Question

(i) Cis - 2 - butene $\to$ trans-2-butene, $△H_1$
(ii) Cis - 2 - butene $\to$ 1 - butene, $△H_2$
(iii) Trans-2-butene is more stable than cis - 2 - butene.
(iv) Enthalpy of combustion of 1 - butene, $△H=-649.8kcal/mol$
(v) $9△H_1+5△H_2=0$
(vi) Enthalpy of combustion of trans 2 - butene, $△H=-647.0kcal/mol.$

The value of $△H_1$ and $△H_2$ in kcal/mol are

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

Answer 1

The correct option is A - 1.0, 1.8

a)
Cis - 2 - butene $\to$ trans-2-butene, $△H_1$
b)
Cis - 2 - butene $\to$ 1 - butene, $△H_2$

Reaction a-b we get this reaction.
1-butene $\to$ trans-2-butene
$△H_{a-b}=△H_1-△H_2=-649.8+647.0=-2.8kcal/mol$

Also,

c)
Enthalpy of combustion of 1 - butene,
$1-butene\to combustion$
$△H=-649.8kcal/mol$

d)
Enthalpy of combustion of trans 2 - butene,
$trans2-butene\to combustion$
$△H=-647.0kcal/mol.$

Reaction c- d we also get the same reaction
1-butene $\to$ trans-2-butene
$△H_{c-d}=-649.8+647.0=-2.8kcal/mol$

since, $Reactio{n}_{a-b}=Reactio{n}_{c-d}$
$△H_{a-b}=△H_{c-d}$
$△H_1-△H_2=-649.8+647.0$
$△H_1-△H_2=-2.8$

also given
$9△H_1+5△H_2=0.$


Solving these two we get
$△H_1=-1.0kcal$
$△H_2=1.8kcal$


solving $△H_1=-1.0kcal/mol;△H_2=1.8kcal/mol$