Question

ΔS would be negative for which of the following reactions? 
(I) \( {CaCO}_3(s) \rightarrow {CaO}(s) + {CO}_2(g) \) 
(II) \( {Ag}^+(aq) + {Cl}^-(aq) \rightarrow {AgCl}(s) \)
(III) \( {N}_2(g) + 3{H}_2(g) \rightarrow 2{NH}_3(g) \)
Choose the correct answer from the codes given below:

• A. I and III only
• B. II and III only
• C. I only
• D. III only
• E. I, II, and III

Hint:

For reactions involving gases, if the number of moles of gas decreases, \( \Delta S \) is negative. If the number of moles of gas increases, \( \Delta S \) is positive.

Answer 1

The Correct Option is B

To determine whether \( \Delta S \) (change in entropy) is positive or negative, we consider the states of the reactants and products:
1. Reaction (I): 
- \( {CaCO}_3(s) \rightarrow {CaO}(s) + {CO}_2(g) \)
- The reaction involves the decomposition of a solid into a solid and a gas. Since gases have higher entropy than solids, \( \Delta S \) is positive.
2. Reaction (II): 
- \( {Ag}^+(aq) + {Cl}^-(aq) \rightarrow {AgCl}(s) \)
- The reaction involves the combination of aqueous ions to form a solid. The disorder decreases as the ions come together to form a solid, so \( \Delta S \) is negative.
3. Reaction (III): 
- \( {N}_2(g) + 3{H}_2(g) \rightarrow 2{NH}_3(g) \)
- The number of moles of gas decreases as reactants (4 moles of gas) combine to form products (2 moles of gas), so \( \Delta S \) is negative.
Thus, \( \Delta S \) will be negative for reactions (II) and (III).