Question

The standard molar enthalpy of vaporisation of benzene $\Delta_{\text {vap }} H^{\circ}$ at $353 \,K$ is $30.8 \,kJ \,mol ^{-1}$. If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of $78 \,g$ of benzene at $353\, K$ in $kJ \,mol ^{-1}$ is

• A. 37.87
• B. 27.87
• C. 33.74
• D. 17.87

Answer 1

Answer: (B)

Given,

Standard molar enthalpy of benzene

$(\Delta H)=30.8\, kJ\, mol ^{-1}$

Temperature $(T)=353 \,K$

Equation

$\underset{\text{(Liquid)}}{C_6H_6} \ce{->[{\Delta}]}\underset{\text{(Gas)}}{C_6H_6}$

Here, change in gaseous mole $\left(\Delta n_{g}\right)=1$

Now, $ \Delta H=\Delta E+\Delta n_{g} R T $

$30.8\, kJ \, mol ^{-1} =\Delta E+(1)(0.082) \times(101.32)(353) 10^{-3} \, kJ \, mol ^{-1} $

$30.8\, kJ\, mol ^{-1} =\Delta E+2934.84 \times 10^{-3} \, kJ\, mol ^{-1} $

$\Delta E =30.8-2.93 $

$ \Delta E =27.87\, kJ \, mol ^{-1} $