The document provides information on Hess's law and how to use standard enthalpy of formation values to calculate enthalpy changes for chemical reactions. It explains that Hess's law states that the enthalpy change for a reaction is independent of pathway and is equal to the sum of enthalpy changes in the stepwise reactions. Standard enthalpy of formation values are given for many substances, and these can be used together with Hess's law to calculate the enthalpy change of a reaction from the standard enthalpies of formation of the products and reactants. Several examples are shown of using this approach to determine enthalpy changes for different reactions.

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Video Tutorial on Hess’s Law, Enthalpy Formation
and Combustion.

2. 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 of its pathway
∆H rxn in series steps = sum of enthalpy changes for individual steps
∆H1
∆H2 ∆H4
∆H1
∆H3
Energy Level or Cycle Diagram to find ΔH
State function - property of system whose
magnitude depend on initial and final state
∆H3
A → D
A → B → C → D
∆H1 = H2 + H3 + H4
∆H A → E is same regardless of 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
Energy Level Diagram
2
2
1
O
∆H2 = H1 - H3 = -394 +283 = -111 kJ mol-1
∆H3 = -283
2
2
1
O
Energy Cycle 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 ????

3. 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 of its pathway
∆H rxn in series steps = sum of enthalpy changes for individual steps
∆H1
∆H2 ∆H4
∆H1
∆H3
Energy Level or Cycle Diagram to find ΔH
State function - property of system whose
magnitude depend on initial and final state
∆H3
A → D
A → B → C → D
∆H1 = H2 + H3 + H4
∆H A → E is same regardless of 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
Energy Level Diagram
Find ∆H cannot be measured
directly/experimentally
∆H2 = H1 - H3 = -395 + 98 = - 297 kJ mol-1
∆H3 = - 98
2
2
1
O
Energy Cycle 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 ?????

4. 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 of its pathway
∆H rxn in series steps = sum of enthalpy changes for individual steps
∆H1
∆H2 ∆H4
∆H1
∆H3
Energy Level or Cycle Diagram to find ΔH
State function - property of system whose
magnitude depend on initial and final state
∆H3
A → D
A → B → C → D
∆H1 = H2 + H3 + H4
∆H A → E is same regardless of 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
Energy Level Diagram
Find ∆H cannot be measured
directly/experimentally
∆H3 = + 9
Energy Cycle 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

5. ∆Hf
θ (reactant) ∆Hf
θ (product)
Using Std ∆Hf
θ formation to 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 formation to 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

6. 2CH4(g) + 4O2(g) → 4H2O + 2CO2(s) ∆Hc
θ = - 890 x 2
= - 1780 kJ mol -1
Std ∆Hf
θ formation to 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 energy released/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 released to surrounding
↓
More ↑ energetically stable (lower in energy)
↓
Do not decompose easily
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 formation vs ∆Hc
θ combustion
∆H Form - std state liquid
∆H Comb - std state liquid
More –ve – more stable
Across Period 3
↓
∆H – more ↑ –ve
↓
Lower in energy
↓
Oxides more stable
Across Period 3
↓
∆H – more ↑ +ve
↓
Higher in energy
↓
Oxides less stable – decompose easily
∆Hf = ∆Hc

7. ∆Hrxn
C2H4 + H2 C2H6
C2H6 + 3.5 O2 2CO2 + 3 H2O
∆Hf
θ (reactant) ∆Hf
θ (product)
Hess’s Law
Std ∆Hf
θ formation to 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
Energy Level 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

8. ∆Hf
θ (reactant) ∆Hf
θ (product)
∆Hf
θ (reactant) ∆Hf
θ (product)
Using Std ∆Hf
θ formation to 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 formation to 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

9. ∆Hc
θ (reactant) ∆Hc
θ (product)
∆Hc
θ combustion to find ∆H formation of hydrocarbon
∆H when 1 mol completely burnt 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 combustion to 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 = formation of H2O
∆Hc = ∆Hf
Combustion C = formation of 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 -12C + 3H2 C2H6
+ 3.5O2 + 3.5O2
x 2 x 3
How enthalpy formation hydrocarbon obtained ?

10. -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 formation of 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
Energy Level Diagram
∆Hc
θ = -395
∆Hc
θ = - 1560
∆Hc
θ = -286
Combusted products
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
Combusted products
2CO2 + 3H2O
∆Hrxn = (- 395 x 2 + -286 x 3) – (-1371 ) = - 275 kJ mol-1
+ 3.5O2 + 3.5O2
+ 3.5O2 + 3.5O2

11. 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 formation of 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
Energy Level Diagram
∆Hc
θ = -395
∆Hc
θ = - 3271
∆Hc
θ = -286
Combusted products
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
Combusted products
6CO2 + 3H2O
∆Hrxn = (- 1300 x 3 ) – (-3270 ) = - 630 kJ mol-1
-3900 - 3270
Combustion
C2H2
+ 7.5O2 + 7.5O2
+ 7.5O2 + 7.5O2

12. 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 formation of hydrocarbon
∆Hrxn
Hess’s Law
Reactants Products
∆Hrxn
θ
∆Hc
θ (product)∆Hc
θ (reactant)
∆Hc
θ -1411 -286 - 1560
Reactants Products
∆Hrxn
θ
Reactants Products
Energy Level Diagram
∆Hc
θ = -1411
∆Hc
θ = - 1560
∆Hc
θ = -286
Combusted products
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
Combusted products
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

13. ∆Hc
Std ∆Hf
θ formation to 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 formation to 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 ∆H rxn 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

14. Diamond unstable respect to graphite
↓
Kinetically stable (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 stability vs Kinetic stability
C(s) + O2(g) → CO2 ∆H = - 394
CO(g) + 1.5O2(g) → CO2 ∆H = - 283
C(s) + 1.5O2(g) → CO ∆H = - 111
Lower in energy ( -ve ∆H)
↓
Thermodynamically more 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
↓
Kinetically stable (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 energetically unstable respect to CO2
↓
Kinetically stable (High Ea)
↓
Wont react spontaneous unless 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

15. ∆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 capacity flask must be determined.
CuSO4 (s) + 5H2O → CuSO4 .5H2O
1. Find heat capacity flask
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 absorb cold 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
T final = 27.5 C
Mass CuSO4 5H2O = 6.24 g (0.025mol)
Mass water = 42.75 g
T initial 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 transfer to 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

16. Time/
m
0 0.5 1 1.5 2 2.5 3 3.5 4
Temp/C 22 22 22 22 22 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
T final = 28 C
Mass CuSO4 5H2O = 6.24 g (0.025 mol)
Mass water = 42.75 g
T initial 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
Data collection
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 23 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

17. ∆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 capacity flask must be determined.
MgSO4(s) + 7H2O → MgSO4 .7H2O
1. Find heat capacity flask
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 absorb cold 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
T final = 35.4 C
Mass MgSO4 7H2O = 6.16 g (0.025mol)
Mass water = 41.8 g
T initial 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

18. ∆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 capacity flask must be determined.
Mg + ½ O2 → MgO
1. Find heat capacity flask
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 absorb cold 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/Conc HCI = 100 g, 0.1M
T initial mix = 22 C
T final = 41 C
Mass MgO = 1 g (o.o248 mol)
Vol/Conc HCI = 100 g, 0.1M
T initial 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

19. 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 capacity of flask must be determined.
1. Find heat capacity flask
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 absorb cold 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 flask, c = 63.5JK-1
Mass KHCO3 = 3.5 g (0.035 mol)
Vol/Conc HCI = 30 g, 2M
T initial mix = 25 C
T final = 20 C
Mass K2CO3 = 2.75 g (0.02 mol)
Vol/Conc HCI = 30 g, 2M
T initial 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 + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(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

20. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com