Calculate ?S° rxn for the following reaction.The S° for each species is shown below the reaction.
C 2 H 2 (g)+ H 2 (g)? C 2 H 4 (g)
S°(J/mol?K)200.9 130.7 219.3

Question

Quizplus · Accepted Answer

ΔS°S1U1B1rxnS1U1B0 = ΣS°S1U1B1(products) - ΣS°S1U1B1(reactants) = [S°(CS1U1B12S1U1B0HS1U1B14S1U1B0(g)) + S°(HS1U1B12S1U1B0(g))] - [S°(CS1U1B12S1U1B0(g)) + S°(HS1U1B12S1U1B0(g))] = [(219.3 J/mol∙K) + (130.7 J/mol∙K)] - [(200.9 J/mol∙K) + (130.7 J/mol∙K)] = 338.4 J/K However, this is the change in entropy at standard conditions and not at the conditions given. To calculate the entropy change at the given conditions, we need to use the equation: ΔS = ΔS°S1U1B1rxnS1U1B0 + ΣnS1U1B1R ln(Q) where Q is the reaction quotient at the given conditions. At equilibrium, Q = K, the equilibrium constant. Since we do not have information about the concentrations of the reactants and products, we cannot directly calculate Q or K. However, we can use the information given in the answer choices to determine the sign and magnitude of ΔS. Since ΔS is a positive value at standard conditions, we can eliminate choices A and E, which have negative values. Next, we need to consider the effect of Q on ΔS. If Q < K, then ln(Q) < 0 and ΔS < ΔS°S1U1B1rxnS1U1B0. If Q > K, then ln(Q) > 0 and ΔS > ΔS°S1U1B1rxnS1U1B0. Since the reaction produces a gas (HS1U1B14S1U1B0(g)), increasing the pressure (or decreasing the volume) would shift the equilibrium towards the side with fewer moles of gas. Therefore, Q will be smaller than K at the given conditions (which are not at standard conditions). This means that ΔS will be smaller than the standard value. The only answer choice that reflects a smaller value than the standard value of 338.4 J/K is C (-112.3 J/K). Therefore, the answer is C.

Calculate ?S°rxn for the Following Reaction

Calculate ?S°_rxn for the Following Reaction