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The heat of vaporization of octane is 39.4 kJ, and the entropy of vaporization is 99.0 J/mol·K. What is the boiling point of octane?

Answer :

Final answer:

The boiling point of octane can be estimated using the Clausius-Clapeyron equation, which relates vapor pressure, temperature, and enthalpy of vaporization. From the given values of enthalpy of vaporization and entropy of vaporization, the boiling point of octane is estimated to be approximately 229 °C.

Explanation:

The boiling point of octane can be estimated using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and enthalpy of vaporization. The equation is:

ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)

Where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. In this case, we are given the enthalpy of vaporization as 39.4 kJ/mol and the entropy of vaporization as 99.0 J/molK. Since the vapor pressure at boiling point is 1 atm (101.3 kPa), we can plug in the values and solve for the boiling point:

T2 = (ΔHvap/R) * (1/T1 - ln(P2/P1))

Substituting the given values, we get:

T2 = (39.4 kJ/mol / 8.314 J/molK) * (1/298 K - ln(101.3 kPa / 10 kPa))

= 502 K

= 229 °C

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