Question

What is the value for ∆Soreaction for the following reaction, given the standard entropy values? 2 Al (s) + 3Cl2(g) 2AlCl3(s)

Answer 1

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

Answer 2

2Al(s) + 3Cl₂ → 2AlCl₃

Forming the equation:
2(28) + 3(223) = ΔH + 2(111)
ΔH = 503 kJ

Writing this outside of the reaction equation:
ΔH = -503 kJ