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Question: Enthalpy of atomization of \( C = a \) Bond enthalpy of \( {H_2} = b \) Enthalpy of formation ...

Enthalpy of atomization of C=aC = a
Bond enthalpy of H2=b{H_2} = b
Enthalpy of formation of CH4=cC{H_4} = c
Enthalpy of formation of C2H6=d{C_2}{H_6} = d
What is the C−CC - C bond energy?
(A) a2−d+3c2\dfrac{a}{2} - d + \dfrac{{3c}}{2}
(B) d−3c2d - \dfrac{{3c}}{2}
(C) a+2d+3ca + 2d + 3c
(D) 2d−3c2d - 3c

Explanation

Solution

Every spontaneous process leads to the formation of new products. Of all the processes known to us, some absorb energy while another result in the evolution of energy. Hence, we always experience a change in enthalpy upon accomplishment of processes. This change in enthalpy can be due to the enthalpy of atomization, solution, etc.

Complete answer:
now we have to find the bond energy of C−CC - C so we will take the help of a reaction and the given information, the reaction is
2C(g)+3H2(g)→C2H6(g)2C(g) + 3{H_2}(g) \to {C_2}{H_6}(g)
Now as we can see in the reaction for making the product several things will happen like the bond between hydrogen will break and there will be formation of bond between carbon – carbon and carbon – hydrogen and the values of these enthalpies are given in the reaction.
Which are,
Enthalpy of atomization of C=aC = a
Bond enthalpy of H2=b{H_2} = b
Enthalpy of formation of CH4=cC{H_4} = c
Enthalpy of formation of C2H6=d{C_2}{H_6} = d
Now the enthalpy of reaction will be ΔHR\Delta {H_R} =bond energy of reactant – bond energy of product
So from this we get the enthalpy of formation of methane which is our c so,
c=a+2bc = a + 2b ⇒\Rightarrow b=c−a2b = \dfrac{{c - a}}{2}
And for our main reaction which is 2C(g)+3H2(g)→C2H6(g)2C(g) + 3{H_2}(g) \to {C_2}{H_6}(g)
The ΔHR\Delta {H_R} will be ΔHR=2a+3b−(bond dissociation enthalpy for C - C bond)\Delta {H_R} = 2a + 3b - (bond{\text{ dissociation enthalpy for C - C bond)}}
Now from here we can find the bond dissociation enthalpy of carbon-carbon bond which is
B.D.E. of C - C = 2a + 3b - dB.D.E.{\text{ }}of{\text{ C - C = 2a + 3b - d}}
Now put the value of b in this equation we will get
B.D.E. of C - C = 2a + 3(c−a2) - dB.D.E.{\text{ }}of{\text{ C - C = 2a + 3}}\left( {\dfrac{{c - a}}{2}} \right){\text{ - d}}
Hence our answer is option A B.D.E. of C - C = a2−d+3c2B.D.E.{\text{ }}of{\text{ C - C = }}\dfrac{a}{2} - d + \dfrac{{3c}}{2} .

Note:
The energy needed to detach one electron from each mole of free gaseous atoms of that element is the first ionization energy of an atom. There are still positive (i.e. endothermic) atomization enthalpies and ionization enthalpies. And also keep in mind that while solving these types of questions you have to take account of every elementary step like from methane there will be formation of ethane, you cannot directly make ethane from carbon and hydrogen.