1. CO(g)
∆H2
∆H2
∆H1 = -394
CO2(g)
C(s) + ½ O2(g)
CO(g) + ½ O2
C(s) + O2(g)
CO2(g)
A E
B
C D
∆H6∆H5
∆H4∆H2
Hess’s Law
C(s) + O2(g) CO2(g)
Overall ∆H rxn is independent ofits pathway
∆H rxn in series steps = sum of enthalpychangesfor individualsteps
∆H1
∆H2 ∆H4
∆H1
∆H3
EnergyLevel or Cycle Diagram to find ΔH
State function- propertyof system whose magnitude
dependon initial and final state
∆H3
A → D
A → B → C → D
∆H1 = H2 + H3 + H4
∆H A → E is same regardlessof its path
Final state
A → E
A → C→ D → E
∆H1 = H2 + H3 + H4
A → E
A → B → E
∆H1 = H5 + H6
Initial stateInitial state
Final state
∆H3 = -283
∆H1 = ∆H2 + ∆H3
EnergyLevel Diagram
2
2
1
O
∆H2 = H1 - H3 = -394 +283 = -111 kJ mol-1
∆H3 = -283
2
2
1
O
EnergyCycle Diagram
∆H1 = -394
∆H2 = H1 - H3 = -394 +283
= -111 kJ mol-1
Path not impt !!!!
∆H1 = ∆H2 + ∆H3
C(s) + O2 → CO2 (g) ∆H1 = -394
C (s) + ½ O2 → CO(g) ∆H2 = ???
CO(g) + ½ O2 → CO2 (g) ∆H3 = -283+
Hess’s Law
Find ∆H cannot be measured
directly/experimentally
C(s) + 1/2O2 → CO(g) ∆H2 ????

2. SO2(g)
∆H2
∆H2
∆H1 = -395
SO3(g)
S(s) + O2(g)
SO2(g) + ½O2
S(s) + 3/2O2(g)
SO3(g)
A E
B
C D
∆H6∆H5
∆H4∆H2
Hess’s Law
S(s) + 3/2O2(g) SO3(g)
Overall ∆H rxn is independent ofits pathway
∆H rxn in series steps = sum of enthalpychangesfor individualsteps
∆H1
∆H2 ∆H4
∆H1
∆H3
EnergyLevel or Cycle Diagram to find ΔH
State function- propertyof system whose magnitude
dependon initial and final state
∆H3
A → D
A → B → C → D
∆H1 = H2 + H3 + H4
∆H A → E is same regardlessof its path
Final state
A → E
A → C→ D → E
∆H1 = H2 + H3 + H4
A → E
A → B → E
∆H1 = H5 + H6
Initial stateInitial state
Final state
∆H3 = -98
∆H1 = ∆H2 + ∆H3
EnergyLevel Diagram
Find ∆H cannot be measured
directly/experimentally
∆H2 = H1 - H3 = -395 + 98 = - 297 kJ mol-1
∆H3 = - 98
2
2
1
O
EnergyCycle Diagram
∆H1 = -395
∆H2 = H1 - H3 = - 395 + 98
= - 297 kJ mol-1
Path not impt !!!!
Hess’s Law
∆H1 = ∆H2 + ∆H3
S(s) + 3/2O2 → SO3 (g) ∆H1 = -395
S(s) + O2 → SO2(g) ∆H2 = ???
SO2(g) + ½O2 → SO3 (g) ∆H3 = -98+
O2
S(s) + O2 → SO2(g) ∆H2 ?????

3. N2(g) + 2O2(g)N2(g) + 2O2(g)
N2O4(g)
2NO2(g)
∆H1
N2O4(g)
∆H1
∆H2 = + 33
2NO2(g)
A E
B
C D
∆H6∆H5
∆H4∆H2
Hess’s Law
N2(g) + 2O2(g) 2NO2(g)
Overall ∆H rxn is independent ofits pathway
∆H rxn in series steps = sum of enthalpychangesfor individualsteps
∆H1
∆H2 ∆H4
∆H1
∆H3
EnergyLevel or Cycle Diagram to find ΔH
State function- propertyof system whose magnitude
dependon initial and final state
∆H3
A → D
A → B → C → D
∆H1 = H2 + H3 + H4
∆H A → E is same regardlessof its path
Final state
A → E
A → C→ D → E
∆H1 = H2 + H3 + H4
A → E
A → B → E
∆H1 = H5 + H6
Initial stateInitial state
Final state
∆H3 = + 9
EnergyLevel Diagram
Find ∆H cannot be measured
directly/experimentally
∆H3 = + 9
EnergyCycle Diagram
∆H2 = + 33
∆H1 = H2 + H3 = -33 + 9
= - 24 kJ mol-1
Path not impt !!!!
Hess’s Law
∆H1 = ∆H2 + ∆H3
N2(g)+ 2O2 → 2NO2(g) ∆H2 = +33
2NO2 → N2+ 2O2 ∆H2 = - 33
N2(g) + 2O2 → N2O4(g) ∆H3 = + 9+
2NO2(g) → N2O4(g) ∆H1 = ?
∆H1 = ∆H2 + ∆H3
∆H1 = H2 + H3 = -33 + 9 = - 24 kJ mol-1
inverse
2NO2(g) → N2O4(g) ∆H = -24

4. ∆Hf
θ (reactant) ∆Hf
θ (product)
Using Std ∆Hf
θ formationto find ∆H rxn
∆H when 1 mol form from its element under std condition
Na(s) + ½ CI2(g) → NaCI (s) ∆Hf
θ = - 411 kJ mol -1
Std Enthalpy Changes ∆Hθ
Std condition
Pressure
100kPa
Temp
298K
Conc 1M All substance
at std states
Hess’s Law
Std ∆Hf
θ formation
Mg(s) + ½ O2(g) → MgO(s) ∆Hf
θ =- 602 kJ mol -1
Reactants Products
O2(g) → O2 (g) ∆Hf
θ = 0 kJ mol -1
∆Hrxn
θ = ∑∆Hf
θ
(products) - ∑∆Hf
θ
(reactants)
∆Hf
θ (products)∆Hf
θ (reactants)
∆Hrxn
θ
Elements
Std state solid gas
2C(s) + 3H2(g)+ ½O2(g) → C2H5OH(I) ∆Hf
θ =- 275 kJ mol -1
1 mole formed
H2(g) + ½O2(g) → H2O(I) ∆Hf
θ =- 286 kJ mol -1
Std state solid gas 1 mol liquid
For element Std ∆Hf
θ formation= 0
Mg(s)→ Mg(s) ∆Hf
θ = 0 kJ mol -1
No product form
Using Std ∆Hf
θ formationto find ∆H of a rxn
Click here chem database
(std formation enthalpy)
Click here chem database
(std formation enthalpy)
C2H4 + H2 C2H6
Find ΔHθ rxn using std ∆H formation
Reactants Products
2C + 3H2
Elements
C2H4 + H2 → C2H6
∆Hrxn
θ
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = Hf
θ C2H6 - ∆Hf
θ C2H4+ H2
= - 84.6 – ( + 52.3 + 0 ) = - 136.9 kJ mol -1

5. 2CH4(g) + 4O2(g) → 4H2O + 2CO2(s) ∆Hc
θ = - 890 x 2
= - 1780 kJ mol -1
Std ∆Hf
θ formationto find ∆H rxn
∆H when 1 mol form from its element under std condition
Na(s) + ½ CI2(g) → NaCI (s) ∆Hf
θ = - 411 kJ mol -1
Hess’s Law
Std ∆Hf
θ formation
Mg(s) + ½ O2(g) → MgO(s) ∆Hf
θ =- 602 kJ mol -1
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(products) - ∑∆Hf
θ
(reactants)
∆Hf
θ (products)∆Hf
θ (reactants)
∆Hrxn
θ
Elements
Std state solid gas 1 mole formed
Total amt energyreleased/absorbed α mol reactants
CH4(g) + 2O2(g) → 2H2O + CO2(s) ∆Hc
θ = - 890 kJ mol -1
ΔH reverse EQUAL in magnitude but opposite sign to ΔH forward
Na+
(g) + CI_
(g) → NaCI(s) ∆Hlatt
θ = - 770 kJ mol -1
NaCI(s) → Na+
(g) + CI_
(g) ∆Hlatt
θ = + 770 kJ mol -1
H2(g) + ½O2(g) → H2O(I) ∆Hf
θ =- 286 kJ mol -1
H2(g) + ½O2(g) → H2O(I) ∆Hc
θ =- 286 kJ mol -1
Compound NaF NaCI NaBr NaI
Hf
θ(kJ mol-1) -573 -414 -361 -288
More ↑ – ve formation
↓
More ↑heat releasedto surrounding
↓
More ↑ energetically stable (lower in energy)
↓
Do not decomposeeasily
Subs Na2O MgO AI2O3
Hf
θ -416 -602 -1670
Subs P4O10 SO3 CI2O7
Hf
θ
-3030 -390 +250
1 mole formed
2 mole formedx 2
О
О
∆Hf
θ formationvs ∆Hc
θ combustion
∆H Form - std state liquid
∆H Comb - std state liquid
More –ve – more stable
Across Period3
↓
∆H – more ↑ –ve
↓
Lower in energy
↓
Oxides more stable
Across Period3
↓
∆H – more ↑ +ve
↓
Higherin energy
↓
Oxides less stable – decomposeeasily
∆Hf = ∆Hc

6. ∆Hrxn
C2H4 + H2 C2H6
C2H6 + 3.5 O2 2CO2 + 3 H2O
∆Hf
θ (reactant) ∆Hf
θ (product)
Hess’s Law
Std ∆Hf
θ formationto find ∆H rxn
Reactants Products
∆Hrxn
θ
∆Hf
θ (product)∆Hf
θ (reactant)
Elements
∆Hf
θ - 85 0 - 393 - 286
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 1644 - ( - 85 )
= - 1559 kJ mol -1
- 85 0 - 858 - 786
x 2 x 3
C2H4 + H2 C2H6
∆Hrxn
θReactants Products
2C + 3H2
∆Hf
θ + 52 o - 85
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 85 - ( + 52 )
= - 137 kJ mol -1
C2H6 + 3.5 O2 2CO2 + 3 H2O
2C + 3H2 + 3.5O2
EnergyLevel Diagram
2CO2 + 3H2O
2C + 3H2 + 3.5O2
C2H6 + 3.5 O2
∆Hf
θ = -85
∆Hf
θ = - 393
∆Hf
θ = 0
2C + 3H2 + 3.5O2
Elements
Reactants
Products
∆Hf
θ = - 286
∆Hrxn= (- 393 x 2 + -286 x 3) – (- 85 + 0) = - 1559 kJ mol-1
x 2
x 3
C2H4 + H2
Reactants
C2H6
∆Hrxn = - 85 – ( + 52 + 0 ) = - 137 kJ mol-1
2C + 2H2 + H2
∆Hf
θ = + 52
∆Hf
θ = 0
∆Hrxn
Products
2C + 3H2
Elements
∆Hf
θ = - 85
Elements

7. ∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hf
θ (reactant) ∆Hf
θ (product)
Using Std ∆Hf
θ formationto find ∆H rxn
Hess’s Law
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(products) - ∑∆Hf
θ
(reactants)
∆Hf
θ (products)∆Hf
θ (reactants)
∆Hrxn
θ
Elements
2H2S + SO2 3S + 2H2O
Find ΔHθ rxn using std ∆H formation
Reactants Products
3S + O2 + 2H2
Elements
∆Hrxn
θ
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 572 - ( - 338 )
= - 234 kJ mol -1
2H2S + SO2 → 3S + 2H2O
∆Hf
θ - 20.6 - 297 0 - 286
- 41.2 - 297 0 - 572
x 2 x 2
Reactants Products
Using Std ∆Hf
θ formationto find ∆H rxn
Reactants Products
∆Hrxn
θ
∆Hf
θ (product)∆Hf
θ (reactant)
4C + 12H2 + 9N2 + 10O2
Elements
∆Hf
θ + 53 - 20 - 393 - 286 0
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 5004 - ( +112 )
= - 5116 kJ mol -1
4CH3NHNH2 + 5N2O4 4CO2 + 12H2O + 9N2
4CH3NHNH2 + 5N2O4 4CO2 + 12H2O + 9N2
+ 212 - 100 - 1572 - 3432 0
x 4 x 5 x 4 x 12
NH4NO3 N2O + 2H2O
∆Hrxn
θReactants Products
N2 + 2H2 + 3/2O2
NH4NO3 N2O + 2H2O
∆Hf
θ - 366 + 82 - 286 x 2
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 488 - ( - 366 )
= - 122 kJ mol -1
Elements

8. ∆Hc
θ (reactant) ∆Hc
θ (product)
∆Hc
θ combustion to find ∆H formationof hydrocarbon
∆H when 1 mol completelyburnt in oxygen under std condition
Std Enthalpy Changes ∆Hθ
Std condition
Pressure
100kPa
Temp
298K
Conc 1M All substance
at std states
Hess’s Law
Std ∆Hc
θ combustion
C(s) + O2(g) → CO2(g) ∆Hc
θ = - 395 kJ mol -1
Reactants Products
∆Hrxn
θ = ∑∆Hc
θ
(reactant) - ∑∆Hc
θ
(product)
∆Hc
θ (product)∆Hc
θ (reactant)
∆Hrxn
θ
Combusted products
1 mole combusted
H2(g) + ½O2(g) → H2O(I) ∆Hc
θ = - 286 kJ mol -1
Std ∆Hc
θ combustionto find ∆H of a rxn
Click here chem database
(std combustion enthalpy)
Click here chem database
(std combustion enthalpy)
Find ΔHθ formation using std ∆H comb
Reactants Products
2CO2 + 3H2O
∆Hrxn
θ
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
∆Hrxn
θ = Hc
θ ( C + H2 )- ∆Hc
θ C2H6
= 2 x (-395) + 3 x (-286) – (-1560)
= - 88 kJ mol -1
H2(g) + ½O2(g) → H2O(I) ∆Hc
θ = - 286 kJ mol -1
C(s) + O2(g) → CO2(g) ∆Hc
θ = - 395 kJ mol -1
Combustion H2 = formationof H2O
∆Hc = ∆Hf
Combustion C = formationof CO2
∆Hc = ∆Hf
Using combustion data
2C(s) + 3H2(g) → C2H6(g) ∆Hf
θ = - 84 kJ mol -1
2C(s) + 3H2(g) + ½O2(g) → C2H5OH(I) ∆Hf
θ = - 275 kJ mol -1
2C + 3H2 C2H6
+ 3.5O2 + 3.5O2
x 2 x 3
How enthalpy formationhydrocarbonobtained ?

9. -790 -858 - 1371
2 C + 3H2 + 3.5O2 C2H5OH
∆Hc
θ (reactant) ∆Hc
θ (product)
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 2 x (-395) + 3 x (-286) – (-1560)
= - 88 kJ mol -1
Reactants Products
2 C + 3H2 + 3.5O2 C2H5OH
2C + 3H2 C2H6
2C + 3H2 C2H6
∆Hc
θ comb to find ∆Hf formationof hydrocarbon
∆Hrxn
Hess’s Law
Reactants Products
∆Hrxn
θ
∆Hc
θ (product)∆Hc
θ (reactant)
∆Hc
θ -395 -286 - 1560
Reactants Products
- 790 -858 - 1560
x 2
∆Hrxn
θ
Reactants Products
EnergyLevel Diagram
∆Hc
θ = -395
∆Hc
θ = - 1560
∆Hc
θ = -286
Combustedproducts
Reactants
Products
∆Hrxn = (- 395 x 2 + -286 x 3) – (-1560 ) = - 88 kJ mol-1
Reactants
∆Hrxn
Products
∆Hc
θ = - 1371
2CO2 + 3H2O
x 3
2C + 3H2
C2H6
2CO2 + 3H2O 2CO2 + 3H2O
x 2 x 3
2 C + 3H2 + 3.5O2
2CO2 + 3H2O
∆Hc
θ -395 -286 - 1371
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 2 x (-395) + 3 x (-286) – (-1371)
= - 275 kJ mol -1
x 2 x 3
C2H5OH
∆Hc
θ = -395
x 2
∆Hc
θ = -286
x 3
2CO2 + 3H2O
Combustedproducts
2CO2 + 3H2O
∆Hrxn = (- 395 x 2 + -286 x 3) – (-1371 ) = - 275 kJ mol-1
+ 3.5O2 + 3.5O2
+ 3.5O2 + 3.5O2

10. 3C2H2 C6H6
∆Hc
θ (reactant) ∆Hc
θ (product)
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 6 x (-395) + 3 x (-286) – (-3271)
= + 49 kJ mol -1
Reactants Products
3 C2H2 C6H6
6C + 3H2 C6H6
6C + 3H2 C6H6
∆Hc
θ comb to find ∆Hf formationof hydrocarbon
∆Hrxn
Hess’s Law
Reactants Products
∆Hrxn
θ
∆Hc
θ (product)∆Hc
θ (reactant)
∆Hc
θ -395 -286 - 3271
Reactants Products
-2370 -858 - 3271
x 6
∆Hrxn
θ
Reactants Products
EnergyLevel Diagram
∆Hc
θ = -395
∆Hc
θ = - 3271
∆Hc
θ = -286
Combustedproducts
Reactants
Products
∆Hrxn = (- 395 x 6 + -286 x 3) – (-3271 ) = + 49 kJ mol-1
Reactants
∆Hrxn
Products
∆Hc
θ = - 3270
6CO2 + 3H2O
x 3
6C + 3H2
C6H6
6CO2 + 3H2O 6CO2 + 3H2O
x 6 x 3
3 C2H2
6CO2 + 3H2O
∆Hc
θ -1300 - 3270
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 3 x (- 1300) – (-3270)
= - 630 kJ mol -1
x 3
C6H6
∆Hc
θ = -1300
x 3
6CO2 + 3H2O
Combustedproducts
6CO2 + 3H2O
∆Hrxn = (- 1300 x 3 ) – (-3270 ) = - 630 kJ mol-1
-3900 - 3270
Combustion
C2H2
+ 7.5O2 + 7.5O2
+ 7.5O2 + 7.5O2

11. Reactants Products
CH2 =CHCH3 + H2 → CH3CH2CH3
CH2 = CHCH3 + H2 CH3CH2CH3
Reactants Products
∆Hc
θ (reactant) ∆Hc
θ (product)
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= ( -1411) + (-286) – (-1560)
= - 137 kJ mol -1
C2H4 + H2 C2H6
∆Hc
θ comb to find ∆Hf formationof hydrocarbon
∆Hrxn
Hess’s Law
Reactants Products
∆Hrxn
θ
∆Hc
θ (product)∆Hc
θ (reactant)
∆Hc
θ -1411 -286 - 1560
Reactants Products
∆Hrxn
θ
EnergyLevel Diagram
∆Hc
θ = -1411
∆Hc
θ = - 1560
∆Hc
θ = -286
Combustedproducts
Reactants
Products
∆Hrxn = (-1411)+ (-286)– (-1560)= - 137 kJ mol-1
Reactants
∆Hrxn
Products
∆Hc
θ = - 2222
2CO2 + 3H2O
C6H6
2CO2 + 2H2O 2CO2 + 3H2O
3CO2 + 4H2O
∆Hc
θ - 2060 -286 - 2222
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= (-2060) + (-286) – (-2222)
= - 124 kJ mol -1
∆Hc
θ = -2060
3CO2 + 3H2O
Combustedproducts
3CO2 + 4H2O
∆Hrxn = (-2060)+ (-286) – (-2222 ) = - 124 kJ mol-1
Combustion
CH2 =CHCH3
C2H4 + H2 C2H6
C2H4 + H2
+ H2O
Combustion
C2H4
CH3CH2CH3
CH2=CHCH3 + H2
∆Hc
θ = -286
+ H2O
+ 3.5O2 + 3.5O2
+ 5O2 + 5O2

12. ∆Hc
Std ∆Hf
θ formationto find ∆H rxn
2H2S + SO2 → 2H2O + 3S ∆Hrxn = ?
∆Hf
θ - 21 x 2 - 297 - 286 x 2 0
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 572 - ( - 339 )
= - 233 kJ mol -1
Reactants Products
IB Question ∆Hf and ∆Hc
1 2
2 Pb(NO3)2 → 2PbO + 4 NO2 + O2 ∆Hrxn = ?
∆Hf
θ - 444 x 2 - 218 x 2 + 34 x 4 0
Std ∆Hf
θ formationto find ∆H rxn
Reactants Products
∆Hrxn
θ = ∑∆Hf
θ
(pro) - ∑∆Hf
θ
(react)
∆Hrxn
θ = - 300 - ( - 888 )
= + 588 kJ mol -1
∆Hc
θ comb to find ∆Hf formation of C6H6 (Benzene)
6C + 3H2 → C6H6 ∆H form= ?
∆Hc
θ -395 x 6 -286 x 3 - 3271
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 6 x (-395) + 3 x (-286) – (-3271)
= + 49 kJ mol -1
Reactants Products
2C + 3H2 + 3.5O2 → C2H5OH ∆Hform= ?
∆Hc
θ -395 x 2 -286 x 3 - 1371
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= 2 x (-395) + 3 x (-286) – (-1371)
= - 275 kJ mol -1
∆Hc
θ comb to find ∆Hf formation of C2H5OH (Ethanol)
Reactants Products
Combustion C / H2
to CO2 and H2O
C2H4 + H2 → C2H6 ∆H rxn = ?
∆Hc
θ comb to find ∆Hrxn of C2H6
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
= ( -1411) + (-286) – (-1560)
= - 137 kJ mol -1
∆Hc
θ -1411 -286 - 1560
Reactants Products
Find ∆Hc using ∆H provided
CH2 = CHCH3 + H2 CH3CH2CH3
3CO2 + 4H2O
CH2 =CHCH3 + H2 → CH3CH2CH3
∆Hc
θ x -286 - 2222
∆Hrxn
θ = ∑∆Hc
θ
(react) - ∑∆Hc
θ
(pro)
- 124 = x + (-286) – (-2222)
x = - 2060 kJ mol -1
Reactants Products
∆H = -2222
∆H = - 124
CH2 = CHCH3 + H2 → CH3CH2CH3 ∆H = -124
CH3CH2CH3 + 5O2 → 3CO2 + 4H2O ∆H = -2222
H2 + ½ O2 → H2O ∆H = -28.6Combustion of
hydrocarbon
Formation Enthalpy
use
NOT Formation
Enthalpy
use

13. Diamondunstablerespect tographite
↓
Kineticallystable (High Ea)
↓
Wont decompose spontaneous
∆H = - 98
∆H = - 187
H2O(I) + 1/2O2(g)
H2O2(I)
H2(g) + O2(g)
∆H2= -111
∆H1 = -394
CO2(g)
C(s) + ½ O2(g)
CO(g) + ½ O2
C(s) + O2(g)
CO2(g)
∆H3 = -283
Energy Level Diagram Energy Cycle Diagram
Energetic stabilityvs Kinetic stability
C(s) + O2(g) → CO2 ∆H = - 394
C(s) + 1.5O2(g) → CO ∆H = - 111
CO(g) + 1.5O2(g) → CO2 ∆H = - 283
Lower in energy ( -ve ∆H)
↓
Thermodynamicallymore stable
∆H = - ve
All are thermodynamically stable (-ve ∆H)
↓
More heat released to surrounding
Lower in energy
↓
Both oxides (CO2/CO) are thermodynamically
↓
Stable with respect to their element (C and O2)
H2(g) + O2(g) → H2O2 ∆H = - 187
H2O2(I) → H2O+ O2 ∆H = - 98
C(diamond) → C (graphite)
C(diamond) → C (graphite) ∆H = - 2
Diamond forever
H2O2 unstable respect to H2O/O2
↓
Kineticallystable (High Ea)
↓
Wont decompose spontaneous
All are thermodynamically stable (-ve ∆H)
↓
More heat released (lower energy)
↓
H2O2 thermodynamically more stable
with respect to its elements H2/O2
↓
H2O2 unstable with respect to H2O and O2
Will decompose to lower energy (stable)
High Activation energy
Kinetically stable
Wont decompose
∆H = - ve
H2O (I) + O2
H2O2(I)∆H = - ve
High Activation energy
Kinetically stable
Wont decompose
C + O2 energeticallyunstablerespect to CO2
↓
Kineticallystable (High Ea)
↓
Wont react spontaneousunless ignited!
C + O2
CO2(g)
High Activation energy
Kinetically stable
Wont decompose
C graphite thermodynamically more
stable with respect diamond
↓
Will diamond decompose to graphite ?
∆H = - ve
Diamond
Graphite

14. ∆H = - 50 – (+12)
= - 62 kJ mol -1
∆H = - 50 kJ mol -1
Cold water = 50g
CuSO4 .100H2O
CuSO4 (s) + 95H2O
→ CuSO4 .100H2O
CuSO4 (s) + 100H2O
→ CuSO4 .100H2O
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1 C
Initial Temp warm water = 41.3 C
Final Temp mixture = 31 C
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
Slow rxn – heat lost huge – flask used.
Heat capacityflask must be determined.
CuSO4 (s) + 5H2O → CuSO4 .5H2O
1. Find heat capacityflask
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorbcold H2O+ Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask CuSO4
Heat capacity flask, c = 63.5JK-1
Mass CuSO4 = 3.99 g (0.025mol)
Mass water = 45 g
T initial flask/water= 22.5 C
Tfinal = 27.5 C
Mass CuSO4 5H2O = 6.24 g (0.025mol)
Mass water = 42.75 g
Tinitial flask/water= 23 C
T final = 21.8 C
2. Find ∆H CuSO4 + 100H2O → CuSO4 .100H2O
3. Find ∆H CuSO4 .5H2O + 95H2O → CuSO4 .100H2O
Hess’s Law
CuSO4 + H2O → CuSO4 .100H2O CuSO4 .5H2O + H2O → CuSO4 .100H2O
Water Flask CuSO4 5H2O
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 5 + 63.5 x 5
∆Hrxn= - 1.25 kJ
∆Hrxn= 42.75 x 4.18 x 1.2 + 63.5 x 1.2
∆Hrxn= + 0.299 kJ
0.025 mol = - 1.25 kJ
1 mol = - 50 kJ mol-1
0.025 mol = + 0.299 kJ
1 mol = + 12 kJ mol-1
CuSO4 (s) + 5H2O → CuSO4 .5H2O
CuSO4 .100H2O
∆H = + 12 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.0 gml-1
Sol diluted Vol CuSO4 = Vol H2O
All heat transferto water + flask
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
limiting
Enthalpy change ∆H - using calorimeter
without temp
correction
Lit value = - 78 kJ mol -1
CONTINUE

15. Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4
Temp/C 22 22 22 22 2
2
27 28 27 26
∆H = - 60 kJ mol -1
CuSO4 .100H2O
CuSO4 (s) + 95H2O
→ CuSO4 .100H2O
CuSO4 (s) + 100H2O
→ CuSO4 .100H2O
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
CuSO4 (s) + 5H2O → CuSO4 .5H2O
Water Flask CuSO4
Mass CuSO4 = 3.99 g (0.025 mol)
Mass water = 45 g
T initial mix = 22 C
Tfinal = 28 C
Mass CuSO4 5H2O = 6.24 g (0.025 mol)
Mass water = 42.75 g
Tinitial mix = 23 C
T final = 21 C
Find ∆H CuSO4 + 100H2O → CuSO4 .100H2O
Find ∆H CuSO4 .5H2O + 95H2O → CuSO4 .100H2O
Hess’s Law
CuSO4 + H2O → CuSO4 .100H2O CuSO4 .5H2O + H2O → CuSO4 .100H2O
Water Flask CuSO4 5H2O
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 6 + 63.5 x 6
∆Hrxn= - 1.5 kJ
∆Hrxn= 42.75 x 4.18 x 2 + 63.5 x 2
∆Hrxn= + 0.48kJ
0.025 mol = - 1.5 kJ
1 mol = - 60 kJ mol-1
0.025 mol = + 0.48 kJ
1 mol = + 19 kJ mol-1
CuSO4 (s) + 5H2O → CuSO4 .5H2O
CuSO4 .100H2O
∆H = + 19 kJ mol -1
∆H = - 60 – (+19)
= - 69 kJ mol -1
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
limiting
Enthalpy change ∆H using calorimeter
Datacollection
Temp correction – using cooling curve for last 5 m
time, x = 2
initial Temp = 22 C
final Temp = 28 C
Extrapolation best curve fit
y = -2.68x+ 33
y = -2.68 x 2 + 33
y = 28 (Max Temp)
Excel plot
CuSO4 + H2O → CuSO4 .100H2O
(Exothermic) Heat released
CuSO4 .5H2O + H2O → CuSO4 .100H2O
(Endothermic) – Heat absorbed
Temp correction – using warming curve for last 5 m
Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4
Temp/C 23 23 23 23 2
3
22 22 22.4 22.7
initial Temp = 23 C
time, x = 2
final Temp = 21 C
Excel plot
Extrapolation curve fit
y = + 0.8 x + 19.4
y = + 0.8 x 2 + 19.4
y = 21 (Min Temp)
Lit value = - 78 kJ mol -1
with temp
correction

16. ∆H = - 113 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50 g
MgSO4 .100H2O
MgSO4.7H2O + 93H2O
→ MgSO4 .100H2O
MgSO4 + 100H2O
→ MgSO4 .100H2O
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1 C
Initial Temp warm water = 41.3 C
Final Temp flask/mixture= 31 C
MgSO4 (s) + 7H2O → MgSO4 .7H2O ∆H = ?
Slow rxn – heat lost huge – flask used.
Heat capacityflask must be determined.
MgSO4(s) + 7H2O → MgSO4 .7H2O
1. Find heat capacityflask
Ti = 23.1 C
Hot water = 50 g
T i = 41.3 C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorbcold H2O+ Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask MgSO4
Heat capacity flask, c = 63.5JK-1
Mass MgSO4 = 3.01 g (0.025mol)
Mass water = 45 g
T initial mix = 24.1 C
Tfinal = 35.4 C
Mass MgSO4 7H2O = 6.16 g (0.025mol)
Mass water = 41.8 g
Tinitial mix= 24.8 C
T final = 23.4 C
2. Find ∆H MgSO4 +100H2O → MgSO4 .100H2O
3. Find ∆H MgSO4 .7H2O + 93H2O → MgSO4 .100H2O
Hess’s Law
MgSO4 + H2O → MgSO4 .100H2O MgSO4 .7H2O + H2O → MgSO4 .100H2O
Water Flask MgSO4 .7H2O
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 11.3 + 63.5 x 11.3
∆Hrxn= - 2.83 kJ
∆Hrxn= 41.8 x 4.18 x 1.4 + 63.5 x 1.4
∆Hrxn= + 0.3 kJ
0.025 mol = - 2.83 kJ
1 mol = - 113 kJ mol-1
0.025 mol = + 0.3 kJ
1 mol = + 12 kJ mol-1
MgSO4 (s) + 7H2O → MgSO4 .7H2O
MgSO4 .100H2O
∆H = + 12 kJ mol -1
∆H = - 113 - 12
= - 125 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.00g ml-1
Sol diluted Vol MgSO4 = Vol H2O
All heat transfer to water + flask
Rxn fast – heat lost to surrounding minimized
Dont need to Plot Temp/time
No extrapolation
Error Analysis
Heat loss to surrounding
Heat capacity sol is not 4.18
Mass MgSO4 ignored
Impurity present
MgSo4 already hydrated
limiting

17. ∆H = - 286 kJ mol -1
∆H = - 442 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50g
Mass cold water add = 50g
Mass warm water add = 50g
Initial Temp flask/cold water = 23.1C
Initial Temp warm water = 41.3C
Final Temp flask/mixture= 31C
Mg(s) + ½ O2 → MgO ∆H = ?
Slow rxn – heat lost huge – flask is used.
Heat capacityflask must be determined.
Mg + ½ O2 → MgO
1. Find heat capacityflask
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorbcold H2O+ Heat absorb Flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask Mg
Heat capacity flask, c = 63.5JK-1
Mass Mg = 0.5 g (0.02 mol)
Vol/ConcHCI = 100 g, 0.1M
T initial mix = 22 C
Tfinal = 41 C
Mass MgO = 1 g (o.o248 mol)
Vol/Conc HCI = 100 g, 0.1M
Tinitial mix= 22 C
T final = 28.4 C
2. Find ∆H Mg + 2HCI → MgCI2 + H2
3. Find ∆H MgO + 2HCI → MgCI2 + H2O
4. Find H2 + ½ O2 → H2O
Hess’s Law
∆H Mg + 2HCI → MgCI2 + H2
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 100 x 4.18 x 19 + 63.5 x 19
∆Hrxn= - 9.11kJ
∆Hrxn= 100 x 4.18 x 6.4 + 63.5 x 6.4
∆Hrxn= -3.1 kJ
0.02 mol = - 9.11 kJ
1 mol = - 442 kJ mol-1
0.0248 mol = - 3.1 kJ
1 mol = - 125 kJ mol-1
∆H = - 125 kJ mol -1
∆H = - 442 – 286 + 125
= - 603 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2O
All heat transfer to water + flask
Rxn fast – heat lost to surrounding minimized
Dont need to Plot Temp/time
No extrapolation
Error Analysis
Heat loss to surrounding
Heat capacity sol is not 4.18
Mass MgO ignored
Impurity present
Effervescence cause loss Mg
+ 2HCI
MgCI2 + H2
HCI Flask MgO
+ ½ O2
MgCI2 + H2O
+ 2HCI
MgCI2 + H2O
∆H MgO + 2HCI → MgCI2 + H2O
Mg + ½ O2 → MgO
MgCI2 + H2
+ 2HCI
MgCI2 + H2O
+ ½ O2
limiting
MgCI2 + H2O
Data given

18. 2KHCO3(s) → K2CO3 + CO2 + H2O
∆H = + 51.4 kJ mol -1
Enthalpy change ∆H for rxn using calorimeter
Cold water = 50g
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1 C
Initial Temp warm water = 41.3 C
Final Temp flask/mixture= 31 C
2KHCO3(s) → K2CO3 + CO2 + H2O ∆H = ?
Slow rxn – heat lost huge – vacuum flask is used.
Heat capacityof flask must be determined.
1. Find heat capacityflask
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorbcold H2O+ Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask KHCO3
Heat capacity vacuum, c = 63.5JK-1
Mass KHCO3 = 3.5 g (0.035 mol)
Vol/ConcHCI = 30 g, 2M
T initial mix = 25 C
Tfinal = 20 C
Mass K2CO3 = 2.75 g (0.02 mol)
Vol/Conc HCI = 30 g, 2M
Tinitial mix= 25 C
T final = 28 C
2. Find ∆H 2KHCO3 + 2HCI → 2KCI + 2CO2 + 2H2O
3. Find ∆H K2CO3 + 2HCI → 2KCI + CO2 + H2O
Hess’s Law
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn= 30 x 4.18 x 5 + 63.5 x 5
∆Hrxn= + 0.9 kJ
∆Hrxn= 30 x 4.18 x 3 + 63.5 x 3
∆Hrxn= -0.56 kJ
0.035 mol = + 0.9 kJ
1 mol = + 25.7 kJ mol-1
0.02 mol = - 0.56 kJ
1 mol = - 28 kJ mol-1
∆H = - 28 kJ mol -1
∆H = +51.4 – (-28)
= + 79 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2O
All heat transfer to water + vacuum
Rxn fast – heat lost to surrounding minimized
Dont need to Plot Temp/time
No extrapolation
Error Analysis
Heat loss to surrounding
Heat capacity sol is not 4.18
Mass of MgO ignored
Impurity present
Effervescence cause loss Mg
+ 2HCI
HCI Flask K2CO3
2KCI + 2CO2 + 2H2O
+ 2HCI
+ 2HCI
limiting
2KHCO3(s) → K2CO3 + CO2 + H2O
2KCI + CO2 + H2O
2KHCO3 +2HCI → 2KCI + 2CO2 + 2H2O K2CO3 +2HCI → 2KCI + CO2 + H2O
x 2
2KCI + 2CO2 + 2H2O 2KCI + CO2 + H2O
+ 2HCI