The molar enthalpy of vaporization of hexane (C6H14) is 28.9 kJ/mol, and its normal boiling point is 68.7 degrees Celsius. What is the vapor pressure of hexane in mmHg at 25 degrees Celsius?
The Correct Answer and Explanation is:
To find the vapor pressure of hexane at 25°C, we can use the Clausius-Clapeyron equation in its two-point form: ln(P2P1)=−ΔHvapR(1T2−1T1)\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} – \frac{1}{T_1} \right)
Given:
- ΔH₍vap₎ = 28.9 kJ/mol = 28900 J/mol
- T₁ = boiling point = 68.7°C = 341.85 K
- T₂ = 25°C = 298.15 K
- P₁ = normal boiling point pressure = 760 mmHg
- R = 8.314 J/mol·K
We are solving for P₂, the vapor pressure at 25°C.
Step-by-step:
ln(P2760)=−289008.314(1298.15−1341.85)\ln\left(\frac{P_2}{760}\right) = -\frac{28900}{8.314} \left( \frac{1}{298.15} – \frac{1}{341.85} \right) ln(P2760)=−3475.2(0.003355−0.002925)\ln\left(\frac{P_2}{760}\right) = -3475.2 \left( 0.003355 – 0.002925 \right) ln(P2760)=−3475.2×0.000430=−1.494\ln\left(\frac{P_2}{760}\right) = -3475.2 \times 0.000430 = -1.494 P2760=e−1.494≈0.2245\frac{P_2}{760} = e^{-1.494} \approx 0.2245 P2=760×0.2245≈170.6 mmHgP_2 = 760 \times 0.2245 \approx \boxed{170.6 \text{ mmHg}}
Explanation
The vapor pressure of a liquid is the pressure exerted by its vapor when the liquid and vapor are in dynamic equilibrium. It depends on the temperature and the enthalpy of vaporization. In this case, we are asked to calculate the vapor pressure of hexane at 25°C, given its molar enthalpy of vaporization (28.9 kJ/mol) and its normal boiling point (68.7°C), where its vapor pressure is 760 mmHg (standard atmospheric pressure).
To solve this, we use the Clausius-Clapeyron equation, which relates the vapor pressures at two temperatures. This thermodynamic relationship is especially useful when one vapor pressure (typically at the boiling point) is known, and we wish to find it at another temperature.
In this formula, we convert all temperatures to Kelvin and ensure the enthalpy is in Joules to match the gas constant R = 8.314 J/mol·K. The equation uses the natural logarithm to express the exponential relationship between vapor pressure and temperature, arising from the energy barrier that molecules must overcome to escape the liquid phase.
After substituting and solving, we find the vapor pressure of hexane at 25°C to be approximately 170.6 mmHg. This lower pressure at room temperature compared to the boiling point pressure (760 mmHg) makes sense, as fewer molecules have sufficient energy to vaporize at this lower temperature. This calculation illustrates how temperature and enthalpy of vaporization critically affect a substance’s volatility and behavior in practical applications like distillation and fuel evaporation.
