Carbon disulfide, CS2, has a normal boiling point of 46°C and a heat of vaporization of 26.8 kJ/mol. What is the vapor pressure of carbon disulfide at 35°C?
The Correct Answer and Explanation is :
To calculate the vapor pressure of carbon disulfide (( P )) at 35°C, we can use the Clausius-Clapeyron equation:
[
\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} – \frac{1}{T_1}\right)
]
Step 1: Known values
- ( \Delta H_{\text{vap}} = 26.8 \, \text{kJ/mol} = 26,800 \, \text{J/mol} )
- ( T_1 = 46^\circ\text{C} = 319.15 \, \text{K} ) (normal boiling point, vapor pressure ( P_1 = 1 \, \text{atm} ))
- ( T_2 = 35^\circ\text{C} = 308.15 \, \text{K} )
- ( R = 8.314 \, \text{J/mol·K} )
Step 2: Rearrange for ( P_2 )
[
\ln(P_2) = \ln(P_1) – \frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} – \frac{1}{T_1}\right)
]
Since ( \ln(P_1) = \ln(1) = 0 ):
[
\ln(P_2) = -\frac{26,800}{8.314} \left(\frac{1}{308.15} – \frac{1}{319.15}\right)
]
Step 3: Calculate the temperature term
[
\frac{1}{308.15} – \frac{1}{319.15} = 0.003245 – 0.003133 = 0.000112 \, \text{K}^{-1}
]
Step 4: Calculate the exponent
[
\ln(P_2) = -\frac{26,800}{8.314} \cdot 0.000112
]
[
\ln(P_2) = -3615.1 \cdot 0.000112 = -0.405
]
Step 5: Solve for ( P_2 )
[
P_2 = e^{-0.405} \approx 0.666 \, \text{atm}
]
Final Answer:
The vapor pressure of carbon disulfide at 35°C is approximately 0.666 atm.
Explanation:
The Clausius-Clapeyron equation relates vapor pressure to temperature, assuming a constant heat of vaporization (( \Delta H_{\text{vap}} )). The equation works well for small temperature ranges and relies on the exponential relationship between pressure and temperature. In this case, we used the normal boiling point (where vapor pressure equals 1 atm) as a reference point. The result shows that as temperature decreases, the vapor pressure also decreases due to lower molecular kinetic energy.