Question

Given the following bond energies
$C-H=414$ $KJ/mol$
$C-Cl=150$ $KJ/mol$
$Cl-Cl=243$ $KJ/mol$
$H-Cl=432$ $KJ/mol$

How much energy would be required in the reaction?
$C{{H}_{4}}(g)+2Cl(g)\xrightarrow[{}]{}C{{H}_{2}}C{{l}_{2}}(g)+2HCl(g)$.
A. $-150KJ$
B. $-130KJ$
C. $-228KJ$
D. $-571KJ$

Answer 1

Hint: In chemistry, the calculation of bond strength in a chemical bond is bond energy, also called mean bond enthalpy or average bond enthalpy here to define bond energy for molecule we can say the larger the average bond energy or per electron-pair bonds of a molecule, the more stable and lower-energy the molecule will be. IUPAC describes bond energy as the average value of the gas-phase bond-dissociation energy within the same chemical species for all bonds of the same kind.

Complete step by step answer:
Here we have reactions and in reactions many bonds get broken and many are formed and during this energy is released or taken. So now let us consider that $e-e$ repel $p-p$ also repel and $e-p$ attract.
Experimentally has been founded that the magnitude of attractive forces is more than the repulsive forces. As we can conclude that when two atoms approach each other then potential energy will decrease so we can say that ultimately a stage will come when the net force of attraction balances the force of repulsion and the system will acquire minimum energy.

Here in the reaction which is given $C{{H}_{4}}(g)+2Cl(g)\xrightarrow[{}]{}C{{H}_{2}}C{{l}_{2}}(g)+2HCl(g)$.Here different elements are there with bond energy which we are going to use to calculate total energy required for the reaction.

In $C{{H}_{4}}$ there are four $C-H$ bonds so we can say that
$C{{H}_{4}}=4(C-H)Bonds$
          $=4(414)$
          $=1656KJ/mol$

In $C{{l}_{2}}$ there is one $Cl-Cl$ bond so we can say that
$C{{l}_{2}}=1(Cl-Cl)Bond$
        $=243KJ/mol$

In$C{{H}_{2}}C{{l}_{2}}$ there are two $C-H$ bonds and two $C-Cl$ bonds so we can say that
$C{{H}_{2}}C{{l}_{2}}=2(C-H)Bonds+2(C-Cl)Bonds$
 $=2(414)+2(150)$
 $=828+300$
 $=1128KJ/mol$

And in $HCl$ there is one $H-Cl$ bond so we can say that
$HCl=1(H-Cl)Bond$
 $=432KJ/mol$
 Till now the values which we have calculated we are going to put in the given reaction

By which energy required in the following reaction will be
$C{{H}_{4}}(g)+2Cl(g)\xrightarrow[{}]{}C{{H}_{2}}C{{l}_{2}}(g)+2HCl(g)$
$=[(1128)+2\times (432)]-[2\times (243)+1656]$
$=1128+864-(486+1656)$
$=1128+864-2142$
$=-150 KJ$

\[\therefore \] We get total energy required for the reaction as\[-150KJ\].
So, the correct answer is “Option A”.

Note: -First calculate each molecule of energy by given bond energy then by using simple calculation add the energy of reactants and less the energy of the product to find the energy required for the following reaction.
-As bond energy is referred to as the sum of all bonds broken less the sum of all the bonds formed in the process/reaction.
-As the change in bond energy is also called as bond enthalpy denoted as ∆H.