Question

Arsine, AsH3, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 torr at -111.95 degree C and 253 torr at -83.6 degree C. Calculate the delta H of vaporization for arsine and its normal boiling point.

Answer 1

Answer : The value of $\Delta H$ 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 $\Delta H$ of vaporization for arsine.

The Clausius- Clapeyron equation is :

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

where,

$P_1$ = vapor pressure of arsine at $-111.95^oC$ = 35 torr

$P_2$ = vapor pressure of arsine at $-83.6^oC$ = 253 torr

$T_1$ = temperature of arsine = $-111.95^oC=273+(-111.95)=161.05K$

$T_2$ = temperature of arsine = $-83.6^oC=273+(-83.6)=189.4K$

$\Delta H_{vap}$ = heat of vaporization = ?

R = universal constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

$\ln (\frac{253}{35})=\frac{\Delta H_{vap}}{8.314}\times (\frac{1}{161.05}-\frac{1}{189.4})$

$\Delta H_{vap}=17694.3J/mol$

Now we have to calculate the normal boiling point of arsine.

The Clausius- Clapeyron equation is :

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

where,

$P_1$ = vapor pressure of arsine at $-111.95^oC$ = 35 torr

$P_2$ = vapor pressure of arsine at normal boiling point = 760 torr

$T_1$ = temperature of arsine = $-111.95^oC=273+(-111.95)=161.05K$

$T_2$ = normal boiling point of arsine = ?

$\Delta H_{vap}$ = 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:

$\ln (\frac{760}{35})=\frac{17694.3J/mole}{8.314J/K.mole}\times (\frac{1}{161.05}-\frac{1}{T_2})$

$T_2=209.9K$

Hence, the value of $\Delta H$ of vaporization for arsine and its normal boiling point is, 17694.3  J/mol and 209.9 K respectively.