Question

Consider the following two reactions:
(i) $Propene+H_2⟶propane;{\Delta H}_1$
(ii) $Cyclopropane+H_2⟶Porpane;{\Delta H}_2$
then ${\Delta H}_2-{\Delta H}_1$ will be:

• A. $0$
• B. $2{BE}_{C-C}-{BR}_{C=C}$
• C. ${BE}_{C=C}$
• D. $2{BE}_{C=C}-{BE}_{C-C}$

Answer 1

Answer: (B)

$2{BE}_{C-C}-{BR}_{C=C}$

Solution:- (B) $\Delta H_2-\Delta H_1=2{BE}_{C-C}-{BE}_{C=C}$

$\text{Propene}+H_2⟶\text{propane};\Delta H_1$

$\text{Cyclopropane}+H_2⟶\text{propane};\Delta H_2.....\left(1\right)$

$\text{Propane}⟶\text{propene}+H_2;-\Delta H_1.....\left(2\right)$

Now $\left(1\right)+\left(2\right)$

$\text{Cyclopropane}+H_2+\text{Propane}⟶\text{propane}+\text{propene}+H_2;\Delta H_2-\Delta H_1$

$\text{Cyclopropane}⟶\text{propene};\Delta H_2-\Delta H_1$

Here,

$\Delta H_2-\Delta H_1=\text{Heat of the reaction}$

Cyclopropane have $3C-C$ bonds while propene have $1C=C$ bond and $1C-C$ bond.

$\therefore \Delta H_2-\Delta H_1=\Delta H_R-\Delta H_P$

$\Rightarrow \Delta H_2-\Delta H_1=3{BE}_{C-C}-({BE}_{C-C}+{BE}_{C=C})$

$\Rightarrow \Delta H_2-\Delta H_1=3{BE}_{C-C}-{BE}_{C-C}-{BE}_{C=C}$

$\Rightarrow \Delta H_2-\Delta H_1=2{BE}_{C-C}-{BE}_{C=C}$