Respuesta :
Answer : The value of [tex]\Delta H[/tex] of vaporization for arsine and its normal boiling point is, 17694.3 J/mol and 209.9 K respectively.
Explanation :
First we have to calculate the value of [tex]\Delta H[/tex] of vaporization for arsine.
The Clausius- Clapeyron equation is :
[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = vapor pressure of arsine at [tex]-111.95^oC[/tex] = 35 torr
[tex]P_2[/tex] = vapor pressure of arsine at [tex]-83.6^oC[/tex] = 253 torr
[tex]T_1[/tex] = temperature of arsine = [tex]-111.95^oC=273+(-111.95)=161.05K[/tex]
[tex]T_2[/tex] = temperature of arsine = [tex]-83.6^oC=273+(-83.6)=189.4K[/tex]
[tex]\Delta H_{vap}[/tex] = heat of vaporization = ?
R = universal constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
[tex]\ln (\frac{253}{35})=\frac{\Delta H_{vap}}{8.314}\times (\frac{1}{161.05}-\frac{1}{189.4})[/tex]
[tex]\Delta H_{vap}=17694.3J/mol[/tex]
Now we have to calculate the normal boiling point of arsine.
The Clausius- Clapeyron equation is :
[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = vapor pressure of arsine at [tex]-111.95^oC[/tex] = 35 torr
[tex]P_2[/tex] = vapor pressure of arsine at normal boiling point = 760 torr
[tex]T_1[/tex] = temperature of arsine = [tex]-111.95^oC=273+(-111.95)=161.05K[/tex]
[tex]T_2[/tex] = normal boiling point of arsine = ?
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 17694.3 J/mol
R = universal constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
[tex]\ln (\frac{760}{35})=\frac{17694.3J/mole}{8.314J/K.mole}\times (\frac{1}{161.05}-\frac{1}{T_2})[/tex]
[tex]T_2=209.9K[/tex]
Hence, the value of [tex]\Delta H[/tex] of vaporization for arsine and its normal boiling point is, 17694.3 J/mol and 209.9 K respectively.
