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  • During which of the following processes, does entropy decrease? (A) Freezing of water to ice at (B) Freezing of water to ice at (C) <i

During which of the following processes, does entropy decrease? (A) Freezing of water to ice at 0^{o}C (B) Freezing of water to ice at -10^{o}C (C) \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) (D) Adsorption of CO(g) on the lead surface. (E) Dissolution of NaCl in water. Choose the correct answer from the options given below :
 
Option: 1 (A), (C) and (E) only
Option: 2 (B) and (C) only
Option: 3 (A) and (E) only  
Option: 4 (A), (B), (C) and (D) only

Answers (1)

best_answer

\text { (A) Water } \stackrel{0^{\circ} \mathrm{C}}{\longrightarrow} \text { ice; } \Delta \mathrm{S}=-\text { ve }

\text { (B) Water } \stackrel{-10^{\circ} \mathrm{C}}{\longrightarrow} \text { ice } ; \Delta \mathrm{S}=-\mathrm{ve}

Freezing of water will decrease entropy as particles will move closer and forces of attraction will increase. This leads to decrease in randomness. So entropy decreases.

\text { (C) } \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{S}=-\mathrm{ve}

No. of molecules decreasing, will lead to decrease in the randomness of gaseous particles.

\text { (D) Adsorption; } \Delta \mathrm{S}=-\mathrm{ve}

Adsorption will lead to decrease in the randomness of gaseous particles.

\text { (E) } \mathrm{NaCl}(\mathrm{s}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) ; \Delta \mathrm{S}=+\mathrm{ve}

Solid to liquid increases in the randomness.

So,

In (A), (B), (C), and (D) processes entropy decreases.

Therefore, the Correct option is (4)

Posted by

Kuldeep Maurya

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