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Question

(i) Cis - 2 - butene trans-2-butene, H1
(ii) Cis - 2 - butene 1 - butene, H2
(iii) Trans-2-butene is more stable than cis - 2 - butene.
(iv) Enthalpy of combustion of 1 - butene, H=649.8 kcal/mol
(v) 9H1+5H2=0
(vi) Enthalpy of combustion of trans 2 - butene, H=647.0 kcal/mol.

The value of H1 and H2 in kcal/mol are

A
- 1.0, 1.8
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B
1.8, -1.0
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C
-5, 9
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D
-2, 3.6
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Solution

The correct option is A - 1.0, 1.8
a)
Cis - 2 - butene trans-2-butene, H1
b)
Cis - 2 - butene 1 - butene, H2

Reaction a-b we get this reaction.
1-butene trans-2-butene
Hab=H1H2=649.8+647.0=2.8 kcal/mol

Also,

c)
Enthalpy of combustion of 1 - butene,
1butenecombustion
H=649.8 kcal/mol

d)
Enthalpy of combustion of trans 2 - butene,
trans2butenecombustion
H=647.0 kcal/mol.

Reaction c- d we also get the same reaction
1-butene trans-2-butene
Hcd=649.8+647.0=2.8 kcal/mol

since, Reactionab=Reactioncd
Hab=Hcd
H1H2=649.8+647.0
H1H2=2.8

also given
9 H1+5 H2=0.


Solving these two we get
H1=1.0 kcal
H2=1.8 kcal


solving H1=1.0 kcal/mol; H2=1.8 kcal/mol

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