Question

The molar enthalpy of vaporization of benzene at its boiling point $ \left( 353\text{ }K \right) $ is $ 30.84\text{ }kJmo{{l}^{-1}} $ . What is the molar internal energy change? For how long would a $ 12\text{ }V $ source need to supply a $ 0.5\text{ }A $ current in order to vaporize $ 7.8\text{ }g $ of the sample at its boiling point?

Answer 1

Hint: We know that a basic concept of thermochemistry is well enough to tackle this kind of problem. The trick is to remember the relation between standard vaporization enthalpy and heat supplied. Here the formula we will use is $ \Delta H=\Delta E+\Delta ngRT $ and $ E=V\times I\times t $

Complete answer:
The basic understanding of the problem is necessary to solve it. We are given data on how much energy is required to vaporize one mole of benzene at its boiling temperature. Having said that, we are asked that if we take $ 7.8\text{ }g $ benzene then how long will it take for an electric heater to vaporize it at its boiling temperature. Product Reactant
Here we have, $ {{C}_{6}}{{H}_{6\left( l \right)}}\to C6{{H}_{6\left( g \right)}} $ and thus we know that $ \therefore \Delta n=\left[ nProduct-nReactant \right]=\left[ 1-0 \right]=1 $
We have given data; $ \Delta H=30.84\dfrac{KJ}{mol}=30840\dfrac{J}{mol}, $ $ Temperature\left( T \right)=353K, $
As we know that, $ \Delta H=\Delta E+\Delta ngRT $
 $ \Rightarrow 30840=\left[ \Delta E+\left( 1\times 8.314\times 353 \right) \right] $
 $ \therefore \Delta E=\left( 30840-2934.842 \right)=\left( 27905.158 \right)=27.9KJ $
Now, we know that; Molecular weight of benzene $ =78g/mol, $ Weight of benzene $ =7.8g $
 $ ~\therefore No.\text{ }of\text{ }moles\text{ }of\text{ }benzene~=\left( \dfrac{7.8}{78} \right)=0.1mole $
Therefore, Energy provided to evaporate $ 0.1~ $ mole of benzene $ (E)=\left( 30.84\times 0.1 \right)=3084J $
As we know that, $ E=V\times I\times t $ and the given data we have is $ V=12~volt, $ $ t=? $ $ I=0.5A, $
Thus by substituting the given values; $ 3084=12\times 0.5\times t $
 $ \Rightarrow t=63084=514s $
Hence the molar internal energy change will be $ 27.9KJ~ $ and for $ t=514s,~ $ a $ 12 $ volt source needs to supply a $ 0.5A~ $ current in order to vaporize $ 7.8g~ $ of the sample at its boiling point.

Additional Information:
The change in free energy occurs when a compound is formed from its elements in their most thermodynamically stable state at standard state conditions i.e. $ 1\text{ }atm. $ In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of irreversible work that may be performed by a thermodynamic system at a constant temperature and pressure. The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamic closed system. This maximum can be attained only in a completely reversible process.

Note:
Remember that it’s worth mentioning again that $ q $ , the heat supplied is the energy that is required to vaporize benzene at its boiling temperature. Benzene melts at and boils at $ 6{}^\circ C $ ; it is a liquid at room temperature. Enthalpy of vaporization is the amount of energy that must be added to the liquid substance, to transform a quantity of that substance into gas. Entropy of vaporization is an increase in entropy upon vaporization of a liquid.