Question

Calculate the ΔS°rxn of the following reaction at 215°C and standard pressure.

Answer 1

Answer: the entropy change associated to the reaction given is -29.9 J/mol.K

Step-by-step explanation:

The question requires us to calculate the entropy change (ΔS) associated to the given reaction at 215 C, given the standard molar entropies (S) of reactants and products.

The entropy change for the reaction (ΔSrxn) can be calculated using a "products minus reactants" rules, as shown below, considering the stoichiometric coefficients of the reaction:

`\Delta S_(rxn)=\sum_^mS^(\circ)(products)-\sum_^nS^(\circ)(reactants)`

where m corresponds to the stoichiometric coefficients of each product, and n corresponds to the stoichiometric coefficients of each reactant.

To apply the equation above, we'll need to use the standard molar entropies of each reactant and product, as well as their stoichiometric coefficients. These information were provided by the question, an can be summarized as:

- PRODUCTS:

CO2: stoichiometric coefficient = 2; standard molar entropy = 213.6 J/mol.K

H2O: stoichiometric coefficient = 2; standard molar entropy = 188.7 J/mol.K

- REACTANTS:

C2H4: stoichiometric coefficient = 1; standard molar entropy = 219.5 J/mol.K

O2: stoichiometric coefficient = 3; standard molar entropy = 205.0 J/mol.K

With the information above, we can calculate the entropy change as:

`\begin{gathered} \Delta S_(rxn)=[(2*213.6\text{ J/mol.K\rparen+\lparen2}*188.7\text{ J/mol.K\rparen}\rbrack-](1*219.5\text{ J/mol.K\rparen+\lparen3}*205.0\text{ J/mol.K\rparen}\rbrack \\ \Delta S_(rxn)=[804.6\text{ J/mol.K}\rbrack-[834.5\text{ J/mol.K}\rbrack \\ \Delta S_(rxn)=-29.9\text{ J/mol.K} \end{gathered}`

Therefore, the entropy change associated to the reaction given is -29.9 J/mol.K.