Question

The enthalpy of vaporization of compound 24.2 KJ/mol At 15.0°C it has a vapor pressure of 160 mmHg , what is its vapor pressure at 25°C

Answer 1

The Clausius-Clapeyron relation can be used to estimate the vapor pressure at 25°C:

$\text{ln}(\frac{p_1}{p_2}) = \frac{\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$ ,

where $\Delta H_{vap}$ is the enthalpy of vaporization, $T$ and $p$ are the temperature and the pressure, respectively and $R$ is the gas constant, 8.314 J mol^-1 K^-1.

Making use of the equation above, we can derive the expression for $p_2$ :

$p_2 = p_1\text{exp}( -\frac{\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1}))$

$p_2 = 160·\text{exp}( -\frac{24.2·10^3}{8.314}(\frac{1}{25+273.15}-\frac{1}{15+273.15}))$

$p_2 =225$ mmHg.

Don't forget to switch from celsius to kelvin (adding 273.15) and also to switch from kJ to J for the enthalpy of vaporization (multiplying by a factor of 10³).

Answer: if the enthalpy of vaporization of compound 24.2 kJ/mol and at 15.0°C it has a vapor pressure of 160 mmHg , its vapor pressure at 25°C is 225 mmHg.