The entropy of vaporization of decane is 86.7 J/mol .K at its normal boiling point of 174.2$\degree$C.What is the enthalpy of vaporization of decane at this temperature?

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Quizplus · Accepted Answer

The enthalpy of vaporization (ΔHvap) can be calculated using the formula ΔHvap = TΔSvap, where T is the temperature in Kelvin and ΔSvap is the entropy of vaporization. Convert 174.2°C to Kelvin (174.2 + 273.15 = 447.35 K) and multiply by 86.7 J/mol·K to get ΔHvap in J/mol, then convert to kJ/mol: (447.35 K * 86.7 J/mol·K) / 1000 = 38.8 kJ/mol.

The Entropy of Vaporization of Decane Is 86 °\degree° CWhat Is the Enthalpy of Vaporization of Decane at This

The Entropy of Vaporization of Decane Is 86 $\degree$ CWhat Is the Enthalpy of Vaporization of Decane at This