Question

The molar heat of vaporization of toluene is $\Delta {{H}_{v}}$. If the vapour pressure at 315K is 60 torr and that at 355 K is 300 torr then $\Delta {{H}_{v}}$=? (log2=0.3)
(A) 37.4 KJ/mole
(B) 3.74 KJ/mole
(C) 37.4 J/mole
(D) 3.74 J/mole

Answer 1

The amount of heat required to evaporate one mole of toluene when pressure and temperature changes accordingly; can be calculated using an equation which includes both pressure and temperature quantities.

Complete step by step solution:
Let us firstly understand the molar heat of vaporisation and its equation,
The molar heat of vaporisation is the energy needed to vaporise one mole of a liquid. The units are usually kilojoules per mole i.e. KJ/mol. Two possible equations can help you determine the molar heat of vaporisation.
To calculate the molar heat of vaporisation, write down the given information, choose an equation that fits the circumstances to be resolved. Then solve the equation using temperature and pressure data.
Steps-
1.Write down your given information.
In order to calculate the molar heat of vaporisation, you should write down the information that the problem provides.
The problem will either provide two pressure or two temperature values, or the molar heat of sublimation, and the molar heat of fusion.
The molar heat of sublimation is the energy needed to sublime one mole of a solid, and molar heat of fusion is the energy needed to melt one mole of a solid.
2.Decide which equation to use.
When calculating the molar heat of vaporisation, you have to decide which equation you will use based on the given information.

If the problem provides two temperature and two pressure values, use the equation $\ln \left( \dfrac{{{P}_{2}}}{{{P}_{1}}} \right)=\dfrac{{{H}_{vap}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$
where, ${{P}_{1}}$ and ${{P}_{2}}$ are the pressure values; ${{T}_{1}}$ and ${{T}_{2}}$ are temperature values
${{H}_{vap}}$ is the molar heat of vaporisation; R is the gas constant.
If the problem provides the molar heat of sublimation and molar heat of fusion,
${{H}_{sub}}={{H}_{fus}}+{{H}_{vap}}$
where, ${{H}_{sub}}$ is the heat of sublimation; ${{H}_{fus}}$ is the heat of fusion.
Illustration-
Given that,
${{P}_{1}}$ = 60 torr
${{P}_{2}}$ = 300 torr
${{T}_{1}}$ = 315 K
${{T}_{2}}$ = 355 K
R = 8.314 J/mol
Now,
The Clausius – Clapeyron equation is given as,
$\ln \left( \dfrac{{{P}_{2}}}{{{P}_{1}}} \right)=\dfrac{{{H}_{vap}}}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$
Substituting the values given in the above equation we get,
$\ln \left( \dfrac{300}{60} \right)=\dfrac{{{H}_{vap}}}{8.314}\left( \dfrac{1}{315}-\dfrac{1}{355} \right)$
Hence, ${{H}_{vap}}$= $\Delta {{H}_{v}}$= 37414 J/mol = 37.41 KJ/mol

Therefore option (A) is correct.

Note: We can solve and calculate $\Delta {{H}_{v}}$ by two methods as described by knowing the given data. Do not mix up any data resulting in difficulties.