Question

The entropy of vaporization of benzene is $85J{K}^{-1}{mol}^{-1}$. When $117$ benzene vaporizes at its normal boiling point, the entropy change of surrounding is:

• A. $-85J{K}^{-1}$
• B. $-85\times 1.5J{K}^{-1}$
• C. $85\times 1.5J{K}^{-1}$
• D. None of these

Answer 1

The correct option is B $-85\times 1.5J{K}^{-1}$

To calculate the moles, we use the formula,

Moles$=\frac{Givenmass}{Molarmass}$

Given mass of Benzene$=117g$

Molar mass of Benzene$=78.11,g/mol$

We get moles$=\frac{117g}{78.11g/mol}=1.49\approx 1.5mole$

For 1 mole of Benzene, the entropy of vaporization is $85J/Kmol$

So, for 1.49 moles of Benzene, the entropy of vaporization will be $=1.5\times 85=127.5J/K$

Now at, boiling point Benzene is in equilibrium with its vapour. Therefore, the two phases are in equilibrium and for such processes

$△S_{system}+△S_{surrounding}=0$

$△S_{surrounding}=-△S_{system}=-127.5J/K$