rm{H}}_8}\left( g \right) + 6{\rm{ }}{{\rm{O}}_2}\left( g \right) \to \,4{\rm{C}}{{\rm{O}}_2}\left( g \right){\rm{ + 4}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)\,\,\,$

Here, products have more entropy than reactants.

Hence, for this reaction, the sign $\Delta {{\rm{S}}^{\rm{o}}}$ is positive.

Consider the following reaction:

${{\rm{I}}_2}\left( s \right) \to {{\rm{I}}_2}\left( g \right)$

Here, products have more entropy than reactants.

Hence, for this reaction, the sign $\Delta {{\rm{S}}^{\rm{o}}}$ is positive.

Ans:

Drag the appropriate items to their boxes as follows: