title: lesson 4 enthalpy of formation and combustion learning objectives: – calculate change in...

Title: Lesson 4 Enthalpy of Formation and Combustion

Learning Objectives:– Calculate change in enthalpy of reactions using enthalpy of

formation or combustion data

Main Menu

How to construct an enthalpy change for the reaction Reactants - the elements that make up the product in their

standard state. Theses need to be balanced (you can use fractions as the co-efficients)

Product – The compound formed from the elements. Enthalpy value will be given in the IB booklet.

Main Menu

Main Menu

Solutions

Main Menu

You need to know the ΔHθf for all the reactants and products that are

compounds.

The ΔHθf for elements is zero – the element is being formed from the

element so there’s no change in enthalpy.

Note: add when going ‘with’ an arrow, subtract when going against an arrow… (The questions will show arrows in a different directions within the cycle)

Standard Enthalpy of Formation

http://www.youtube.com/watch?v=c8Adft3M8mg





Main Menu

8 of 36 © Boardworks Ltd 2009

Using enthalpies of formation


Enthalpies of formation calculations

Calculate the enthalpy change of the combustion of ethanol

Ene

rgy

of t

he s

yste

m

Reaction path

C2H5OH + 3O2

2C + 31/2O2 + 3H2

2CO2 + 3H2O

Compound ΔHf

C2H5OH -277CO2 -394H2O -286

C2H5OH + 3O2 → 2CO2 + 3H2O

Ene

rgy

of t

he s

yste

m

Reaction path

C2H5OH + 3O2

2C + 31/2O2 + 3H2

2CO2 + 3H2O

Compound ΔHf

C2H5OH -277CO2 -394H2O -286

-277

3x -286

2 x -394


-1369KJmol-1

Ene

rgy

of t

he s

yste

m

Reaction path

C2H5OH + 3O2

2C + 31/2O2 + 3H2

2CO2 + 3H2O

Compound ΔHf

C2H5OH -277CO2 -394H2O -286

3x -286

2 x -394

-277


ΔHc = ∑ΔHofProducts - ∑ΔHofReactants = ( + ) – ( )-2772 x 394 3x -286

Ene

rgy

of t

he s

yste

m

Reaction path

CaO + H2O

What are the intermediates?

Ca(OH)2

Calculate the enthalpy change of the reaction below

CaO + H2O → Ca(OH)2

Ene

rgy

of t

he s

yste

m

Reaction path

CaO + H2O

Ca(s) + O2(g) + H2(g)

Ca(OH)2



Compound ΔHf

CaO -635Ca(OH)2 -986H2O -286

Now work out ∆Hf

Ene

rgy

of t

he s

yste

m

Reaction path

CaO + H2O

Ca(s) + O2(g) + H2(g)

Ca(OH)2



Compound ΔHf

CaO -635Ca(OH)2 -986H2O -286

∆Hf = - ((-635) + (-286)) + (-986) = -65KJmol-1

-65KJmol-1

-286-986

-635

This way is just following the direction of the arrows of the Hess cycle (elements towards reactants and products). Using the equation will give you the same answer!

CaO + H2O

Ca(s) + O2(g) + H2(g)

Ca(OH)2



Compound ΔHf

CaO -635Ca(OH)2 -986H2O -286

This can be represented as a Hess cycle

-65KJmol-1

-286-986

-635

4CH3NHNH2 + 5 N2O4

4C(s) + 9N2(g) + 12H2(g) + 10O2(g)

4CO2 + 12 H2O + 9N2

Calculate the enthalpy change of the reaction below Compound ΔHf

CH3NHNH2 +54.0N2O4 -20.0CO2 -393H2O -286

∆H = ∆H2 -∆H1

∆H

∆H2

∆H1

4CH3NHNH2 + 5 N2O4

4C(s) + 9N2(g) + 12H2(g) + 10O2(g)

4CO2 + 12 H2O + 9N2



Compound ΔHf

CH3NHNH2 +54.0N2O4 -20.0CO2 -393H2O -286

∆H1 = (5 x -20) + (4 x 54) = +116KJmol-1

∆H2 = (12 x -286) + (4 x -393) = -5004KJmol-1

∆H = ∆H2 -∆H1

∆H = -5004 – (-116)∆H = -4888KJmol-1

5 x -2012 x -286

4 x 54 4 x -393

∆H

SO2(g) + 2H2S(g)

3S(s) + 2H2(g) + O2(g)

3S(s) + 2H2O(l)

ΔHθr

Route 1

Route

2

ΔHθf(reactants)

ΔHθf(products)

REACTANTS PRODUCTS

ELEMENTS

Step 1: Write the balanced equation for the reaction. This will be ΔHθ

r

Step 2: Under the equation write a list of the elements present. This must be balanced

STANDARD ENTHALPY OF FORMATION

SO2(g) + 2H2S(g)

3S(s) + 2H2(g) + O2(g)

3S(s) + 2H2O(l)

ΔHθr

Route 1

Route

2

ΔHθf(reactants)

ΔHθf(products)

REACTANTS PRODUCTS

ELEMENTSStep 3:

ΔHθf[SO2(g)] = -297 kJmol-1

ΔHθf[H2S(g)] = -20.2 kJmol-1

ΔHθf[H2O(l)] = -286kJmol-1

Using Hess’ Law; Route 1 = Route 2


ΔHθf values give

the enthalpy change going from the element to the compound

SO2(g) + 2H2S(g)

3S(s) + 2H2(g) + O2(g)

3S(s) + 2H2O(l)

ΔHθr

Route 1

Route

2

ΔHθf(SO2) +

2 x ΔHθf(H2S)

REACTANTS PRODUCTS

ELEMENTSStep 3:






ΔHθf values give

the enthalpy change going from the element to the compound

3 x ΔHθf(S) +

2 x ΔHθf(H2O)

SO2(g) + 2H2S(g)

3S(s) + 2H2(g) + O2(g)

3S(s) + 2H2O(l)

ΔHθr

Route 1

Route

2

-297 + (2 x -20.2)

REACTANTS PRODUCTS

ELEMENTSStep 3:






ΔHθf(s)

is zero because its an element

(3 x 0) + (2 x -286)

Step 4:

ΔHθf(SO2) + ΔHθ

f(H2S) + ΔHθr = 3ΔHθ

f(S) + 2ΔHθf(H20)

-297 + (2 x -20.2) + ΔHθr = (3 x 0) + (2 x -286)

ΔHθr = (3 x 0) + (2 x -286) – (-297 + (2 x 20.2))

= -234.6 kJmol-1

SO2(g) + 2H2S(g)

3S(s) + 2H2(g) + O2(g)

3S(s) + 2H2O(l)

ΔHθr

Route 1

Route

2

-297 + (2 x -20.2)


(3 x 0) + (2 x -286)


Using enthalpies of combustion


Enthalpies of combustion calculations

TASKS:

1. Complete the ‘Enthalpy of Formation’ worksheet

2. Complete the Hess’s Law cut and stick exercise

Terry’s Cutting tip: Keep the left column and top row in one piece

HESS’S LAW AND ENTHALPY OF FORMATION

1 One mole of carbon burns to give one mole of carbon dioxide, releasing 393.5 kJ. One mole of carbon burns to give one mole of carbon monoxide, releasing 110.5 kJ. Calculate the energy from burning one mole of carbon monoxide.

By Hess’s Law: (SUM)HPRODUCTS - (SUM)HREACTANTSH == (-393.5)-(-100.5)

CO(g) + ½O2(g) CO2(g)

= -393.5

H

= - 283.0 kJ mole-1

= -110.5

Alternative route

C(s) + O2(g)

Elements here because H

f given

Hf[CO(g)] H

f[CO2(g)]

Hf[CO2(g)] = -393.5 kJ mole-1 and H

f[CO(g)] = -110.5 kJ mole-1

What info do you have?Which equation to use?

2 Use the Hf values given to calculate H of :

CH3COCH3(l) + 4O2(g) 3CO2(g) + 3H2O(l)

Hf[CO2(g)] = - 394 kJmol-1

Hf[CH3COCH3(l)] = - 248 kJmol-1

Hf[H2O(l)] = - 286 kJmol-1

= 3(-394) + 3(-286)

HCH3COCH3(l) + 4O2(g) 3CO2(g) + 3H2O(l)

= -248

H =

+ 3Hf[H2O(l)]

= - 1792 kJ mole-1

Elements here because H

f given

3C(s) + 3H2(g) + 4.5O2(g)

Hf[CH3COCH3(l)] 3H

f[CO2(g)]

= (3(-394) + 3(-286)) – (-248)

(SUM)HPRODUCTS - (SUM)HREACTANTS


3 Calculate Hf [CH4(g)], given

Hf [CO2(g)] = -393.5 kJ mole-1

Hf [H2O(l)] = -285.8 kJ mole-1

Hc[CH4(g)] = -890.3 kJ mole-1

HC(s) + 2H2(g) CH4(g)

= (-393.5) + 2(-285.8)

= - 890.3

H = (-393.5 + 2(-285.8)) – (-890.3) = - 74.8 kJ mole-1

Oxides here because H

C given

CO2(g) + 2H2O(l)

also = HC [C(s)]

also = HC [H2(g)]

HC [C(s)]

+ 2HC [H2(g)] H

c[CH4(g)]

= (SUM)HREACTANTS- (SUM)HPRODUCTS


4 Calculate HR for : C2H4(g) + H2(g) C2H6(g), given

Hc [C2H4(g)] = - 1410.8 kJ mole -1

H c [H2(g)] = - 285.8 kJ mole -1

H c [C2H6(g)] = - 1559.7 kJ mole -1

H C2H4(g) + H2(g) C2H6(g)

H =

= (-1410.8) + (-285.8)

= -1559.7

= - 136.9 kJ mole-1

= ((-1410.8) + (-285.8)) -(-1559.7)

2CO2(g) + 3H2O(l)Oxides here because H

C given

Hc [C2H4(g)]

+ H c [H2(g)]

H c [C2H6(g)]

+ 3.5O2(g) + 3.5O2(g)

(SUM)HREACTANTS- (SUM)HPRODUCTS


NB H values calculated from bond energies are

1. Average values used

2. Gaseous state may not apply.

Also, you may care to remember :

Hr = H

f [PRODUCTS] - Hf [REACTANTS]

APPROXIMATE because:

Hr = H

C [REACTANTS] - HC [PRODUCTS]

Hr = E [REACTANTS] - E [PRODUCTS]

Main Menu

Key Points

∆Hor is the enthalpy change of a reaction for

molar quantities under standard conditions

∆Hor can be calculated from enthalpies of

combustion/formation using Hess Cycles With formation, the arrows point up With combustion, the arrows point down

title: lesson 4 enthalpy of formation and combustion learning objectives: – calculate change in...

Documents

enthalpy change

enthalpy of reactions

enthalpy value

h2o 9n2

combustion of ethanol

hf14 energy

hess cycle elements

combustion of ethanolhc