Question

The entropy of vaporisation of benzene is 85 J K⁻¹ mol⁻¹. When 117 g of benzene vaporises at its boiling point, what is the entropy change of the surroundings if the process is at equilibrium?

पर्याय

• −85 J K⁻¹

• −85 × 1.5 J K⁻¹

• 85 × 1.5 J K⁻¹

• 42.5 J K⁻¹

Answer 1

−85 × 1.5 J K⁻¹

Explanation:

Δ_vapH of benzene for 1 mole = 85 J K⁻¹ mol⁻¹

117 g of C₆H₆ = `(117  g)/(78  g  mol^-1)` = 1.5 moles

∴ Δ_vapH for 117 g of C₆H₆ = −85 × 1.5 J K⁻¹