Question

Standard molar.Enthalpiesbof.Formation of cyclohexane and benźene at 25 c

Answer 1

Answer:

Cyclohexene (l)+{H}_{2}(g)\rightarrow(l)+H

2

(g)→ Cyclohexene(l);\Delta H=-119kJ(l);ΔH=−119kJ

Enthalpy of formation of cyclohexane (l)=-156kJ\quad { mol }^{ -1 }(l)=−156kJmol

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So, enthalpy of formation of cyclohexane =-156-(119)kJ=−156−(119)kJ

=-37kJ\quad { mol }^{ -1 }=−37kJmol

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Since, \Delta {H}_{cyclohexane}ΔH

cyclohexane

is -156kJ\quad { mol }^{ -1 }−156kJmol

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, we can say that for every double bond the energy decreases by an amount +119kJ\quad { mol }^{ -1 }+119kJmol

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and therefore for the introducton of three double bonds (present in benzene ring) the energy required

3\times 119kJ\quad { mol }^{ -1 }=357kJ\quad { mol }^{ -1 }3×119kJmol

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=357kJmol

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Hence, theoritical \Delta { H }_{ f }ΔH

f

for benzene=(357-156)kJ\quad { mol }^{ -1 }=(357−156)kJmol

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=201kJ\quad { mol }^{ -1 }=201kJmol

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Resonance energy== Theoretical \Delta { H }_{ f }-ΔH

f

−Observed \Delta { H }_{ f }ΔH

f

(201-49)kJ\quad { mol }^{ -1 }=152kJ\quad { mol }^{ -1 }(201−49)kJmol

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=152kJmol

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