1. Tardigrade
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  3. Chemistry
  4. The molar enthalpy of vaporization of C 6 H 6 at its boiling point (353 K ) is 7.4 k Cal / mol. The molar internal energy change of vaporization is:

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Q. The molar enthalpy of vaporization of $C _{6} H _{6}$ at its boiling point $(353\, K )$ is $7.4 \,k \,Cal / mol$. The molar internal energy change of vaporization is:

Thermodynamics

Solution:

$C _{6} H _{6}(l) \rightleftharpoons C _{6} H _{6}( g )$

$\Delta H =\Delta E +\Delta n _{g} RT$

$7.4=\Delta E +1 \times 2 \times 10^{-3} \times 353$

$7.4=\Delta E +0.706$

$7.4-0.706=\Delta E$

$6.694 \,k\,Cal / mol =\Delta E$