The ΔHvap of a certain compound is 46.94 kJ·mol–1 and its ΔSvap is 94.51 J·mol–1

Chemistry
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The ΔHvap of a certain compound is 46.94 kJ·mol–1 and its ΔSvap is 94.51 J·mol–1·K–1. What is the boiling point of this compound?

Apr 8th, 2015

T = delta H / delta S

Remember that we need to make sure that both number have similar units.  In this case delta H is in kJ and delta S in J.  Divide the delta S value by 1000 to make sure it is in kJ

delta S = 94.51 / 1000 = 0.09451 kJ / mol*K

T = 46.94 / 0.09451 = 496.67 K

T = 223.67 deg C

Apr 8th, 2015

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