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The ΔHvap of a certain compound is 45.76 kJ·mol–1 and its ΔSvap is 84.68 J·mol–1·K–1. What...

The ΔHvap of a certain compound is 45.76 kJ·mol–1 and its ΔSvap is 84.68 J·mol–1·K–1. What is the boiling point of this compound?

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Expert Solution

Answer – Given, ∆Hvap = 45.76 kJ·mol-1 , ∆Svap = 84.68 J.mol-1.K-1

boiling point of this compound, T = ?

We know the formula

∆Svap = ∆Hvap / T

So, T = ∆Hvap /∆Svap

          = 45760 J.mol-1 / 84.68 J.mol-1.K-1

           = 540.4 K

In the degree Celsius boiling point of this compound is T = 540.4 K – 273.15 K

                                                                                       = 267.3oC

queen_honey_blossom answered 1 year ago
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