Question

$\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}=\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}+H-H\to H-\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}-\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}-H$

From the following bond energies:

$H-H$ bond energy : $431.37kJmo{l}^{-1}$

$C=C$ bond energy : $606.10kJmo{l}^{-1}$

$C-C$ bond energy : $336.49kJmo{l}^{-1}$

$C-H$ bond energy : $410.50kJmo{l}^{-1}$

Enthalpy for the reactions, will be?

• A. $+553.0kJmo{l}^{-1}$
• B. $+1523.6kJmo{l}^{-1}$
• C. $-243.6kJmo{l}^{-1}$
• D. $-120.0kJmo{l}^{-1}$

Answer 1

The correct option is D $-120.0kJmo{l}^{-1}$

$\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}=\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}+H-H\to H-\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}-\stackrel{\underset{|}{H}}{\underset{\stackrel{|}{H}}{C}}-H$

Given:-

${B.E.}_{H-H}=431.37KJ/mol$

${B.E.}_{C=C}=606.10KJ/mol$

${B.E.}_{C-C}=336.49KJ/mol$

${B.E.}_{C-H}=410.50KJ/mol$

$\therefore$ Enthalpy for the given reaction $=$ Bond dissociation energy of reactant $-$ Bond dissociation energy of product

$\therefore {\Delta H}_{reaction}=({B.E.}_{C=C}+4{B.E.}_{C-H}+{B.E.}_{H-H})-(6{B.E.}_{C-H}+{B.E.}_{C-C})$

$\Rightarrow {\Delta H}_{reaction}=(606.1+(4\times 410.5)+431.37)-(6\times 410.5+336.49)$

$\Rightarrow {\Delta H}_{reaction}=(606.1+1642+431.37)-(2463+336.49)$

$\Rightarrow {\Delta H}_{reaction}=2679.47-2799.49=-120.02KJ/mol$

Hence enthalpy for the given reaction will be $-120KJ/mol$.