Question
In: Chemistry
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?
Solutions
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 2 years agoRelated Solutions
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