Question

Methylamine has a vapor pressure of 344torr at −25∘C and a boiling point of −6.4∘C . Find ΔHvap for methylamine.

Answer 1

The Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R * (1/T₂ - 1/T₁)

-25 °C = 248 K and pressure is 344 torr
-6.4 °C = 266.6 K and pressure is 760 torr (atmospheric)

Using the equation:

ln(760/344) = ΔH/62.364 * (1/248 - 1/266.6)
ΔH = 175.7 kJ/mol

Answer 2

The value of $\Delta {H_{{\text{vap}}}}$ for methylamine is $\boxed{176550.25{\text{ kJ/mol}}}$.

Further Explanation:

Clausius Clapeyron Relation

It describes the relationship between pressure and temperature of the substance when two phases of the substance exist in equilibrium. It is used in chemical engineering, meteorology, and climatology.

The expression for Clausius Clapeyron equation is as follows:

$\ln \dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}} = \dfrac{{\Delta {H_{{\text{vap}}}}}}{{\text{R}}}\left( {\dfrac{1}{{{{\text{T}}_{\text{2}}}}} - \dfrac{1}{{{{\text{T}}_{\text{1}}}}}} \right)$                                                      

...... (1)

Here,

${{\text{P}}_{\text{1}}}$ is the initial pressure.

${{\text{P}}_{\text{2}}}$ is the final pressure.

${{\text{T}}_1}$ is the initial temperature.

${{\text{T}}_{\text{2}}}$ is the final temperature.

R is universal gas constant.

$\Delta {H_{{\text{vap}}}}$ is the heat of vaporization.

The temperatures are to be converted into K. The conversion factor for this is,

${\text{1 }}^\circ {\text{C}} = 273.1{\text{5 K}}$

So the temperature $\left( {{{\text{T}}_{\text{1}}}} \right)$ can be calculated as follows:

$\begin{aligned}{{\text{T}}_1} &= \left( { - 6.4 + 273.15} \right)\;{\text{K}}\\&= 26{\text{6}}{\text{.75 K}}\\\end{aligned}$  

So the temperature $\left( {{{\text{T}}_{\text{2}}}} \right)$ can be calculated as follows:

$\begin{aligned}{{\text{T}}_2}&=\left( { - 25 + 273.15} \right)\;{\text{K}}\\&= 248.15{\text{ K}}\\\end{aligned}$  

Rearrange equation (1) to calculate  .

$\Delta {H_{{\text{vap}}}} = \dfrac{{\left( {\ln \dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}} \right)\left( {\text{R}} \right)}}{{\left( {\dfrac{1}{{{{\text{T}}_{\text{2}}}}} - \dfrac{1}{{{{\text{T}}_{\text{1}}}}}} \right)}}$                                                     ...... (2)

The pressure corresponding to the boiling point is 760 torr.

Substitute 760 torr for ${{\text{P}}_{\text{1}}}$, 344 torr for ${{\text{P}}_{\text{2}}}$, 266.75 K for ${{\text{T}}_{\text{1}}}$, 248.15 K for ${{\text{T}}_{\text{2}}}$, $62.364{\text{ L}} \cdot {\text{torr}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ for R in equation (2).

 $\begin{aligned}\Delta {H_{{\text{vap}}}}&= \dfrac{{\left( {\ln \dfrac{{{\text{760 torr}}}}{{{\text{344 torr}}}}} \right)\left( {62.364{\text{ L}} \cdot {\text{torr}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}} \right)}}{{\left( {\dfrac{1}{{{\text{248}}{\text{.15 K}}}} - \dfrac{1}{{266.75{\text{ K}}}}} \right)}} \\&=176550.25{\text{ kJ/mol}}\\\end{aligned}$

Learn more:

1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201
2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: P1, P2, T1, T2, R, 176550.25 kJ/mol, K, 344 torr, 760 torr, methylamine, 266.75 K, 248.15 K, conversion factor, pressure, temperature, universal gas constant.