Question

liquid has an enthalpy of vaporization of 30.8 kJ/mol. At 273 K it has a vapor pressure of 102 mmHg. What is the normal boiling point of this liquid? (R = 8.31 J/(K.mol)a. 257kb. 238kc.273kd.292ke.320k

Answer 1

Option E is correct. 320K

Explanations:

In order to get the normal boiling point of the liquid, we will use the Clausius-Clapeyron equation expressed according to the equation:

`\ln ((P_2)/(P_1)_{})=-(\triangle H_(vap))/(R)\cdot((1)/(T_2)-(1)/(T_1))`

where:

P1 is the vapor pressure of that substance at T1

P2 is the vapor pressure of that substance at T2

ΔHvap is the enthalpy of vaporization.

R is the gas constant - usually expressed as 8.314J/Kmol

Given the following parameters:

Substitute the given parameters into the formula to get T2

Simplify the result to get the value of T2

`\begin{gathered} \ln (7.4509)=-3,706.3778((1)/(T_2)-(1)/(273)) \\ 2.0083=(-3,706.3778)/(T_2)+13.5765 \\ (-3,706.3778)/(T_2)=2.0083-13.5765 \\ (-3,706.3778)/(T_2)=-11.5682 \\ -11.5682T_2=-3,706.3778 \\ T_2=(-3,706.3778)/(-11.5682) \\ T_2=320.39K \\ T_2\approx320K \end{gathered}`

Hence the normal boiling point of this liquid is approximately 320K