Time (mins)Time (mins) TempTemp ObservationsObservations

55 6464

1010 8585

10.310.3 7070

1111 6060

11.311.3 5656

1212 5555

12.312.3 5454

1313 5454

13.313.3 5454

1414 5353

14.314.3 5353 solidsolid

1515 5050

15.315.3 4646

Thermochemical principles includes:

transfer of heat between the system and the surroundings

the enthalpy change for any process is the sum of the enthalpy changes for the steps into which the process can be divided

definition of the following terms: cH, fH, rH, vapH and fusH.

Heat

Heat is the transfer of energy from regions of high temperature to regions of low temperature.

As temperature is a measure of the movement of molecules (average kinetic energy), heat moves from regions of high thermal motion to a region where there is less thermal motion.

Enthalpy

When considering the heat change in chemical reactions we use a quantity called

enthalpy (H) which is a measure of the chemical potential energy stored in the

bonds of the substances involved.

Enthalpy

A change in enthalpy of a system is equal to the heat released or absorbed at constant

pressure. It is measured in kilojoules, kJ (or joules, J).

The change in enthalpy is given by rH, where

rH = Hfinal – H initial

= H products - H reactants

Enthalpy

Exothermic and Endothermic processes

For an exothermic reaction, heat is released and

rH < 0In this case the products are more stable because they contain less enthalpy than

they started with.

Enthalpy, H rH < 0

Products

This means the temperature of all species and the surroundings increases. All combustion reactions are exothermic.

This is shown in an energy profile diagram

Enthalpy

The chemical potential energy is not lost but is converted into increased kinetic energy of

all the particles (both reactants and products) and also transferred to the

surroundings.

An endothermic reaction absorbs heat from the surroundings because the enthalpy of the products is more than reactants and rH > 0. This means the

temperature of the system gets colder.

Dissolving may be endothermic or exothermic depending on the nature of the solute and solvent e.g. dissolving NH4NO3(s) in water is endothermic while dissolving conc H2SO4 is exothermic.

Exercise:

In the space below draw a labelled diagram for an endothermic reaction, showing reactants, products and rH.

Enthalpy, H rH > 0Products

REACTANTS

PRODUCTS

Enthalpy of physical changes A heating curve of a substance shows the variation of temperature of a sample as it is heated. gas b.p. Temperature / oC liquid m.p. solid Time

At both its melting point and boiling point, even though the sample is being heated, temperature remains constant. All heat being added is used to break the forces holding the molecules in their solid or liquid states, rather than being converted into kinetic energy.

Enthalpy of physical changes A heating curve of a substance shows the variation of temperature of a sample as it is heated. gas b.p. Temperature / oC liquid m.p. solid Time

Could you explain looking at the graph why a drink stays cold while it contains ice cubes?

Molar heat of Fusion (∆fusH)

This is the energy required to change one mole of a substance from a solid to a liquid at the melting point.

H2O (solid at 0 oC) H2O (liquid at 0 oC) ∆fusH=6.0 kJmol-1

The ∆fusH gives an indication of the strength of forces holding the particles together in the solid phase

The melting and boiling points of a substance indicate the strength of the forces between the particles in the solid or liquid phases

Molar heat of vaporisation (∆vapH)

This is the energy required to change one mole of a substance from a liquid to a gas at the boiling point.

H2O (liquid at 100 oC) H2O (gas at 100 oC) ∆vapH = 41.0kJmol-1

The ∆vapH gives an indication of the strength of forces holding the particles together in the liquid phase

Enthalpies

We can’t measure enthalpies of reactants and products directly all we can measure are enthalpy changes.

Data books list the standard enthalpies of combustion and /or the standard enthalpy of formation for compounds.

These figures can be used to calculate the enthalpy change for other reactions

(∆cH 0) – Standard enthalpy (heat) of combustion

The standard enthalpy of combustion is the enthalpy change when one mole of the substance is completely burnt with all reactants and products in their standard states

Standard states are the states at room temperature (25oC) and pressure - that is O2 is a gas and water is H2O (l)

Enthalpy Definitions

C2H6(g) + 3 ½ O2(g) 2CO2(g) + 3H2O(l) ∆cH 0 = -1557kJmol-1

Heats of combustion are always exothermic

(∆fH 0) – Standard enthalpy (heat) of formation

The standard enthalpy of formation of a compound is the enthalpy change when one mole of the substance is formed from its elements, with all the reactants and products in their standard states. 2C(s) + 3 H2(g) + ½ O2 (g) C2H5 OH(l)∆fHo= -278kJmol-1

Do you think these enthalpies of formation look strange if so why?

Data tables list heats of formation for compounds not elements - why? Think standard states

The standard enthalpy of formation for all elements is zero

Remember fH0 is the standard reaction enthalpy for the formation of one mole of substance from its elements in their most stable form at standard state. The units of fH0 are kJ mol-1 and the equation it refers to must have only one mole of product.

Note:

If 4C(s) + 6H2(g) + O2(g) 2C2H5OH (l) rH0 = -555 kJ

Then f H0 (C2H5OH, l) = -555 2 = - 277.5 kJ mol-1

Exercise - Write equations for the reactions which have the following enthalpies of combustion:

cH0 (H2, 25 oC) = -286 kJ mol-1

cH0 (CH3OH, 25 oC) = -726 kJ mol-1

H2(g) + ½ O2(g) H2O(l) ∆cH = -286kJmol-1

CH3OH(g) + 1 ½ O2(g) CO2(g) + 2H2O(l) ∆cH = -726kJmol-1

Exercise 1 - Write balanced equations for the formation reactions from their elements

(a) C6H12O6(s)

6C(s) + 6H2(g) + 3O2(g) C6H12O6 (s)

(b) CO(g)

C(s) + ½ O2(g) CO (g)

(c) HCl(g)

½ H2 (g) + ½ Cl2(g) HCl (g)

Note : We use mole fractions where necessary to give one mole of the compound

StartercH0 means the e_______ of c__________the 0 means s_______ c________

standard conditions - refers to____ mole at ____degrees celsius and ___ atmosphere of p______

ENERGY CALCULATIONS

Calculations involving thermochemical principles may include:

relating enthalpy changes to heat and mass, and use of specific heat capacity of water

application of ΣrH = Σ fH (products) - Σ fH (reactants)

application of Hess’s Law

use of average bond energies in enthalpy change calculations.

Measuring enthalpies of reaction – calorimetry

To measure enthalpy changes, the reaction is carried out in an insulated container (such as a polystyrene cup) and the temperature change is measured.

Using this temperature change, ΔT, and the value of the specific heat capacity, c, the amount of energy transferred to the mass m of substance can be calculated using the expression

E = m c ΔT

If the reaction takes place in an aqueous solution, and the solutions are placed in an insulated container then we can assume:

c =The specific heat capacity of the solutions is 4.18 J oC-1 g-1. The densities of these aqueous solutions can be taken as 1.0 g mL-1 so that 100 mL would have a mass of 100 g.

Calorimetry

25.0 mL of 1.0 mol L-1 HCl is placed in a polystyrene cup and its temperature is measured as 21.0 oC.

Following the addition of 25.0 mL of 1.0 mol L-1 NaOH solution, the mixture is stirred and the final temperature recorded as 27.8 oC.

Calculate the molar enthalpy change, ΔrH in kJ mol-1, for this

reaction.

(Note: If the temperature increases then the reaction is exothermic and ΔrH will be negative)

Exercise

E = m c ΔT

= 50g X 4.18 J oC-1 g-1 x 6.8 oC

= 1421.2 J

If the reaction takes place in aqueous solution, and the solutions are placed in an insulated container, the specific heat capacity of the solutions is 4.18 J oC-1 g-1.

The densities of these aqueous solutions can be taken as 1.0 g mL-1 so that 100 mL would have a mass of 100 g.

E = m c ΔT

= 50g X 4.18 J oC-1 g-1 x 6.8 oC

= 1421.2 J

Energy per mole

mol of NaOH (n) = 0.025L x 1molL-1 = 0.025 mol

1

1-

l56.848kJmo

56848Jmol

0.025mol1421.2J

nE

reactionH

ΔrH

Calculating Enthalpy From Heats of Formation

Given the enthalpies of formation of all the compounds in a given reaction, we can calculate the enthalpy of reaction using the following formula

ΣrH = ΣnfH (products) - ΣnfH (reactants)

(the symbol Σ means sum of)

Calculating Enthalpy From Heats of Formation

ΣrH = ΣnΔ fH (products) – ΣnΔ fH (reactants)

These calculations are straight forward providing you remember the following:

The formula only works if the heats of formation for all the compounds are used

Set your work out very carefully, paying particular attention to the + and - signs

Always check your answers!

What does the “n” stand for?

Example

Using the standard heats of formation of CO2(g), H2O(l), and C6H12O6(s), calculate the standard enthalpy of combustion of glucose.

fHo(CO2, g) = -394 kJ mol-1

fHo(H2O, l) = -286 kJ mol-1

fHo(C6H12O6, s) = -1268 kJ mol-1

fHo(CO2, g) = -394 kJ mol-1

fHo(H2O, l) = -286 kJ mol-1

fHo(C6H12O6, s) = -1268 kJ mol-1

Note - Start by writing an equation for the combustion of 1 mole of glucose.

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

rHo = nfHoproducts - nfHo

reactants

rHo = ( 6 x -394 + 6 x -286) - ( -1268 + 0)

= - 2812 kJ mol-1

Ethanoic acid can be formed by the oxidation of ethanol.

C2H5OH(l) + O2(g) CH3COOH(l) + H2O(l)

This reaction occurs when wine goes sour. By calculating the standard reaction enthalpy for this oxidation reaction, decide whether the reaction is exothermic or endothermic given the following data:

fHo(H2O, l) = -286 kJ mol-1

fHo(CH3COOH, l) = -485 kJ mol-1

fHo(C2H5OH, l) = -278 kJ mol-1

rHo = nfHoproducts - nfHo

reactants

rHo = (- 485 + - 286) - (-278)

= - 493 kJ mol-1 Therefore exothermic

Exercises

1. a) Using the standard heats of formation given calculate the enthalpy change for the following reaction.

2SO2(g) + O2(g) 2SO3(g)

fHo(SO2, g) = -297 kJ mol-1

fHo(SO3, g) = -396 kJ mol-1

rHo = nfHoproducts - nfHo

reactants

rHo = (2 x -396) - (2 x -297)

rHo = (2 x -396) - (2 x -297)

= - 198 kJ mol-1

b) What is the enthalpy change when 100 g of SO2 burn to form SO3?

2SO2(g) + O2(g) 2SO3(g)

fHo(SO2, g) = -297 kJ mol-1 fHo(SO3, g) = -396 kJ mol-1

rHo = nfHoproducts - nfHo

reactants

rHo = (1.563 x -396) - (1.563 x -297)

= - 154 .7 kJ

mol563.1

64gmolg100

)n(SO 1-2

= - 155 kJ (3SF)

Calculate fHo(PCl5, s) from the following information.

fHo(PCl3, l) = -320 kJ mol-1

PCl3(l) + Cl2(g) PCl5(s) rH = -124 kJ mol-1

rHo = nfHoproducts - nfHo

reactants

(-124) = fHo - (-320)

(-124) = fHo + 320

This is the unknown

fHo = - 444 kJ mol-1

Tricky enthalpy calculations

1. Calculate the final temperature when 50g of water at 70 C is mixed with 150g of water at 40 C ?

Energy loss by hot water = Energy gained by cool water

E lost = E gained (conservation of energy)

∆H(loss) = ∆H (gained)

m c ∆T = m c ∆T m c ( 70 – T) = m c (T - 40)

Hot water will cool

Cool water will heat up

Tricky enthalpy calculations

m c ∆T = m c ∆T

m c ( 70 – T) = m c (T - 40)

Hot water will cool Cool water will heat up

Solve for T

CT

T

TT

TxxTxx

o5.47

701204

120370

)40(18.4150)70(18.450

Hess’s Law

If an overall reaction can be broken down into a series of two or more steps, then the corresponding overall enthalpy of reaction is the sum of the enthalpies of the individual reaction steps.

None of the steps need to be a reaction that can be carried out in the laboratory.

Another way of saying this is that the energy difference depends only on the difference in energy between the reactants and products, not on the reaction path taken.

The process of photosynthesis is an endothermic process in which energy from the sun is trapped and stored in the bonds of glucose.

6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) rH0 = +2808kJ mol-1

It is however easier to measure the enthalpy change for the reverse reaction, the combustion of glucose (i.e. the process of respiration).

C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l) rH0 = -2808kJ mol-1

Remember if you reverse the reaction reverse the sign of rH0

Reaction 1, H1

Reaction 2, H2

Reaction 3, H3

Enthalpy,

H

rH total = rH1 - rH2 + rH3

products

reactants

rH total

the energy difference depends only on the difference in energy between the reactants and products, not on the reaction path taken.

Hess’s Lawillustrated

As a result of this principle of conservation of energy we can say:

Chemical reactions and their corresponding H values can be added , subtracted and multiplied as if they were algebraic equations

Another illustration of Hess’s law is:

C(s)

+ O2

CO2 (g)

+ ½ O2

In practice it is impossible to measure the value of H o3 (the heat of formation of CO(g)) directly as some CO2 gas is always formed as a by product.

However we can calculate it using Hess’s law.

H o 1= H o2 + H o3

H o1H o2

H o3+ ½ O2 (g) CO (g)

Using algebra and the following info find H o3

H o1 = -393kJmol-1 and H o2 = -282kJmol-1

Answer

H o 3 = H o1 - H o2

= -393 – (-282)

= -111kJ mol-1

C(s)

+ O2

CO2 (g)

+ ½ O2H o1

H o2

H o3+ ½ O2 (g) CO (g)

How’d ya go?

Solving Problems Using Hess’s Law

1.Write the data in the form of equations

2.Rewrite the equations to give the desired species on the correct side of the equation. If the reaction must be reversed (perhaps because we require a species to be a reactant and not a product) then the sign of the H must also be reversed.

3.Add the equations, and the H together

4.Check your working!!

Hang on to your hats!

Here we go………

Hess’s Law Calculation

Calculate the heat of formation of CS2(l) given that the heats combustion of carbon, sulfur and carbon disulfide. Sulfur burns to SO2 in oxygen

-393, -297 and -1007 kJ mol -1 respectively

Write the data given in the form of equations

(1) C (s) + O2(g) CO2 (g) H = -393 kJ mol -1

(2) S (s) + O2 (g) SO2 (g) H = -297 kJ mol -1

(3) CS2(s) + 3O2 (g) CO2 (g) + 2SO2 (g) H = -1007 kJ mol -1

The equation we are looking for is C(s) + 2S(s) CS2 (l)

Now rewrite the equations to give the desired species on the correct side of the equation

Hess’s Law Calculation

The equation we are looking for is C(s) + 2S(s) CS2 (l)

Rewrite the equations to give the desired species on the correct side of the equation

(1) C (s) + O2(g) CO2 (g) H = -393 kJ mol -1

(2) 2S (s) + 2O2 (g) 2SO2 (g) H = ( 2 x -297) = - 594 kJ mol -1

(3) CO2 (g) + 2SO2 (g) CS2(s) + 3O2 (g) H = +1007 kJ mol -1Now add equations and the H values together cancelling where appropriate

(1) C (s) + O2(g) CO2 (g) H = -393 kJ mol -1

(2) 2S (s) + 2O2 (g) 2SO2 (g) H = -594 kJ mol -1

(3) CO2 (g) + 2SO2 (g) CS2(s) + 3O2 (g) H = +1007 kJ mol -

1

C(s) + 2S(s) CS2 (l) H = +20 kJ mol -1

Therefore the fH o =(CS2(l)) is +20 kJ mol -

1

Now add

Important• All booklets are to be handed in worksheets 1

(8),2 (9) ,3 (10) , 4 (11) and 5 (12) Change

• All worksheets must be completed and marked before next Monday as all booklets are handed in on day of test (Monday)

• Read pathfinder thermochemistry page 54 to help understanding and bestchoice for extra practice

Hess’s Law calculation

2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l)

C(s) + O2(g) → CO2(g) ∆H = –393 kJ mol–1

H2(g) + ½O2(g) → H2O(l) ∆H = –285 kJ mol–1

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H = –1364 kJ mol–1

Calculate the enthalpy change for this reaction:

given:

2 C atoms are required as a reactant.

C(s) + O2(g) → CO2(g) ∆H = –393 kJ mol–1

2

2 2 × 2

3 H2 molecules are required as a reactant.

H2(g) + ½O2(g) → H2O(l) ∆H = –285 kJ mol–1

3

1 3 × 3

C2H5OH is a product.

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)→ ∆H = 1364 kJ mol–1

+

Remember to change the sign of the ∆H when reversing the equation.

2C(s) + 2O2(g) → 2CO2(g) ∆H = –393 kJ mol–1 × 2

3H2(g) + 1½O2(g) → 3H2O(l) ∆H = –285 kJ mol–1 × 3

2CO2(g) + 3H2O(l) → C2H5OH(l) + 3O2(g) ∆H = +1364 kJ mol–1 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l) ∆H = –277 kJ mol–1

Before adding these three equations, cancel out the terms which appear on both sides of the arrows.

2C(s) +

3H2(g) + ½O2(g) → C2H5OH(l)

Use the bracket keys on your calculator to add up the ∆H values.

It’s very easy to make errors in the exam, so do each sum twice.

You may use either of the two techniques to work out the enthalpy of a reaction

ie

Hess’s law

Or

rHo= nfHoproducts - nfHo

reactants

Very important homework

Read chapter 13 page 49

Complete questions 2 and 3 page 52

(relating ∆fusH to intermolecular forces is a favourite of examiners!)

Complete All 3 worksheets ie 10,11 & 12 in preparation for test

Bond EnergiesBond energy is a measure of the intramolecular bond strength in a covalent bond

AB (g) A(g) + B(g)

Note that these energy equations have their species as gases not as species in their standard states

Bond energy is the energy required to break one mole of a particular bond when the reactants and products are in the gaseous state

Bond Energies

The exact value of the C H bond depends on the environment it is in:

H

H H

H

C

H

Cl H

H

C

H

H H

CH3

C

So the bond energies given in data books are average bond energies

Bond Energy CalculationsBond Energy Calculations

A chemical reaction is a series of bond breaking processes (∆H positive)

And bond making processes (∆H is negative)

We can estimate the enthalpy of a chemical reaction by adding the positive bond energies for those bonds which are broken to the negative bond energies for those bonds which are made.

Enthalpy calculation using bond energiesCalculate the heat of reaction for the following

C2H6(g) + Cl2(g) C2H5Cl(g) + HCl (g)

Given the following bond energies:

C H 413 kJ mol-1 Cl Cl 242 kJ mol-1 H Cl 431 kJ mol-1

C Cl 339 kJ mol-1

Calculate the heat of reaction for the following

C2H6(g) + Cl2(g) C2H5Cl(g) + HCl (g)

Given the following bond energies:

C H 413 kJ mol-1 Cl Cl 242 kJ mol-1 H Cl 431 kJ mol-1 C Cl 339 kJ mol-1

This kind of problem is simple once you write out the equation showing structural formula with every bond shown

C C

H

H

H

H

H

H + Cl

Cl C C

H

H

H

H

H

Cl +

H Cl

C H +413 kJ mol-1

Cl Cl + 242 kJ mol-1

H Cl - 431 kJ mol-1

C Cl - 339 kJ mol-1

C C

H

H

H

H

H

H + Cl

Cl C C

H

H

H

H

H

Cl +

H Cl

Bond breaking (∆H +) Bond making (∆H -)

∆H = +413 + 242 - 339 - 431

= -115kJmol-1

rHo = Ebonds broken - E bonds formed

If you use all positive values you can

use

Note

In multiple bonds such as

O O we do not double the bond energy for a single bond (O O).

The double bond is a different kind of bond from a single bond

Enthalpy calculation using bond energiesCalculate the heat of reaction for the following

CH4(g) + 2O2(g) CO2 (g) + 2H2O (g)

Given the following bond energies:

C H = 413 kJ mol-1

H O = 463 kJ mol-1

O O = 498 kJ mol-1

C O =805 kJ mol-1

Enthalpy calculation using bond energies

Calculate the heat of reaction for the following

CH4(g) + 2O2(g) CO2 (g) + 2H2O (g)

4 x C H 4 x 413 kJ mol-1

4 x H O 4 x - 463 kJ mol-1

2 x O O 2 x 498 kJ mol-1

2 x C O 2 x -805 kJ mol-1

C

H

H

H

H +O O

COH O

O +O O

HH O H

Bonds Broken Bonds Formed

+2648 kJ mol-1 -3466 kJ mol-1

Write out structural formula

E C - H = 413 kJ mol-1

E C - Br = 285 kJ mol-1

E H - Br = 366 kJ mol-1

5 b. Bromine reacts with methane as shown below

CH4 (g) + Br2 (g) CH3Br (g) + HBr (g) rH0 = - 45kJ mol-1Given this equation and the bond enthalpies given above to calculate the bond enthalpy for Br-Br

A slightly tricky example

4 x C H 4 x 413 kJ mol-1

4 x H O 4 x - 463 kJ mol-1

2 x O O 2 x 498 kJ mol-1

2 x C O 2 x -805 kJ mol-1

Bonds Broken Bonds Formed

+2648 kJ mol-1 -3462 kJ mol-1

rHo = Hbonds broken + H bonds formed

rHo =

= - 814 kJmol -1

(+2648 kJ mol-1) +

(-3462 kJ mol-1)

Therefore Reaction is exothermic

C H + 413 kJ mol-1

Br Br ? H Br - 366 kJ mol-1

C Br - 285 kJ mol-1

C

H

H

H

H + Br Br C

H

H

H

H Br

Bond breaking (∆H +) Bond making (∆H -)

Br +

CH4 (g) + Br2 (g) CH3Br (g) + HBr (g) rH0 = - 45kJ mol-1

∆H = +413 + Br-Br - 285 - 366

-45 = +413 + Br-Br - 651 -458 = Br-Br - 651

193 kJmol-1= Br-Br

It can be seen that:

H o 1 = H o2 + H o3

NaOH(s)

+ HCl (aq)

NaCl (aq) + H2O(l)

+ HCl (l)H o1H o2

H o3 NaOH(aq)

Hesses Law of Summation page 150

What kind of question can you expect in the exam?

Here’s a question from the 2000 exam

Don’t worry if the question isn’t in a form that you recognize immediately-

Remember always keep cool and think carefully!

Are you ready for a toughie?

Q5a. Shorter covalent bonds are stronger than longer covalent bonds as illustrated by the following bond enthalpies:

E C - H = 413 kJ mol-1

E C - Br = 285 kJ mol-1

E H - Br = 366 kJ mol-1

Circle the value below that is more likely to be the bond enthalpy for C- Cl

346 kJ mol-1

243 kJ mol-1

Justify your choice

Hint: compare the radius of the Cl atom with the Br atom by looking back on the handout chart on atom radius

E C - H = 413 kJ mol-1

E C - Br = 285 kJ mol-1

E H - Br = 366 kJ mol-1

Circle the value below that is more likely to be the bond enthalpy for C- Cl

346 kJ mol-1

243 kJ mol-1 Justify your choice

Ans: C- Cl bond is stronger because the Cl atom has a smaller radius than Br so there is stronger attraction between the nucleus and the bonded electrons / C-Cl bond is shorter than the C-Br bond

(Note : discussions involving electronegativity difference or bond polarity were considered to be irrelevant)

Note

Correct answer must include units for full credit

ethanalpropanal

butanalethanoic acid

∆vapH / kJ mol–1 26 30 34 52

QUESTION FOUR: ENTHALPY OF VAPORISATIONUse the following information to answer the question

below.

Discuss the trend in vapH of the compounds in the table above in terms of the attractive forces between the particles and the factors affecting those forces.

EvidenceEvidence

Ethanal, propanal and butanal are all aldehydes. Ethanal, propanal and butanal are all aldehydes. They all have weak intermolecular or permanent They all have weak intermolecular or permanent dipole forces attracting molecules together. dipole forces attracting molecules together.

vapvapHH increases from ethanal to butanal due to increases from ethanal to butanal due to increasing M / electron numbers. The greater the M / increasing M / electron numbers. The greater the M / electron numbers the greater the strength of electron numbers the greater the strength of intermolecular or temporary dipole forces.intermolecular or temporary dipole forces.

Ethanoic acid has an H atom bonded to an O atom Ethanoic acid has an H atom bonded to an O atom so can form H-bonding between molecules.so can form H-bonding between molecules.

H bonding is a stronger intermolecular attraction H bonding is a stronger intermolecular attraction compared to temporary and permanent dipole compared to temporary and permanent dipole forces between the aldehydes.forces between the aldehydes.

AchievedAchieved

Identifying H-bonding with ethanoic acid and any weak intermolecular attraction involved.

OR

Identifying of H-bonding in ethanoic acid and explaining that there is a difference in electronegativity between O and H.

OR

Relating increase in MW to increase in vapH.

MeritMerit

Identifying H-bonding with ethanoic acid and any weak intermolecular attraction involved.

AND

Explaining presence of

H-bonding in ethanoic acid

using a difference in electronegativity.

AND ONE OF

H-bond stronger than other intermolecular forces of attraction.

OR

Explaining trend in vapH of aldehydes by relating molecular

mass to strength of intermolecular forces.

ExcellenceExcellence

Comprehensive analysis and explanation of trends in the table.

Explain temporary dipoles and the effect of increasing mass.

Explain H bonding and how it occurs in ethanoic acid.

Compare the strength of H bonding as weak intermolecular attractions.

Explain how these attractions affect vapH

Q7 (2000)

The contact process for the preparation of sulfuric acid begins with the conversion of sulfur to sulfur dioxide, followed by oxidation of the sulfur dioxide to sulfur trioxide

(1) S (s) + O2(g) SO2 (g) fHo (SO2) = -297 kJ mol -1

(2) 2 SO2 (g) + O2 (g) 2SO3 (g) rHo = -191 kJ mol -1

rHo = nfHoproducts - nfHo

reactants

exothermic

a. Is reaction 2 exothermic or endothermic?b. Use the information above to calculate fHo

(SO3,g)

This is what we want to find out

rHo = nfHoproducts - nfHo

reactants

-191 = 2fHoproducts - (-594)

Therefore fHo (SO3 g) = - 392.5kJ mol-1

(1) S (s) + O2(g) SO2 (g) -297

(2) 2 SO2 (s) + O2 (g) 2SO3 (g) rHo = -191 kJ mol -1

2 2 2 2 x = -594 kJ mol -1

-191 = 2fHoproducts + 594

(- 594 to bs) -785 = 2fHoproducts

(divide by 2) -392.5 = fHoproducts

(1) SO2 (s) + ½ O2 (g) SO3 (g) rHo = -95.5 kJ mol -1

(2) S (s) + O2(g) SO2 (g) rHo = -297 kJ mol -1

Or you could have used Hesses law adjusting each reaction to give 1 mole of product where necessary :

S (s) + 1½ O2 (g) SO3 (g) fHo = -392.5 kJ mol -1

You will know you are right when you end up with the balanced equation for the formation of SO3

Quite satisfying really don’t you think?

QUESTION ONE: SMELTING ZINC

(a) The smelting of zinc ores involves the reaction of Zn and ZnS with oxygen gas according to the following equations:

Zn(s) + ½ O2(g) ZnO(s) rH ° = – 348 kJ mol–1

ZnS(s) + 1 ½ O2(g) ZnO(s) + SO2(g) rH ° = – 441 kJ mol–1

Using the following information and the data above,

S(s) + O2(g) SO2(g) fH ° = – 297 kJ mol–1

calculate the value of the enthalpy change, r  H °, for the reaction:

Zn(s) + S(s) ZnS(s)

Reverse this reaction

EvidenceEvidence1(a) S + O2 SO2 fHo = –297 kJ mol–1

SO2 + ZnO ZnS + 1½ O2 fHo = +441 kJ mol–1

Zn + ½O2 ZnO fHo = –348 kJ mol–1

By Hess’s law of heat summation – adding these equations and the enthalpies gives

Zn(s) + S(s) ZnS(s) :

rHo = –297 + 441 – 348 = –204 kJ mol–1

AchievedAchieved MeritMerit

Correct process

with one error.

Correct value

rHo with units.

Q 1 (b)

Using the result of the calculation in part (a) above, describe, with a reason, whether the heat of formation of ZnS is endothermic or exothermic.

rHo = –297 + 441 – 348 = –204 kJ mol–1

EvidenceEvidence AchievedAchieved

1(b) 1(b)

Reaction is Reaction is exothermic since exothermic since rHo is is negative. negative.

Explanation links Explanation links answer to sign of answer to sign of rHo

calculated.calculated.

QUESTION TWO: FUEL CELLS

A fuel cell, such as that used on a space-craft, is similar to a battery. An example is the fuel cell that ‘burns’ hydrogen and oxygen to produce water and energy.

The overall equation for the reaction is

2H2(g) + O2(g) 2H2O(l ) rH ° = – 572 kJ mol–1

(a)If the water produced is in the gas phase, the equation for the reaction is

2H2(g) + O2(g) 2H2O(g)

Use the following bond enthalpies to calculate rH ° for this reaction.

Bond Bond Enthalpy kJ mol1

H–H 436

O=O 498

O–H 460

EvidenceEvidence

2(a)

= (2 × 436 + 498) – (4 × 460)

= +1370 – 1840 = – 470 kJ mol-1

rHo = Ebonds broken - E bonds formed

AchievedAchieved MeritMerit

Correct process

with one error.

Correct value

rHo with units.

2(b)Write an equation for which the enthalpy change is equal to vapH ° (H2O).

Answer For Achieved

H2O(l) H2O(g)

Correct equation showing states and 1 mole

2(c) By considering the nature of the reaction in part (b), describe why it is an endothermic change.

Evidence2(c) Energy must be absorbed to break the attractions (hydrogen bonds) holding the molecules together in the liquid state. AchievedRecognition that bond breaking is endothermic or that energy has to be put in to overcome the intermolecular attractive forces in liquid state.

(d) Using the information in parts (a) to (c) above, calculate the value of vapH ° (H2O).

2(d) 2H2O( l ) 2H2(g) + O2(g) rH = +572 kJ 2H2(g) + O2(g) 2H2O(g) rH = - 470 kJ

2H2O(l) 2H2O(g) rH = +102 kJ

Therefore H2O(l) H2O(g) rH = +51 kJ

vapHo(H2O) = +51.0 kJ mol–1 Some recognition of the fact that vapHo can be related to the enthalpies of the two reactions at the top of the page.

AchievedAchieved MeritMerit ExcellenceExcellence

Some recognition of the fact that vapHo can be related to the enthalpies of the two reactions at the top of the page.

Appropriate calculation with one error.

Correct Correct answer answer

with units of with units of kJ kJ

molmol–1–1. .

The diagram to the right shows a simple calorimeter.

It can be used to measure the enthalpy of combustion of ethanol, C2H5OH.

(a) If 1.00 g of ethanol is burned in the spirit burner, the temperature of the 200 g of water is found to increase from 22°C to 40°C. Using these results, calculate the experimental value of ∆cH (C2H5OH, ).

M(C2H5OH, ) = 46 g mol–1

Specific heat capacity of water = 4.18 J g–1 °C–1

EvidenceEvidence AchievemenAchievementt

MeritMerit ExcellencExcellencee

E = 200 × 4.18 × E = 200 × 4.18 × 18 = 15 048 18 = 15 048 joules joules

= 15.048 kJ = 15.048 kJ released released

= 0.0217 = 0.0217 mol mol

One step One step

calculated calculated

correctly. correctly.

OneOne

error in error in

calculatiocalculation.n.

cHcH

calculated calculated

correctly correctly

including including

NegativeNegative

sign & sign &

CorrectCorrect

units of units of

kJ molkJ mol–1–1..

14600.1

)( gmol

gethanoln

1693

0217.0048.15

kJmol

molkJ

cH

The experimental value of ∆cH(C2H5OH, ) calculated above, is found to be only about half the ‘accepted’ value. Use the following data to calculate ∆cH °(C2H5OH, ).

C2H5OH() + 3O2(g) 2CO2(g) + 3H2O()

∆fH °(C2H5OH, ) = –277 kJ mol–1

∆fH °(H2O, ) = –286 kJ mol–1

∆fH °(CO2, g) = –394 kJ mol–1

Evidence

rHo = nfHoproducts - nfHo

reactants

= (2 × –394 + 3 × –286) – (–277)

= –1369 kJ mol–1

AchievementAchievement MeritMerit

One error in One error in calculation.calculation.

Answer calculated Answer calculated

correctly including correctly including

Negative sign & Negative sign &

Correct units of Correct units of

kJ molkJ mol–1–1..

(c) Give two reasons why the experimental value for the enthalpy of combustion of ethanol calculated in part (a) is so much less than the ‘accepted’ value calculated in part (b).

Evidence3(c) 1. Heat is lost to the surroundings/lack of insulation.

2. Some of the ethanol that is burned undergoes incomplete combustion that releases less energy. 3. Experiment not carried out under standard conditions.

AchievementAchievement MeritMerit

One correct reason given.

A physical and a chemical reason given.

NCEA Level 3 (Chemistry) 2004 — page 5 Students were able to interpret data and explain it in terms of the nature of the solid and the forces of attraction between the particles. At this level, candidates were able to write a response which linked their understanding of thermochemical principles to the question, rather than simply write all they knew about the topic.

Many candidates who completed other parts of the paper to at least merit level omitted Question Three (a), suggesting that they were unfamiliar with the concept of specific heat capacity. This is one of the thermochemical principles included in the explanatory notes for this standard. Candidates need to be reminded that in the case of fusH (H2O), it is

simply the melting of water and NOT the reaction to produce H2 and O2. A very high proportion of the candidates in this paper gave answers that conveyed the latter misconception.

1. Calculate the enthalpy change, ΔrH° for the decomposition of sodium chlorate.

NaClO3(s) NaCl(s) + 3/2O2(g)

ΔfH° (NaClO3(s)) = –359 kJ mol1.

ΔfH° (NaCl(s)) = –411 kJ mol-1.

EvidenceEvidence AA MM

ΔrH° = [ΔfH°(NaCl) + ΔfH°(O2] – [ΔfH°(NaClO3)]

= – 411 + 0 – [–359]

= – 411 + 359

= – 52 kJ mol–1.

Correct Correct method method with with one one error.error.

CorrecCorrect t answeanswer and r and unitsunits

2. Calculate the enthalpy change for the oxidation of ammonia.

4NH3(g) + 5O2(g) 6H2O(g) + 4NO(g)

given:

N2(g) + 3H2(g) 2NH3(g) ΔH = –92 kJ mol–1

2H2(g) + O2(g) 2H2O(g) ΔH = –484 kJ mol–1

N2(g) + O2(g) 2NO(g) ΔH = +180 kJ mol–1

EvidenceEvidence

4NH3 2N2 + 6H2 ΔH = +184 kJ mol–1

6H2 + 3O2 6H2O ΔH = –1452 kJ mol–1

2N2 + 2O2 4NO ΔH = +360 kJ mol–1

4NH3(g) + 5O2(g) 6H2O(g) + 4NO(g) ΔH= –908 kJ mol–1.

OR

ΔrH° = [6 ΔfH°(H2O) + 4 ΔfH°(NO)] – [4 ΔfH°(NH3) + 5ΔfH°(O2)]

= (–1452 + 360) – (–184 + 0)

= –1092 + 184

= – 908 kJ mol–1.This was an AME question

Use the bond energy data provided below calculate the energy change for the reaction:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Bond C–H C=O O=O H–O

Average bond energies in kJ mol–1

413 745 498 463

EvidenceEvidence AA MM

Bonds broken: Bonds formed:

4 C–H = 4(413) 2 C=O = 2(– 745)

2 O=O = 2(498) 4 O–H = 4(– 463)

+2648 –3342

Overall –3342 + 2648 = – 694 kJ.mol–1

Correct Correct method method with with one one error.error.

CorrecCorrect t answeanswer and r and unitsunits

Explain why the reaction

H2O(g) H2O(l) is exothermic

Energy is released when bonds are formed (between molecules).

Explain in full what is meant by ΔcH° (S(s)) = –297 kJ mol–1.

When one mole of sulfur is burned (completely) to produce SO2, 297 kJ of heat energy is released, when starting and

finishing conditions are 25 °C and 101.3 kPa (or at standard conditions).

Explain why ΔcH° (S(s)) = ΔfH

° (SO2(g)).

ΔfH°(SO2(g)) is the heat energy released or

absorbed when 1 mole of SO2 is formed from its

elements under standard condition. Burning sulfur in oxygen produces sulfur dioxide from its elements so the two energy changes represent the same process.

Sulfuric acid is made by dissolving SO3 in water. Calculate the final

temperature of a solution of H2SO4 made by dissolving 0.03 moles

of SO3(g) in 250mL of water at 20 °C given:

ΔH = m c Δt

Where: m = mass, t = temperature c = specific heat capacity of water, 4.18 J g–1 °C–1.

H2O(l) + SO3(g) H2SO4(l) ΔrH = –814 kJ mol–1.

Molar masses H = 1, S = 32, O = 16 g mol–1

cm

Ht

18.4250

24420

ΔH = 0.03 x 814 = 24.42 kJ = 24420 J

23.36 °C

Final temperature = 23.36 + 20

= 46.36 °C AME