The problem in the image is about finding the boiling point of hexane at a pressure of 0.5 atm, given the following information:

• Molar heat of vaporization of hexane ($\Delta H_{vap}$) = 30.8 kJ/mol
• Boiling point at 1 atm = 68.90°C

To solve this, we use the Clausius-Clapeyron equation:

$\ln \left(\frac{P_1}{P_2}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Where:

• $P_1$ = 1 atm (initial pressure)
• $P_2$ = 0.5 atm (final pressure)
• $T_1$ = 68.90°C = 342.05 K (initial boiling temperature)
• $T_2$ = final boiling temperature in Kelvin
• $R$ = 8.314 J/mol·K (gas constant)

Steps to solve:

1. Convert $\Delta H_{vap}$ to J/mol: $30.8 \text{ kJ/mol} = 30,800 \text{ J/mol}$.
2. Rearrange the Clausius-Clapeyron equation to solve for $T_2$.
3. Solve for $T_2$.

Let's proceed with the calculations.

$\ln \left(\frac{1}{0.5}\right) = \frac{30,800}{8.314} \left(\frac{1}{T_2} - \frac{1}{342.05}\right)$

Calculating this will give the new boiling point $T_2$. Let's do the math.The boiling point of hexane at a pressure of 0.5 atm is approximately $92.29^\circ \text{C}$.

Would you like more details or have any questions about this process?

Here are 5 related questions to deepen your understanding:

1. How does the Clausius-Clapeyron equation describe the relationship between pressure and boiling point?
2. Why do we need to convert temperatures to Kelvin in these calculations?
3. What would happen to the boiling point if the pressure decreased even further, say to 0.2 atm?
4. How does the heat of vaporization affect the change in boiling point with pressure?
5. What are the practical applications of using the Clausius-Clapeyron equation in chemistry?

Tip: When using the Clausius-Clapeyron equation, always ensure that the pressures used are in the same units, and temperatures are converted to Kelvin.