The normal boiling point of C₂Cl₃F₃ is 47.6 °C and its molar enthalpy of vaporization is 27.49 kJ/mol.What is the change in entropy in the system in J/K when 28.6 grams of C₂Cl₃F₃ vaporizes to a gas at the normal boiling point?

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First, we need to calculate the number of moles of C₂Cl₃F₃: n = m/MW = 28.6 g / 102.96 g/mol = 0.278 mol Then, we use the formula ΔS = qrev / T to calculate the change in entropy, where qrev is the heat absorbed during a reversible process (in this case, the molar enthalpy of vaporization) and T is the boiling point in Kelvin: ΔS = qrev / T = ΔHvap / T = 27.49 kJ/mol / (47.6 + 273.15) K = 0.0913 kJ/(mol*K) = 91.3 J/(mol*K) Finally, we multiply ΔS by the number of moles to get the total change in entropy: ΔStotal = ΔS x n = 91.3 J/(mol*K) x 0.278 mol = 25.4 J/K Since we want the answer in J/K and not kJ/mol, we divide by the number of grams instead of the molar mass: ΔStotal = 25.4 J/K / 28.6 g = 0.888 J/(g*K) To get the answer in the requested format, we round to one decimal place: Answer: D The change in entropy in the system is 0.9 J/(g*K), which is equivalent to 13.1 J/K for the given amount of C₂Cl₃F₃ that vaporizes. The answer is D.

The Normal Boiling Point of C2Cl3F3 Is 47

The Normal Boiling Point of C₂Cl₃F₃ Is 47