\(NH_{3(g)} + HCl_{(g)} → NH_4Cl_{(s)}\)
The entropy change in the system above is
- positive
- zero
-
negative
- indeterminate
The correct answer is: C
Explanation
\(NH_{3(g)} + HCl_{(g)} → NH_4Cl_{(s)}\)
Reactant = 2 moles of gaseous molecules
Product = 1 mole of gaseous molecule
Entropy change = 1 - 2 = -1 (negative)
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