Question

The entropy of vaporization of benzene is $65{k^{ - 1}}J/mol.$ If 117g benzene vaporizes at its normal boiling point, calculate the entropy change of the surrounding.

Answer 1

Hint: We know that the formula of entropy change is ,
Entropy change of the surrounding = Number of moles × entropy change of vaporization
To solve the question, we first need to find the mole of the benzene for the given mass and already we have entropy of vaporization of for one mole i.e. $85J{k^{ - 1}}mo{l^{ - 1}}.$

Complete step-by-step answer:
Given mass is 117g and the molar mass of benzene can be calculated as follows:
$\begin{array}{c}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_6} = 6 \times {\rm{Atomic}}\;{\rm{mass}}\;{\rm{of}}\;{\rm{carbon}} + 12 \times {\rm{Atomic}}\;{\rm{mass}}\;{\rm{of}}\;{\rm{carbon}}\\ = 6 \times 12\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6 \times 1\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\\ = 72\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 6\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\\ = 78\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\end{array}$
Amount of benzene = 117gram.
From above calculation, molar mass of benzene ${{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}$ is, =$78\;g\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
So, number of moles $ = \left( {\dfrac{{117g}}{{78.11g/mol}}} \right) = 1.49 \approx 1.5\;mole.$
So, the number of mole of benzene is = 1.5 mole.
Now, we have given in the question, the entropy of vaporization for 1 mole is 85J/kmol.
So, for 1.49 or 1.5 moles of benzene, the entropy of vaporization will be $ = \left( {1.15 \times 85} \right) = 127.5J/k.$

Hence, the required entropy change of surrounding $ = \left( { - 127.5\,J/k} \right).$

Note: We calculate the entropy change of surrounding with the help of the product of entropy of vaporization of benzene and number of moles, at boiling point benzene is in equilibrium with its vapour. Therefore, the two phases are in equilibrium and for such process is
$\Delta {S_{system}} + \Delta {S_{surrounding}} = 0$
$\Delta {S_{system}} = - \Delta {S_{surrounding}} = - 127.5J/k.$