IB Chemistry on Bond Enthalpy and Bond Dissociation Energy

1. 4 (C-H) bond - ∆H = 1662
1 (C-H) bond – ∆H = 1662/4
∆H = 414
∆Hat C = + 715
∆Hat H = + 218
H
H
H
C(g) + 4H (g)
CH4 (g)
∆Hrxn
CH4 (g)
∆H(a)H = + 218
∆H(a)C = + 715
∆H (f) = -75
C(s) + 2H2 (g)
Bond Enthalpy/BE
Average Energyneed to breakbond in gaseousstate
BE dependon molecularenvironwhichbond exist.
↓
Average BE values are used/ kJ mol-1
↓
Only approximation !!!!!!!
Average Energy414 kJ
neededbreakbond
Energyneededseparatedatoms(gas state)
↓
Measure actualstrength chemicalbond
↓
∆Hθ when bond is cleavedby homolysis
Bond DissociationEnergy/BDE
Click here database
(average BE)
Average only
ActualEnergy need to break chemical bond
• .• .
Homolysis break
ActualEnergy
neededbreakbond
CH
H
How average BE (C –H),
CH4 calculated?
Actual energy
Vs
Hess’s Law
∆H = + 715 + 218 + 75
∆H = + 1662 kJ mol-1
C(g) + 4H (g)
H
C
H
H H
∆H atomization - 1 mol gaseousatom form
from its element under std condition
Bond DissociationEnergy
H
C
Breaking CH3 – H = 435
Breaking CH2 – H = 459
Breaking CH – H = 422
Breaking C – H = 338
H
H
H
H
HH C
C C
H
Breaking C-H in
diff environment
Break CH3 – H = 435 Break CH2 – H = 459
Break CH – H = 422 Break C – H = 338
H
All BE C –H bond diff
Average BE vs BDE
AverageBE =414
Gaseous
state

2. ∆H(a)C = + 715
C(g) + 4CI (g)
Average Energyneed to breakbond in gaseousstate
4 (C-CI) bond- ∆H = 1300
1 (C-CI) bond– ∆H = 1300/4
∆H = 325
∆Hat C = + 715
∆Hat CI = + 121
CI
CI
CI
CCI4 (g)
∆Hrxn
CCI4 (g)
∆H(a)CI = + 121
∆H (f) = -105
C(s) + 2CI2 (g)
Bond Enthalpy/BE
BE dependon molecularenvironwhichbond exist.
↓
Average BE values are used/ kJ mol-1
↓
Only approximation !!!!!!!
Average Energy 325 kJ
neededbreakbond
Energyneededseparatedatoms(gas state)
↓
Measure actualstrength chemicalbond
↓
∆Hθ when bond is cleavedby homolysis
Bond DissociationEnergy/BDE
Click here database
(average BE)
Average only
ActualEnergy need to break chemical bond
• .• .
Homolysis break
Actual Energy
neededbreakbond
CCI
CI
How average BE (C –CI),
CCI4 calculated?
Actual energy
Vs
Hess’s Law
∆H = + 715 + 121 + 105
∆H = + 1300 kJ mol-1
C(g) + 4CI (g)
CI
C
CI
CI CI
∆H atomization - 1 mol gaseousatom form
from its element under std condition
Bond Dissociation Energy
CI
C
Breaking CCI3 – CI = 402
Breaking CCI2 – CI = 420
Breaking CCI – CI = 388
Breaking C – CI = 350
CI
CI
CI
CI
CICI C
C C
CI
Breaking C-CI in
diff environment
Break CCI3 – CI = 402 Break CCI2 – CI = 420
Break CCI – CI = 388 Break C – CI = 350
CI
All BE C –CI bond diff
Average BE vs BDE
AverageBE = 325
Gaseous
state

3. H
Bond Enthalpy/BE
Average Energy414 kJ
neededbreakbond
Bond DissociationEnergy/BDE
Click here database
Average only
• .• .
Homolysis breakActualEnergy
neededbreakbond
Actual energy
vs
Bond DissociationEnergy
H
CCH3 – H = 435
CH2 – H = 459
CH – H = 422
C – H = 338
H
H
H
H
HH C
C C
H
Breaking C-H in
diff environment
Break CH3 – H = 435 Break CH2– H = 459
Break CH – H = 422 Break C – H = 338
H
All BE C –H bond diff
Gaseous
state
Average BE vs BDE
Average BE (C –CI), CCI4 calculated?
Average BE = 414
Limitation using Average BE
• Ave BE only approximation and NOT accurate
• ∆Hc and ∆f more accurate !
• Ave BE used for similar bond within identicalmolecule
H
│
H – C – H
│
H
Diatomic molecule
2 atom with 1 bond/1 type bond exist
↓
No other molecule have same bond
↓
H - H, CI - CI, N = N, Br- Br,
H – CI, H – Br, H – I
↓
They do not have average BE.
I
│
H – C – H
│
H
Molecules> 1 bonds,similar (C – H ) bond
averageBE ( C- H ) for differentmolecule
Br
│
H – C – H
│
H
O
║
C – C – H
Similar bond (C – H) in different molecule
↓
Average BE (C- H) + 414 is used
+ 414 + 414 + 414 + 414
Similar bond (C – CI) in different molecule
↓
Average BE (C- CI) + 325 is used
H
│
H – C – CI
│
H
I
│
H – C – CI
│
H
Br
│
H – C – CI
│
I
O
║
C – C – CI
Molecules> 1 bonds,similar (C – CI ) bond
averageBE ( C- CI ) for differentmolecule
+ 325 + 325 + 325 + 325

4. Average Energyneed to breakbond in gaseousstate
Average BE ( Breaking certain bond)
CH4 + CI2 → CH3CI + HCI ∆H=?
C H H H H CI CI
CH4 + CI2 → CH3CI + HCI ∆H=?
∆Hrxn
Bond Enthalpy/BE
Average Energy
neededbreakbond
Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 1890 – 2006
= - 116 kJ mol-1
∆Hθ = ∑ Bond broken ∆Hθ = ∑ Bond form
H
│
H – C – H
│
H
CI - CI
∆H determinationCH4 + 2O2 → CO2 + 2H2O
Average BE ( Breaking all bond)
4 (C-H) → + 414 x 4
1 (CI - CI) → + 242
Total + 1890
3 (C- H) → - 412 x 3
1 (C – CI) → - 339
1 (H - CI) → - 431
Total - 2006
Using 2 method
+ HCI
+
H
│
H – C – CI
│
H
H
│
H – C – H
│
H
CH3 H CI CI
∆Hθ = ∑ Bond broken
1 (C-H) → + 414
1 (CI - CI) → + 242
Total + 656
CI - CI+
∆Hrxn
∆Hθ = ∑ Bond form
1 (C – CI) → - 339
1 (H - CI) → - 431
Total - 770
H
│
H – C – CI
│
H
+ H - CI
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 656 – 770
= - 114 kJ mol-1
SAME
Gaseous
state
Same number/type
bonds broken

5. Average Energyneed to breakbond in gaseousstate
∆Hf
θ + 52 o - 85
∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hrxn
Bond Enthalpy/BE
Average EnergykJ
neededbreakbond
Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 2696 – 2820
= - 124 kJ mol-1
∆Hθ = ∑ Bond broken
= Energy Absorbed
∆Hθ = ∑ Bond form
= Energy Released
H H
│ │
H – C – C – H
│ │
H H
H H
│ │
H - C=C - H + H - H
∆H determinationC2H4 + H2 → C2H6
C C H H H H H H
Average BE Std ∆Hf
θ formationto find ∆H rxn
1 (C=C) → + 612
4 (C-H) → + 414 x 4
1 (H-H) → + 436
Total + 2696
1 (C-C) → - 346
6 (C-H) → - 414 x 6
Total - 2820
C2H4 + H2 C2H6
Reactants Products
2C + 3H2
∆Hrxn
θ
C2H4 + H2 C2H6
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 85 - ( + 52 )
= - 137 kJ mol -1
C2H4 + H2 → C2H6 ∆H = ?C2H4 + H2 → C2H6 ∆H = ?
Using 2 method
Different
Estimate Accurate
Gaseous
state

6. Average Energyneed to breakbond in gaseousstate
∆Hf
θ + 52 o - 85
∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hrxn
Bond Enthalpy/BE
Average Energy
needed break bond Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 2696 – 2820 = - 124 kJ mol-1
∆Hθ = ∑ Bond broken ∆Hθ = ∑ Bond form
H H
│ │
H – C – C – H
│ │
H H
H H
│ │
H - C=C - H + H - H
∆H determination C2H4 + H2 → C2H6
C C H H H H H H
Average BE Std ∆Hf
θ formation
1 (C=C) → + 612
4 (C-H) → + 414 x 4
1 (H-H) → + 436
Total + 2696
1 (C-C) → - 346
6 (C-H) → - 414 x 6
Total - 2820
C2H4 + H2 C2H6
Reactants Products
2C + 3H2
∆Hrxn
θ
C2H4 + H2 C2H6
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 85 - ( + 52 )
= - 137 kJ mol -1
C2H4 + H2 → C2H6 ∆H = ?C2H4 + H2 → C2H6 ∆H = ?
Using 3 method
∆Hc
θ comb
C2H4 + H2 → C2H6 ∆H = ?
C2H4 + H2 C2H6
Reactants Products
2CO2 + 3H2O
∆Hrxn
θ
+ 3.5O2 + 3.5O2
∆Hc
θ (product)∆Hc
θ (reactant)
C2H4 + H2 C2H6
∆Hc
θ -1411 -286 - 1560
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= ( -1411) + (-286) – (-1560)
= - 137 kJ mol -1
Reactants Products
Gaseous
state
∆Hf
θ form and ∆Hc
θ comb more accurate

7. Average Energyneed to breakbond in gaseousstate
CH4 + 2O2 CO2 + 2H2O
CH4 + 2O2 → CO2 + 2H2O ∆H=?
C H H H H O O O O
CH4 + 2O2 → CO2 + 2H2O ∆H=?
∆Hf
θ - 74 o - 393 - 286 x 2
∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hrxn
Bond Enthalpy/BE
Average Energy
neededbreakbond
Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 2652 – 3466
= - 814 kJ mol-1
∆Hθ = ∑ Bond broken
= Energy Absorbed
∆Hθ = ∑ Bond form
= Energy Released
H
│
H – C – H
│
H
O = O
O = O
∆H determinationCH4 + 2O2 → CO2 + 2H2O
Average BE Std ∆Hf
θ formationto find ∆H rxn
4 (C-H) → + 414 x 4
2 (O=O) → + 498 x 2
Total + 2652
2 (C=O) → - 804 x 2
4 (O-H) → - 463 x 4
Total - 3466
CH4 + 2O2 CO2 + 2H2O
Reactants Products
C +2O2 +2H2
∆Hrxn
θ
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 965 - (- 74 )
= - 891 kJ mol -1
Using 2 method
+
Different
Estimate
Accurate
H – O – H
H – O – H0 = C =O +
Gaseous
state

8. Average Energyneed to breakbond in gaseousstate
Reactants Products
3C2H2 → C6 H6 ∆H=?
C C C C C C H H H H H H
∆Hf
θ + 228 x 3 + 49
∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hrxn
Bond Enthalpy/BE
Average Energy
neededbreakbond
Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 5001 – 5364
= - 364 kJ mol-1
∆Hθ = ∑ Bond broken
∆Hθ = ∑ Bond form
∆H determination3C2H2 → C6 H6
Average BE Std ∆Hf
θ formationto find ∆H rxn
6 (C-H) → + 414 x 6
3 (C=C) → + 839 x 3
Total + 5001
6 (C - H) → - 414 x 6
3 (C - C) → - 346 x 3
3 (C=C) → - 614 x 3
Total - 5364
Reactants Products
6C + 3H2
∆Hrxn
θ
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = + 49 - (+ 684 )
= - 635 kJ mol -1
Using 2 method
Different
Estimate
Accurate
3 C2H2 C6H6
3C2H2 → C6 H6 ∆H=?
3C2H2 C6H6
Gaseous
state

9. Average Energyneed to breakbond in gaseousstate
3C2H2 C6H6
Reactants Products
3C2H2 C6H6
3C2H2 → C6H6 ∆H = ?
∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hrxn
Bond Enthalpy/BE
Average Energy
neededbreakbond Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken ∆Hθ = ∑ Bond form
Average BE Std ∆Hf
θ formation
Reactants Products
∆Hrxn
θ
3C2H2 → C6H6 ∆H = ?
Using 3 method
∆Hc
θ comb
Reactants Products
∆Hrxn
θ
+ 7.5O2 + 7.5O2
∆Hc
θ (product)∆Hc
θ (reactant)
Reactants Products
∆H determination3C2H2 → C6 H6
3C2H2 → C6H6 ∆H = ?
C C C C C C H H H H H H
6 (C-H) → + 414 x 6
3 (C=C) → + 839 x 3
Total + 5001
6 (C - H) → - 414 x 6
3 (C - C) → - 346 x 3
3 (C=C) → - 614 x 3
Total - 5364
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 5001 – 5364
= - 364 kJ mol-1
3 C2H2 C6H6
6C + 3H2
∆Hf
θ + 228 x 3 + 49
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = + 49 - (+ 684 )
= - 635 kJ mol -1
3 C2H2 C6H6
6CO2 + 3H2O
∆Hc
θ -1300 x 3 - 3270
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 3 x (- 1300) – (-3270)
= - 630 kJ mol -1
Gaseous
state
∆Hf
θ form and ∆Hc
θ comb more accurate

10. Average BE
3C(g) + 6H(g) → CH2 = CH – CH3 ∆H =?
C C C H H H H H H
∆Hrxn
Bond Enthalpy/BE
∆Hθ = ∑ Bond broken - ∑ Bond form
= 0 – 3444
= - 3444 kJ mol-1
∆Hθ = ∑ Bond broken
= Energy Absorb
∆Hθ = ∑ Bond form
= Energy Released
Find ∆H rxn, 3C(g) + 6H(g) → CH2 = CH – CH3
AverageBE
6 (C - H) → - 414 x 6
1 (C – C) → - 356
1 (C = C) → - 598
Total - 3444
No bond broken
Find ∆H rxn, 3C(g) + 6H(g) + 2Br(g)→ CH2 = CH – CH3
3C(g) + 6H(g) + 2Br(g) → CH2 = CH – CH3
H H
│ │
C = C – C - H
│ │ │
H H H
No bond broken
∆Hθ = ∑ Bond broken
= Energy Absorb
C C C H H H H H H Br Br
∆Hθ = ∑ Bond form
= Energy Released
∆Hrxn
Gaseous
state
Gaseous
state
6 (C - H) → - 414 x 6
2 (C – C) → - 356 x 2
2 (C - Br) → - 284 x 2
Total - 3770
H H
│ │
H - C - C – C - H
│ │ │
Br Br H
∆Hθ = ∑ Bond broken - ∑ Bond form
= 0 – 3770
= - 3770 kJ mol-1
Bond Enthalpy/BE
Average Energyneed to breakbond in gaseousstate
Average Energy
needed break bond
Gaseous
state
Click here database
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release

11. N2H4 + 2F2 N2 + 4HF
N2H4(g) + 2F2(g) → N2(g) + 4HF (g) ∆H=?
N N H H H H F F F F
∆Hf
θ - 51 o 0 - 273 x 4
∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hrxn
Bond Enthalpy/BE
Average Energy
needed break bond
Click here database
Average only
Bond breaking/making involve energy
Bond Breaking
Heat energy absorb
Bond Making
Heat energy release
∆Hθ = ∑ Bond broken - ∑ Bond form
= + 2031 – 3204
= - 1173 kJ mol-1
∆Hθ = ∑ Bond broken
∆Hθ = ∑ Bond form
H H
│ │
N – N
│ │
H H
F – F
F – F
Average BE Std ∆Hf
θ formationto find ∆H rxn
1 (N - N) → + 163
2 (F - F) → + 158 x 2
4 (N- H) → + 388 x 4
Total + 2031
1 (N N) → - 944
4 (H - F) → - 565 x 4
Total - 3204
N2H4 + 2F2 N2 + 4HF
Reactants Products
N2 + 2F2 + 2H2
∆Hrxn
θ
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 1092 - (- 51)
= - 1041 kJ mol -1
Using 2 method
+
N2H4(g) + 2F2(g) → N2(g) + 4HF (g) ∆H=?
∆H determinationN2H4(g) + 2F2 → N2 + 4HF
≡
N N≡
H – F
H – F
H – F
H – F
+
Different
Estimate
Accurate
Gaseous
state
Average Energyneed to breakbond in gaseousstate