Question

calculate the entropy for the reaction below. 2al(s) + 3cl₂(g) → 2alcl₃(s) ∑ products − ∑ reactants compound | s° (j/mol k) al(s) | 28.3 cl₂(g) | 223.1 alcl₃(s) | 109.3 δsᵣₓₙ = ? j/k enter either a + or - sign and the magnitude. hint: include the moles for the reactants and products in your calculation.

Explanation:

Step1: Calculate entropy of products

For \(2\ce{AlCl3}(s)\), use \(n = 2\) and \(S^\circ = 109.3\ \text{J/mol K}\).
\(\sum S_{\text{products}} = 2\times109.3 = 218.6\ \text{J/K}\)

Step2: Calculate entropy of reactants

For \(2\ce{Al}(s)\) (\(n = 2\), \(S^\circ = 28.3\)) and \(3\ce{Cl2}(g)\) (\(n = 3\), \(S^\circ = 223.1\)):
\(\sum S_{\text{reactants}} = (2\times28.3) + (3\times223.1) = 56.6 + 669.3 = 725.9\ \text{J/K}\)

Step3: Calculate \(\Delta S_{\text{rxn}}\)

\(\Delta S_{\text{rxn}} = \sum S_{\text{products}} - \sum S_{\text{reactants}} = 218.6 - 725.9 = -507.3\ \text{J/K}\)

Answer:

-507.3