the enthalpy of vaporization of a liquid is 30 kilo joule per
mole and entropy of vaporization is 50 j.k per mole what is boiling point of the liquid
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400k
ΔG⁰= ΔH⁰ - TΔS⁰
Given:
ΔH vap= 40.8
ΔG⁰= 0 at equilibrium
so we get
ΔS vap = ΔH vap /T
= 40.8/373
= 40.8 x 1000/ 373
= 109.38 J mol-1 K-1
ΔG⁰= ΔH⁰ - TΔS⁰
Given:
ΔH vap= 40.8
ΔG⁰= 0 at equilibrium
so we get
ΔS vap = ΔH vap /T
= 40.8/373
= 40.8 x 1000/ 373
= 109.38 J mol-1 K-1
simransimmi:
is it answer of my question??
the answer is 400k
109.38
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