Answer: -503 J/K
Explanation:
The balanced chemical reaction is,
\(2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3\)
The expression for enthalpy change is,
\(\Delta S=\sum [n\times S^0(product)]-\sum [n\times \Delta S^0(reactant)]\)
\(\Delta S=[(n_{AlCl_3}\times S_{AlCl_3})]-[(n_{Cl_2}\times S_{Cl_2})+[(n_{Al}\times S_{Al})]\)
where,
n = number of moles
Now put all the given values in this expression, we get
\(\Delta S=[(2\times 111)]-[(3\times 223)+(2\times 28)]\)
\(\Delta S=-503J/K\)
Therefore, the entropy change for this reaction is, -503J/K
