Problem: Calculate the boiling point of benzene (C6H6). Round to nearest degree Celsius.

FREE Expert Solution

The vaporization of benzene involves the following reaction


C6H6(l)  C6H6 (g)


Svap = HvapTb


where

ΔSvap = entropy of vaporization

ΔHvap = enthalpy of vaporization

Tb = normal boiling point


Step 1: For ΔH˚vap:

We can use the following equation to solve for ΔH˚vap:

H°vap = H°f, prod - H°f, react


Note that we need to multiply each ΔH˚f by the stoichiometric coefficient since ΔH˚f is in kJ/mol. Also, note that ΔH˚f for elements in their standard state is 0. 


We are given:

ΔH°, C6H6 (l) = 49.1 kJ/mol

ΔH°, C6H6 (g) = 82.9 kJ/mol


So the enthalpy of vaporization


H°vap = H°f, prod - H°f, reactH°vap =H°, C6H6 (g) - H°, C6H6 (l)H°vap = 82.9 kJmol - 49.1 kJmol


ΔH°vap = 33.8 kJ/mol



Step 2: For ΔS˚vap:

We can use the following equation to solve for ΔS˚vap:

S°vap = S°f, prod - S°f, react


Note that we need to multiply each S˚ by the stoichiometric coefficient since S˚ is in J/mol • K. We also need to convert ΔS˚rxn from J/mol • K to kJ/ mol • K so our units remain consistent.


We are given:

S°, C6H6 (l) = 173.4 J/mol-K

S°, C6H6 (g) = 269.2 J/mol-K


So the enthalpy of vaporization


S°vap = S°f, prod - S°f, reactS°vap =S°, C6H6 (g) - S°, C6H6 (l)S°vap = 269.2 Jmol·K - 173.4 Jmol·KS°vap = 95.8 Jmol·K×1 kJ103 J


ΔS°vap = 0.0958 kJ/mol-K



Step 3: We can derive the normal boiling from:


Svap = HvapTbSvap ×Tb= HvapTb×TbSvap ×TbSvap=HvapSvapTb = HvapSvap


We substitute the values we calculate

ΔHvap = 33.9 kJ/mol

ΔSvap= 0.0958 kJ/mol-K

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Problem Details

Calculate the boiling point of benzene (C6H6). Round to nearest degree Celsius.

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