Question

(a) Find ΔS°rxn for the formation of Cu2₂O(s) from its elements.

Answer 1

$-75.6 Jmol^{-1}K^{-1}$ is the ΔS°rxn for the formation of Cu₂O(s) from its elements.

WHAT IS  ΔS°rxn FOR THE REACTION ?

The direction of spontaneous change is the direction in which total entropy increases. Total entropy change, also called the entropy change of the universe, is the sum of the entropy change of a system and of its surroundings

Standard entropy change of below reaction (ΔS°$_{rxn}$)

$2Cu(s) + \frac{1}{2} O_2(g) --------- > Cu_2O$      ∴ ΔH°$_{rxn}$ = - 168.6 KJ/mol

Now, $\Delta S^o_{rxn} = S^o_{(products)} - S^o_{(reactants)}$

                   $= S^o_{Cu_2 O(s)} - [2 * S^o_{Cu(s)} + \frac{1}{2}(S^o_{O_2(g)}]$

                   $= (93.1 J/molK)- [2 * 133.1 + \frac{1}{2}(205.0)]J/molK$

                   $= - 75.6 J/molK$

Thus, -75.6 J/mol K is the ΔS°rxn for the formation.

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