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 state Initial state

Final state

∆H3 = -283

∆H1 = ∆H2 + ∆H3

Energy Level Diagram

22

1O

∆H2 = H1 - H3 = -394 +283 = -111 kJ mol-1

∆H3 = -283

22

1O

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 ????

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 state Initial 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

22

1O

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 ?????

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 state Initial 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

∆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

∆Hrxnθ = ∑∆Hf

θ(products) - ∑∆Hf

θ(reactants)

∆Hf θ (products) ∆Hf

θ (reactants)

∆Hrxnθ

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

∆Hrxnθ

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = Hf

θ C2H6 - ∆Hf

θ C2H4+ H2 = - 84.6 – ( + 52.3 + 0 ) = - 136.9 kJ mol -1

2CH4(g) + 4O2(g) → 4H2O + 2CO2(s) ∆Hcθ = - 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

∆Hrxnθ = ∑∆Hf

θ(products) - ∑∆Hf

θ(reactants)

∆Hf θ (products) ∆Hf

θ (reactants)

∆Hrxnθ

Elements Std state solid gas 1 mole formed

Total amt energy released/absorbed α mol reactants

CH4(g) + 2O2(g) → 2H2O + CO2(s) ∆Hcθ = - 890 kJ mol -1

ΔH reverse EQUAL in magnitude but opposite sign to ΔH forward

Na+(g) + CI_

(g) → NaCI(s) ∆Hlattθ = - 770 kJ mol -1

NaCI(s) → Na+(g) + CI_

(g) ∆Hlattθ = + 770 kJ mol -1

H2(g) + ½O2(g) → H2O(I) ∆Hf θ =- 286 kJ mol -1

H2(g) + ½O2(g) → H2O(I) ∆Hcθ =- 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 formed x 2

О

О

∆Hfθ 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

∆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

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 1644 - ( - 85 )

= - 1559 kJ mol -1

- 85 0 - 858 - 786

x 2 x 3

C2H4 + H2 C2H6

∆Hrxnθ

Reactants Products

2C + 3H2

∆Hf θ + 52 o - 85

Reactants Products

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 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

∆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

∆Hrxnθ = ∑∆Hf

θ(products) - ∑∆Hf

θ(reactants)

∆Hf θ (products) ∆Hf

θ (reactants)

∆Hrxnθ

Elements

2H2S + SO2 3S + 2H2O

Find ΔHθ rxn using std ∆H formation

Reactants Products

3S + O2 + 2H2

Elements

∆Hrxnθ

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 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

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 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

∆Hrxnθ

Reactants Products

N2 + 2H2 + 3/2O2

NH4NO3 N2O + 2H2O

∆Hf θ - 366 + 82 - 286 x 2

Reactants Products

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 488 - ( - 366 )

= - 122 kJ mol -1

Elements

∆Hc θ (reactant) ∆Hc

θ (product)

∆Hcθ 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) ∆Hcθ = - 395 kJ mol -1

Reactants Products

∆Hrxnθ = ∑∆Hc

θ(reactant) - ∑∆Hc

θ(product)

∆Hcθ (product)

∆Hc θ (reactant)

∆Hrxnθ

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

∆Hrxnθ

∆Hrxnθ = ∑∆Hc

θ(react) - ∑∆Hc

θ(pro)

∆Hrxnθ = 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) ∆Hcθ = - 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 -1 2C + 3H2 C2H6

+ 3.5O2 + 3.5O2

x 2 x 3

How enthalpy formation hydrocarbon obtained ?

-790 -858 - 1371

2 C + 3H2 + 3.5O2 C2H5OH

∆Hc θ (reactant) ∆Hc

θ (product)

∆Hrxnθ = ∑∆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

∆Hcθ 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

∆Hrxnθ

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

∆Hrxnθ = ∑∆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

3C2H2 C6H6

∆Hc θ (reactant) ∆Hc

θ (product)

∆Hrxnθ = ∑∆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

∆Hcθ 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

∆Hrxnθ

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

∆Hrxnθ = ∑∆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

Reactants Products

CH2 =CHCH3 + H2 → CH3CH2CH3

CH2 = CHCH3 + H2 CH3CH2CH3

Reactants Products

∆Hc θ (reactant) ∆Hc

θ (product)

∆Hrxnθ = ∑∆Hc

θ(react) - ∑∆Hc

θ(pro)

= ( -1411) + (-286) – (-1560) = - 137 kJ mol -1

C2H4 + H2 C2H6

∆Hcθ 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

∆Hrxnθ

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

∆Hrxnθ = ∑∆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

∆Hc

Std ∆Hf θ formation to find ∆H rxn

2H2S + SO2 → 2H2O + 3S ∆Hrxn = ? ∆Hf

θ - 21 x 2 - 297 - 286 x 2 0

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 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

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = - 300 - ( - 888 )

= + 588 kJ mol -1

∆Hcθ comb to find ∆Hf formation of C6H6 (Benzene)

6C + 3H2 → C6H6 ∆H form= ?

∆Hc θ -395 x 6 -286 x 3 - 3271

∆Hrxnθ = ∑∆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

∆Hrxnθ = ∑∆Hc

θ(react) - ∑∆Hc

θ(pro)

= 2 x (-395) + 3 x (-286) – (-1371) = - 275 kJ mol -1

∆Hcθ comb to find ∆Hf formation of C2H5OH (Ethanol)

Reactants Products

Combustion C / H2

to CO2 and H2O

C2H4 + H2 → C2H6 ∆H rxn = ?

∆Hcθ comb to find ∆H rxn of C2H6

∆Hrxnθ = ∑∆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

∆Hrxnθ = ∑∆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 = -28.6 Combustion of

hydrocarbon

Formation Enthalpy

use

NOT Formation Enthalpy

use

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 C(s) + 1.5O2(g) → CO ∆H = - 111 CO(g) + 1.5O2(g) → CO2 ∆H = - 283

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

∆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

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

∆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

∆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

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 vacuum, 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 + 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