Question

The energy required to vaporize one mol of benzene at its boiling point is \[31.2kJ\]. For how long a \[100W\] electric heater has to be operated in order to vaporize a \[100g\] benzene at its boiling temperature?
A.\[ \approx 743\sec \]
B.\[400\sec \]
C.\[3.9 \times {10^4}\sec \]
D.\[ \approx 371\sec \]

Answer 1

Hint: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.
Formula used:
\[Power = \dfrac{{Energy}}{{Time}}\] & \[q = n \times {\vartriangle _{vap}}{H^0}\] where \[q\] is the heat supplied, \[n\] is the number of moles & \[{\vartriangle _{vap}}{H^0}\] is the standard vaporization enthalpy.

Complete step by step 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 \[100g\] benzene then how long will it take for an electric heater of \[100W\] to vaporize it at its boiling temperature.

-We know that \[Power = \dfrac{{Energy}}{{Time}}\]& here the power is known, that is \[100W\]& we have to calculate the time. Hence, our first job is to get the energy and then substitute it in the equation.

-From the formula \[q = n \times {\vartriangle _{vap}}{H^0}\], we can see that \[q\] is the heat supplied, and this is the energy that we are looking for(Heat is a form of energy)

-To do that we need the number of moles of benzene
It is given that \[100g\] benzene is there
Dividing it by its molecular weight, we get the number of moles as \[\dfrac{{100}}{{78}}moles\]
And since \[{\vartriangle _{vap}}{H^0}\]is given to us as \[31.2kJ\]
Hence, \[q = \dfrac{{100}}{{78}} \times 31200 = 40000J\]
Now substituting it in the formula of \[Power = \dfrac{{Energy}}{{Time}}\]we get,
\[Time = \dfrac{{40000}}{{100}} = 400\sec \]

Hence, the answer to this question is option B.


Note:
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 $6^oC$ and boils at $80^oC$; it is a liquid at room temperature. If both the normal melting point and the normal boiling point are above room temperature, the substance is a solid.