Methane (CH4) has a boiling point of -164 C at 1 atm and a vapor pressure of 42.8 atm at -100 C. What is the heat of vaporization of CH4?
The Correct Answer and Explanation is:
To calculate the heat of vaporization (ΔHvap) of methane (CH₄), we use the Clausius-Clapeyron equation, which relates the vapor pressure and temperature to the heat of vaporization: ln(P2P1)=−ΔHvapR(1T2−1T1)\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} – \frac{1}{T_1}\right)
Given:
- P1=1.00 atmP_1 = 1.00\ \text{atm} at T1=−164∘C=109 KT_1 = -164^\circ C = 109\ \text{K}
- P2=42.8 atmP_2 = 42.8\ \text{atm} at T2=−100∘C=173 KT_2 = -100^\circ C = 173\ \text{K}
- R=8.314 J/mol\cdotpKR = 8.314\ \text{J/mol·K}
Step-by-step calculation:
- Convert temperatures to Kelvin: T1=−164+273=109 K,T2=−100+273=173 KT_1 = -164 + 273 = 109\ \text{K},\quad T_2 = -100 + 273 = 173\ \text{K}
- Plug into the Clausius-Clapeyron equation:
ln(42.81.00)=−ΔHvap8.314(1173−1109)\ln\left(\frac{42.8}{1.00}\right) = \frac{-\Delta H_{\text{vap}}}{8.314} \left(\frac{1}{173} – \frac{1}{109}\right) ln(42.8)≈3.755\ln(42.8) \approx 3.755 (1173−1109)=0.00578−0.00917=−0.00339 K−1\left(\frac{1}{173} – \frac{1}{109}\right) = 0.00578 – 0.00917 = -0.00339\ \text{K}^{-1} 3.755=−ΔHvap8.314×(−0.00339)3.755 = \frac{-\Delta H_{\text{vap}}}{8.314} \times (-0.00339) ΔHvap=3.755×8.3140.00339≈31.230.00339≈9211 J/mol≈9.21 kJ/mol\Delta H_{\text{vap}} = \frac{3.755 \times 8.314}{0.00339} \approx \frac{31.23}{0.00339} \approx 9211\ \text{J/mol} \approx \boxed{9.21\ \text{kJ/mol}}
Explanation
The heat of vaporization is the energy required to convert one mole of a substance from liquid to gas at constant pressure. Methane (CH₄), being a small nonpolar molecule, has relatively weak intermolecular forces (London dispersion forces), resulting in a low boiling point of -164 °C.
To determine its heat of vaporization, we used the Clausius-Clapeyron equation, which describes how vapor pressure varies with temperature for phase changes. The equation is derived from thermodynamic principles and assumes that ΔHvap remains constant over the temperature range considered — a reasonable approximation here.
In this case, we are given two vapor pressures at two different temperatures: 1 atm at -164 °C and 42.8 atm at -100 °C. After converting temperatures to Kelvin and plugging values into the Clausius-Clapeyron equation, we solved for ΔHvap using natural logarithms and algebra.
The result is approximately 9.21 kJ/mol, indicating the amount of energy needed to vaporize one mole of liquid methane at its boiling point. This relatively low value aligns with methane’s weak intermolecular forces and low molecular weight.
Understanding ΔHvap is essential in areas like natural gas processing, cryogenics, and climate science. Methane is a potent greenhouse gas, and its evaporation behavior influences atmospheric concentrations and energy balance. Accurately determining its thermodynamic properties helps in designing storage systems and understanding its environmental impact.
