Calculate the change in entropy of the following equation:

FeCl3(s) + 3Na(s) --> Fe(s) + 3NaCl(s)
  1. 52.1 J/K
  2. -52.1 J/K
  3. 64.1 J/K
  4. -64.1 J/K
  5. -11.6 J/K
Explanation
Answer: B. Solution is as follows:

DSrxn° = DSproducts° - DSreactants°

Students must look up the DS° for each product and reactant. The numbers are as follows:

DS° (FeCl3(s)) = 142.3 J/mol K
DS° (Na(s)) = 51.3 J/mol K
DS° (Fe(s)) = 27.2 J/mol K
DS° (NaCl(s)) = 72.3 J/mol K

Therefore DSrxn° = [(1mol * 27.2 J/mol K) + (3mol * 72.3 J/mol K) ] - [ (1mol * 142.3 J/mol K) +(3mol * 51.3 J/mol K)]

DSrxn° = -52.1 J/K

***mole amounts come from the stoichiometry
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