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The entropy of vaporization of benzene is 85 \[J{{K}^{-1}}mo{{l}^{-1}}\]. When 117g benzene vaporizes at its normal boiling point, calculate the entropy change of entropy.

Last updated date: 13th Jun 2024
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Hint: For two phases which is in equilibrium, entropy is
\[\Delta {{S}_{vaporisation}} = 85J{{K}^{-1}}mo{{l}^{-1}}\]
Entropy is the degree of disorder of a given system

Complete Solution :
-So to solve this question let’s take and write the data provided in this question.
It is given that, entropy of vaporization of benzene is, \[\Delta {{S}_{vaporisation}} = 85J{{K}^{-1}}mo{{l}^{-1}}\]
And then it is given that we are vaporizing 117g of benzene at its normal boiling point.

Now we have to find 117g of benzene is equal to how much of mole, for that we use the equation,
i.e. \[Mole=\dfrac{117g}{78.11g/mol}=1.49mol\]

So now we got, how much moles are equal to 117g.
So for 1 mole of benzene the entropy of vaporization is equal to 85\[J{{K}^{-1}}mo{{l}^{-1}}\]

- Now we have to find for 1.49 moles how much is the entropy of vaporization will be,
\[\Delta {{S}_{vaporisation}} = 1.49\times 85\]\[\Delta {{S}_{vaporisation}} = 1.49\times 85\]
\[\Delta {{S}_{vaporisation}} = 126.65J/K\]

- At the boiling point the benzene will be in equilibrium with its vapor phase and since the two phases are in equilibrium the entropy of the surrounding can be called by the relation,
\[{{\Delta }_{system}}+{{\Delta }_{surrounding}} = 0\] \[{{\Delta }_{system}}+{{\Delta }_{surrounding}}=0\]
\[{{\Delta }_{system}}=-{{\Delta }_{surrounding}}=-126.65J/K\]

Note: This problem can be solved using a direct equation, after converting given mass into the moles, substitute the values in,
\[{{\Delta }_{system}}=-{{\Delta }_{surrounding}}={{\Delta }_{vaporisation}}\times Moles\]\[{{\Delta }_{system}}=-{{\Delta }_{surrounding}}={{\Delta }_{vaporisation}}\times Moles\]

- This equation will directly yield you the answer.
If we do not know the molecular mass of the compound and know its formulae then add the mass of each atom to get the molecular mass.

- For example if we don’t know the molecular mass of benzene, the molecular formula of benzene is ${{C}_{6}}{{H}_{6}}$
Mass of carbon is 12, so $6\times 12=72$
Mass of Hydrogen is 1, so $6\times 1= 6$ $6\times 1= 6$
So total 72 + 6 = 78, which is around the actual molecular mass of benzene which is 78.11g/mol