Question

The process with negative entropy change is;

• A. Dissociation of ${\mathrm{CaSO}}_4\left(\mathrm{s}\right)\mathrm{to}\mathrm{CaO}\left(\mathrm{s}\right)\mathrm{and}{\mathrm{SO}}_3\left(\mathrm{g}\right)$
• B. Dissolution of iodine in water
• C. Synthesis of ammonia from ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$
• D. Sublimation of dry ice

Answer 1

The correct option is C

Synthesis of ammonia from ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$

The explanation for the correct option:

Option (C): Synthesis of ammonia from ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$

Entropy change

• Entropy change is the measure of change of randomness or disorder in a thermodynamic system.
• It is represented as $∆{\mathrm{S}}_{\mathrm{system}}=S_{product}-S_{reac\tan t}$

where, $∆\mathrm{S}$ is entropy change.

• Its SI unit is joules per kelvin $\left({\mathrm{JK}}^{-1}\right)$
• As the disorder increases, entropy increases.
• The entropy of the gas is higher than liquid and the entropy of liquid is higher than solid.
• $$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right) \\ 4\mathrm{moles}2\mathrm{moles} \end{gathered}$$
• Here, $$\begin{gathered} ∆\mathrm{S}=2-4 \\ =-2 \end{gathered}$$
• That is, $∆\mathrm{S}=-\mathrm{ve}$
• $∆\mathrm{S}={\mathrm{S}}_{\mathrm{reactant}}>{\mathrm{S}}_{\mathrm{product}}$
• So, entropy decreases, and this is a negative entropy change.

The explanation for the incorrect options:

Option (A): Dissociation of ${\mathrm{CaSO}}_4\left(\mathrm{s}\right)\mathrm{to}\mathrm{CaO}\left(\mathrm{s}\right)\mathrm{and}{\mathrm{SO}}_3\left(\mathrm{g}\right)$

• Here, a solid is converted into solid and gas.
• $$\begin{gathered} ∆\mathrm{S}={\mathrm{S}}_{\mathrm{product}}>{\mathrm{S}}_{\mathrm{reactant}} \\ ∆\mathrm{S}=+\mathrm{ve} \end{gathered}$$
• Thus, entropy increases, and this is a positive entropy change.

Option (B): Dissolution of iodine in water

• Here, solid is converted into liquid.
• $$\begin{gathered} ∆\mathrm{S}={\mathrm{S}}_{\mathrm{product}}>{\mathrm{S}}_{\mathrm{reactant}} \\ ∆\mathrm{S}=+\mathrm{ve} \end{gathered}$$
• Thus, entropy increases, and this is a positive entropy change.

Option (D): Sublimation of dry ice

• Here, ice is converted into gas.
• $$\begin{gathered} ∆\mathrm{S}={\mathrm{S}}_{\mathrm{product}}>{\mathrm{S}}_{\mathrm{reactant}} \\ ∆\mathrm{S}=+\mathrm{ve} \end{gathered}$$
• Thus, entropy increases, and this is a positive entropy change.

Hence, the correct option for the process with negative entropy change is an option (C): Synthesis of ammonia from ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$