Question:
Find \(\Delta n\) when 1 mol of each \(\text{NH}_{3(g)}\) and \(\text{HCl}_{(g)}\), reacts to form solid \(\text{NH}_4\text{Cl}\).
Find \(\Delta n\) when 1 mol of each \(\text{NH}_{3(g)}\) and \(\text{HCl}_{(g)}\), reacts to form solid \(\text{NH}_4\text{Cl}\).
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Count only gaseous moles: products minus reactants.
Updated On: Oct 1, 2026
- \(1\)
- \(-2\)
- \(-1\)
- \(-3\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
\(\Delta n_g\) is the change in moles of gas: moles of gaseous products minus moles of gaseous reactants. Solids and liquids are not counted.
Step 2: Detailed Explanation
The reaction is:
\[ \text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)} \]
Gaseous reactants = 1 + 1 = 2 mol. Gaseous products = 0 (the product is solid).
\[ \Delta n = 0 - 2 = -2 \]
The value +1 or -1 would come from counting the solid or miscounting a reactant.
Final Answer:
\(\Delta n = -2\), option (B). \[ \boxed{-2} \]
\(\Delta n_g\) is the change in moles of gas: moles of gaseous products minus moles of gaseous reactants. Solids and liquids are not counted.
Step 2: Detailed Explanation
The reaction is:
\[ \text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)} \]
Gaseous reactants = 1 + 1 = 2 mol. Gaseous products = 0 (the product is solid).
\[ \Delta n = 0 - 2 = -2 \]
The value +1 or -1 would come from counting the solid or miscounting a reactant.
Final Answer:
\(\Delta n = -2\), option (B). \[ \boxed{-2} \]
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