Question

The entropy of vaporisation of benzene is $85 \text{ J K}^{-1} \text{ mol}^{-1}$. When 117 g of benzene vaporises at its boiling point, what is entropy change of surrounding if process is at equilibrium?

• A. $-85 \text{ J K}^{-1}$
• B. $-85 \times 1.5 \text{ J K}^{-1}$
• C. $85 \times 1.5 \text{ J K}^{-1}$
• D. $42.5 \text{ J K}^{-1}$

Hint:

Equilibrium rule: $\Delta S_{\text{total}} = 0$ System $\uparrow$ $\Rightarrow$ Surroundings $\downarrow$

Answer 1

The Correct Option is B

Concept: At equilibrium: \[ \Delta S_{\text{system}} + \Delta S_{\text{surrounding}} = 0 \Rightarrow \Delta S_{\text{surrounding}} = -\Delta S_{\text{system}} \]

Step 1: Calculate moles of benzene.
Molar mass of benzene = 78 g/mol \[ n = \frac{117}{78} = 1.5 \text{ mol} \]

Step 2: Entropy change of system. \[ \Delta S_{\text{system}} = n \times \Delta S_{\text{vap}} = 1.5 \times 85 \]

Step 3: Entropy change of surroundings. \[ \Delta S_{\text{surrounding}} = -1.5 \times 85 \]

Step 4: Conclusion.
Thus, entropy change of surrounding is $-85 \times 1.5$. Final Answer: Option (B)