Test: Bond Dissociation Enthalpy - NEET MCQ

∆H = (B.E)_reactant - (B.E)_product
117.36 = 716.682 + 2(278.805) - ∆H_2C=S
∆H _2C=S = 1156.932
∆H_C=S = 1156.932/2 = 578.74 kJ mol^-1

In CH4 there are 4-C-H bonds. 
Let C-H bond energy be X. So 4X = 360. or X = 90. 
Now, in C2H6 there are 6 C-H bonds and 1C-C bond. 
Let C-C B. E be y. 
So 6X + Y = 620. or Y = 620-6×90 = 80.

The correct answer is option B
During the polymerization of ethylene, one mole of ethylene breaks i.e. one C = C double bond breaks and the two CH₂- groups are linked with C-C single bonds thus forming three single bonds. 

But in the whole unit of polymer, number of single C-C bonds formed/mole of ethylene is 2.

Energy due to formation of 2 C-C single bonds = 2 X 331 = 662 kJ/mol. 

 Energy due to dissociation of 1  C = C double bond = 590 kJ/mol. 

Thus, the enthalpy of polymerisation per mole of ethylene at 298 K = (590-662) = -72 kJ/mol

C₆H₁₀(l) + H₂(g) → C₆H₁₂(l) , ∆hydrogenationH° = -119 kJ mol-1
6C(s) + 6H₂(g) → C₆H12(l) , ∆_fH° = -156 kJ mol-1
6C(s) + 3H₂(g)  → C₆H₆(l) , ∆fH° = +49 kJ mol-1
C6H6(l) → C₆H₁₂(l) , ∆hydrogenationH°
(∆_{hydrogenation}H°)_Actual = -156-49 = -2055 kJ mol-1
Hypothetical heat of        = (heat of hydrogenation = 3 × (Heat of hydrogenation 
hydrogenation of benzene   of cyclohexatriene) of cyclohexene)
= 3(-119) = -357 kJmol-1
Here, we have assumed that there are only double bonds in benzene(Kekule structure) and in hydrogenating it, energy released will be roughly 3 times of hydrogenating cyclohexene
Clearly, heat of hydrogenation, for theoretical benzene is more negative and thus it is less stable. So, resonance energy should be negative(since actual benzene is stable)
Resonance energy = -357 -(-205) = -152 kJ mol^-1

Energy of ethene =x₁
Number of ethene in benzene=3.
Therefore, Observed energy of hydrogenation of benzene =3x₁
Calculated energy of hydrogenation of benzene =x₂ (Given)
Therefore, 
Resonance energy = observed - calculated =3x₁−x₂

3O2(g) → 2O3(g)
∆H = 3(B.E)_O2 - 2(B.E)_O3
= 3(498.8) - 2(302.3)
= 287.2 kJ ml-1

S(g) + 6F(g) → SF₆(g)
∆rH = ∑∆Hreactant -  ∑∆H_product
= (275+6 80) - (-1100) 
= 1855
So,  bond energy of (S—F) = 1855/6 = 309.16 kJ mol^-1

= 946 + 249 - 607 - 418 = 1195 - 1025 = 170 kJ mo^l-1 
Resonance energy = Observed heat of formation – calculated heat of formation 
= 82 -170 = 88 kJ mol^-1

Required reaction= Heat of formation of water – H2(g) +1/2O₂ (g)= H₂O(l)
WRITE THE EQUATION PROPERLY FOR EACH EQUATION
 rearranging the reactions we have to get the above equation
B.E. of H--H= H₂(g) → 2H(g)..........X1
B.E of O₂ = O₂(g) ---> 2O(g)...........X2
B.E. OF O-H = O-H(aq)--> O(g)+H(g) …......X3
LATENT HEAT OF VAPOURISATION OF WATER = H₂O(l)----> H₂O(g)..........X4
THUS REVERSING EQTN X₃ AND X₄ , HALVING EQUATION X₂ AND DOUBLING EQUATION X₃ WE WILL GET THE DESIRED EQUATION.

For C₃H₈ : 3C + 4H₂ → C₃H₈ ; ∆H₁
For C₃H₈ : 2C + 3H₂ → C₂H₆ ; ∆H₂
∆H₁ = -[2(C-C)+8(C-H)]+[3C_s→g+4(H-H)]     -(I)
∆H₂ = -[1(C-C)+6(C-H)]+[2C_s→g+3(H-H)]     -(2)
Let bond energy of C-C be x kcal and bond energy of C-H be y  kcal.
From eqn I and 2
 ∆H₁ = -[2x+8y]+[3×172+4×104]     -(3)
∆H₂ = -[x+6y]+[2×172+3×104]     -(4)
Also given, C+O₂ → CO₂; ∆H = -94.0 kcal     -(5)
H₂ + ½ O2 → H₂O ; ∆H = -68.0 kcal      -(6)
C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O ; ∆H = -372 kcal     -(7)
C₃H₈ + 5O₂  → 3CO₂ + 4H₂O ; ∆H = -530 kcal-(8)
By inspection method,2 × (5) + 3 × (6) - (7) gives
2C + 3H₂ → C₂H₆ ; ∆H₂ = -20 kcal
And  3×(5) + 4×(6) -(8)
3C + 4H₂ → C₃H₈ ; ∆H₃ = -20 kcal 
By eqn. (3), (4), (9) and (10)
x+6y = 676
2x+8y = 956
x = 82 kcal and y = 99 kcal

For C₃H₈ : 3C + 4H₂ → C₃H₈ ; ∆H₁
For C₃H₈ : 2C + 3H₂ → C₂H₆ ; ∆H₂
∆H₁ = -[2(C-C)+8(C-H)]+[3C_s→g+4(H-H)]     -(I)
∆H₂ = -[1(C-C)+6(C-H)]+[2C_s→g+3(H-H)]     -(2)
Let bond energy of C-C be x kcal and bond energy of C-H be y  kcal.
From eqn I and 2
 ∆H₁ = -[2x+8y]+[3×172+4×104]     -(3)
∆H₂ = -[x+6y]+[2×172+3×104]     -(4)
Also given, C+O₂ → CO₂; ∆H = -94.0 kcal     -(5)
H₂ + ½ O₂ → H₂O ; ∆H = -68.0 kcal      -(6)
C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O ; ∆H = -372 kcal     -(7)
C₃H₈ + 5O2  → 3CO₂ + 4H₂O ; ∆H = -530 kcal     -(8)
By inspection method,  2 × (5) + 3 × (6) - (7) gives
2C + 3H₂ → C₂H₆ ; ∆H₂ = -20 kcal
And  3×(5) + 4×(6) -(8)
3C + 4H₂ → C₃H₈ ; ∆H₃ = -20 kcal 
By eqn. (3), (4), (9) and (10)
x+6y = 676
2x+8y = 956
x = 82 kcal and y = 99 kcal

Let us assume the structures as given below and calculate bond energies,
Let us assume the given structures of C₆H₆ are:

A + 4H2  ---> n - hexane
H =5101+4(435) -7451 = -610KJ/mol
B+4H2 ---> n - hexane
H =5138+4(435)-74551 =-573KJ/mol
Hence our assumed structures are correct.
 

Number of bonds between C and C are 3
So the correct answer is 3.

∆H = ∑(B.E)_reactant - ∑(B.E)_product
For CH₄
C(g) + 4H(g) → CH₄
∆H = (718.4) + 4(217.95) - (-74.9)
4 C-H  = 1665.1
C-H = 416.3
For C₂H₆
2C(g) + 6H(g) → C₂H₆(g)
∆H = 2(718.4) + 6(217.95) - (-84.7)
C-C+ 6C-H = 2828.5
C-C = (2828.5 - 2497.8) = 331.4 
From here, we can see that option c is correct.