To find the boiling point of 1-pentanol, we can use the relationship between the heat of vaporization ([tex]\(\Delta H_{\text{vap}}\)[/tex]) and the entropy of vaporization ([tex]\(\Delta S_{\text{vap}}\)[/tex]), given by the equation:
[tex]\[ T_b = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}} \][/tex]
where:
- [tex]\(\Delta H_{\text{vap}}\)[/tex] is the heat of vaporization,
- [tex]\(\Delta S_{\text{vap}}\)[/tex] is the entropy of vaporization,
- [tex]\(T_b\)[/tex] is the boiling point in Kelvin.
Given:
- [tex]\(\Delta H_{\text{vap}} = 55.5 \text{ kJ/mol} = 55500 \text{ J/mol} \)[/tex] (since 1 kJ = 1000 J),
- [tex]\(\Delta S_{\text{vap}} = 148 \text{ J/K·mol} \)[/tex].
Let's calculate the boiling point in Kelvin:
[tex]\[ T_b = \frac{55500 \text{ J/mol}}{148 \text{ J/K·mol}} = 375.0 \text{ K} \][/tex]
Next, we need to convert the boiling point from Kelvin to Celsius. The conversion formula is:
[tex]\[ T_C = T_K - 273.15 \][/tex]
where:
- [tex]\(T_C\)[/tex] is the temperature in Celsius,
- [tex]\(T_K\)[/tex] is the temperature in Kelvin.
Substituting [tex]\( T_K = 375.0 \text{ K} \)[/tex]:
[tex]\[ T_C = 375.0 \text{ K} - 273.15 = 101.85 \text{°C} \][/tex]
Since we are asked for the approximate boiling point, we should round [tex]\(101.85 \text{°C}\)[/tex] to the nearest whole number, which gives us:
[tex]\[ T_C \approx 102 \text{°C} \][/tex]
Given the options available:
A. [tex]\(375^{\circ} C\)[/tex]
B. [tex]\(100^{\circ} C\)[/tex]
C. [tex]\(25^{\circ} C\)[/tex]
D. [tex]\(0^{\circ} C\)[/tex]
The closest and most reasonable approximation is:
[tex]\[ \boxed{100^{\circ} C} \][/tex]
