Respuesta :
a. Using the Clausius-Clapeyron equation
ln P1/P2 = (-Hvap / R) (1/T1 - 1/T2)
P1 = 197 mm Hg
P2 = 448 mm Hg
T1 = 23 C = 296 K
T2 = 45 C = 318 K
R = 8.314 J/mol-K
Solving for Hvap
Hvap = 29225.43 J/mol
b. For normal boiling point
P1 = 197 mm Hg
P2 = 760 mm Hg (normal pressure)
T1 = 23 C = 296 K
T2 = ?
R = 8.314 J/mol-K
Hvap = 29225.43 J/mol
ln P1/P2 = (-Hvap / R) (1/T1 - 1/T2)
Use the equation to solve for T2
Answer:
a. [tex] \Delta H_{vap} = 29222.56 J/mol [/tex]
b. normal boiling point: 61 C
Explanation:
Data
[tex]T_1 = 23\, C = 296 K[/tex]
[tex]T_2 = 45\, C = 318 K[/tex]
[tex]P_1 = 197\, mm Hg[/tex]
[tex]P_2 = 448\,mm Hg[/tex]
R = 8.314 J/mol K, universal gas constant
[tex]\Delta H_{vap}[/tex]: heat of vaporization
From Clausius-Clapeyron equation:
[tex] ln\frac{P_1}{P_2} = \frac{-\Delta H_{vap}}{R} (\frac{1}{T_1} - \frac{1}{T_2}) [/tex]
[tex] ln\frac{197}{448} = \frac{-\Delta H_{vap}}{8.314} (\frac{1}{296} - \frac{1}{318}) [/tex]
[tex] -0.8215 = \frac{-\Delta H_{vap}}{8.314} (\frac{11}{47064}) [/tex]
[tex] \Delta H_{vap} = 0.8215 \times 8.314 \times \frac{47064}{11} [/tex]
[tex] \Delta H_{vap} = 29222.56 J/mol [/tex]
We can use this result to compute the normal boiling point as follows:
[tex] ln\frac{197}{760} = \frac{-29222.56}{8.314} (\frac{1}{296} - \frac{1}{T_2}) [/tex]
[tex] -1.35 \times \frac{8.314}{-29222.56 = (\frac{1}{296} - \frac{1}{T_2}) [/tex]
[tex] \frac{1}{T_2} = \frac{1}{296} -3.84\cdot 10^{-4} [/tex]
[tex] T_2 = (3 \cdot 10^{-3})^{-1} [/tex]
[tex] T_2 = 334\, K = 61\, C [/tex]