enthalpy - handouts

Chains, Energy and Resources 2.3.1. Enthalpy Changes

ENTHALPY CHANGES sections (a) and (h) Answer the following questions in your group 1. What type of energy changes takes place when (the first part is answered for you)

a) A candle is burned in air chemical energy gives heat and light

b) A match is struck ……………………………………………………………………………. c) A rocket firework is lit …………………………………………………………………………….

2. In each of the following test tube reactions list the observations that would indicate to you that a reaction had occurred.

a. Adding silver nitrate solution to hydrochloric acid

………………………………………………………………………………………………………..………

b. Magnesium metal added to hydrochloric acid

…………………………………………………………………………………………………………………

c. Water added to quicklime, CaO

…………………………………………………………………………………………………………………

3. Which reactions, in solution, are quicker - ionic or covalent?

……………………………………………………………………………………………………………………...

4. Does the rate of a chemical reaction increase or decrease when the temperature of

the reaction is increased?

……………………………………………… Why? …………………………………………………………………………………………………………….

……………………………………………………………………………………..…………………….

5. Explain what the terms exothermic and endothermic mean.

exothermic …………………………………………………………..………………………………… endothermic ………………………………………………………………..……………………………

6. What does the symbol ∆H represent?

GMA/CER Enthalpy notes 2008 1


…………………………………………………………………………………………………………..………...

7. What do you understand by the statement, “∆H is negative”?

…………………………………………………………………………………………………………..………... 8. Would ∆H be positive or negative when water is added to quicklime?

…………………………………………………………………………………………………………..………... 9. What is a bond energy?

…………………………………………………………………………………………………………..………... 10. How does bond length relate to bond energy?

…………………………………………………………………………………………………………..………... 11. Complete the following using these words so the sentences make sense

endothermic, exothermic, greater, lower, releases, requires

Bond breaking is an ____________ process and ____________ energy

Bond making is an ____________ process and ____________ energy

So bond making produces ____________ stability and a ____________ energy state.

12. All bonds are the same strength. true false 13. More energy is released when strong bonds are

formed than when weak bonds are are formed. true

14. So long as particles collide they will react. true

Explain your answer: ……………………………………………… …………………………………………………………………………………

15. Complete the definition of the term bond energy/bond e

The bond enthalpy is the enthalpy ________ required to

__________________ in a ________ molecule.

GMA/CER Enthalpy notes 2008

false

false

………………………

……………………

nthalpy

________ and

…………….

…..………...

separate

2


SIMPLE ENTHALPY PROFILE DIAGRAMS sections (d) and (h) These are visual representations of the energy needed to break bonds and the energy released when bonds are made. The difference between the two values is the enthalpy change, ∆H.

Note the arrows point in the direction of the enthalpy value:

Energy given out or exothermic value arrows point downwards Energy taken in or endothermic value arrows point upwards



ENTHALPY CHANGES sections (b) and (c) Write equations, with state symbols, for the following exothermic reactions

1) methane burning in oxygen 2) petrol (assume this is octane) under going combustion in a car engine 3) the complete combustion of ethanol, C2H5OH (l) 4) oxidation of glucose, C6H12O6, in the body

Write equations, with state symbols, for the following endothermic reactions

1) heating of limestone 2) photosynthesis in plants involving the reation of carbon dioxide and water, in sunlight and

with a chlorophyll catalyst, to form glucose and oxygen

Note: Photosynthesis is a redox reaction.

Two half-equations can be written to summarise the reactions of water and carbon dioxide:

a) 2H2O O2 + 4H+ + 4e-

b) 4e- + 4H+ + CO2 (CH2O) + H2O

Reaction a) is an oxidation, reaction b) is a reduction.

Combining the two gives the overall equation for photosynthesis.



BOND ENERGY sections (h) and (i) For a chemical reaction to take place bonds must be broken. In ionic reactions the electrostatic forces between oppositely charged ions must be overcome. In reactions involving covalent molecules the attractive force between the positive nuclei and negative bonding pair electrons must be broken. Bond Energy measures the strength of a bond, often written as BE(X-Y), where X-Y is the bond in question and all measurements are taken in the gaseous state.

∗ Bond Energy is the energy needed to break a mole of the same type of bond in gaseous molecules. The units of BE are kJ mol-1.

As energy needs to be supplied to break the bond these values are endothermic that is a positive value. For example in the reaction

H2(g) ⇒ 2H(g) ; BE (H-H) = + 436 kJ mol-1

There is only one possible H - H bond that can be broken, similarly in a chlorine molecule. However, as most of the time there are many bonds between the same atoms in different molecules, for example C-C in different alkanes, we use average values of Bond Energy to calculate approximate values for the enthalpy changes in chemical reactions.

∗ Average bond energy is the energy to break one mole of a particular bond found in one mole of gaseous molecules, giving gaseous atoms.

Average bond energies are really an estimated amount of energy required to break a bond and this value is assumed to be the same in all molecules containing that bond. However the value actually depends on the chemical environment of the bond which is why we use average bond energy values. For example to break the C-H bond in methane we actually have 4 different energies CH4(g) -----------------> CH3 (g) + H (g) ∆Ho = + 425 kJ mol-1 CH3 (g) -----------------> CH2 (g) + H (g) ∆Ho = + 475 kJ mol-1

CH2 (g) -----------------> CH (g) + H (g) ∆Ho = + 409 kJ mol-1

CH (g) -----------------> C (g) + H (g) ∆Ho = + 355 kJ mol-1

Average bond energy is 416 kJ mol-1 for the C-H bond in methane These values are really called standard molar enthalpy change of bond dissociation and they refer to a specific bond in a specific molecule. Note we assume that the same numerical value is needed to form the bond. Table of common bond energies



C-H 413 kJ mol-1 C-O 358 kJ mol-1 H-H 436 kJ mol-1 Cl-H 432 kJ mol-1 C-C 347 kJ mol-1 C=O 749 kJ mol-1 O-H 464 kJ mol-1 Cl-Cl 243 kJ mol-1 C=C 611 kJ mol-1 O-O 144 kJ mol-1 C-Cl 346 kJ mol-1 O=O 496 kJ mol-1

This idea of averaging different bond energies in many molecules is why tables of bond energies can vary as they are really average bond energies. Note there are several bond energy values for the different Carbon-Carbon bonds and this is explained below: A Carbon-Carbon single bond, called a sigma or σ bond, the average BE (C-C) = + 347 kJ mol-1.

This bond is found in ethane, propane, butane etc. and is slightly different in each as we will see shortly. This bond consists of a σ bond only, with the length of the bond being 0.154nm. The bond length is a measure of the distance between two nuclei of adjacent atoms

A Carbon-Carbon double bond, called a pi or π bond, the average BE (C=C) = + 612 kJ mol-1.

This bond consists of a σ bond and a π bond, with the length of this bond being shorter at 0.134nm. As this is a shorter distance the double bond is stronger than the single bond with the atoms closer together.

However, because the double bond is two bonds of unequal strength you can work out the value of the π bond on its own.

π bond = BE (C=C) - BE (C-C) = …………….. kJ mol-1. This means that the …………….. bond is easier to break than the ………….. bond. We will see shortly in reactions of a group of organic compounds called alkenes that the π bond is often broken while leaving the σ bond intact. Note that as the bond length decreases the bond energy increases. This indicates the strength of the forces between the atoms in a covalent bond. Generally values of bond energies for covalent molecules are several hundred kJ mol-1. A similar explanation is found with the oxygen-oxygen bonds and you need to be careful the correct values are chosen when using these in calculations.



BOND ENERGIES AND ENTHALPY CHANGES sections (i) and (j) Bond energies can be used to calculate the overall energy/enthalpy changes for reactions and below is an example for you to follow. Example: Calculate the enthalpy change for the following reaction C2H6 (g) + Cl2 (g) ⎯⎯⎯⎯→ C2H5Cl (g) + HCl (g) Step 1. Draw out molecules, remember that carbon can form four bonds.

CH C H

H H

H H

Cl Cl ClC C

H H

H H

ClH H+ +

Step 2. Imagine all bonds in reactants are broken Break 6 C-H bonds 6* 413 = 2478 kJ mol-1

Break 1 C-C bond 1* 347 = 347 Break 1 Cl-Cl bond 1* 243 = 243 Total BREAK +3068 kJ mol-1 Note: The BREAK value is an endothermic value so it is a positive value Step 3. Imagine all bonds in products are made Make 5 C-H bonds 5* 413 = 2065 kJ mol-1

Make 1 C-C bonds 1* 347 = 347 Make 1 * C-Cl bond 1* 346 = 346 Make 1 * H-Cl bond 1* 432 = 432 Total MAKE +3190 kJ mol-1

Note: The MAKE value is an exothermic value so it is actually a negative value Step 4. Enthalpy change, ∆H = BREAK - MAKE = 3068 - 3190 = -122 kJ mol-1

with this formula use the numerical values only

Now try numbers 1 to 5 of the calculation sheet.



CALCULATIONS INVOLVING BOND ENERGIES sections (i) and (j) 1) CH3CH=CHCH3 (g) + H2 (g) ⎯⎯⎯→ CH3CH2CH2CH3 (g)

C C

C C

H

H

H H

H

H

H

H

+ HH

C

CC

C

H

HH

H

HH

H

HH

H

2) CH3OH (g) ⎯⎯⎯→ C (g) + 4H (g) + O (g) known as atomisation

C OH

H

H

H 3) (CH3)2C=O (g) + H2 (g) ⎯⎯⎯→ (CH3)2 CH-OH (g)

+ H H C

CC

O

H

H

H

H

H

HH

HC C

C

OH

H

H

HH

H

4) CH2=CH2 (g) + Cl2 (g) ⎯⎯⎯→ Cl CH2-CH2 Cl (g)

H

H H

H

+ Cl Cl C CH

Cl

H

H

H

Cl 5) CH2=CH2 (g) + HCl (g) ⎯⎯⎯→ CH3-CH2 Cl (g)

H

H H

H

+ H Cl C CH

H

H

H

H

Cl



MORE CALCULATIONS INVOLVING BOND ENERGIES 6) Atomisation of CH3CH2CH2CH2CH3 (g) 7) (CH3)2C=O (g) + HC≡N (g) ⎯⎯⎯→ (CH3)2 C (OH) C≡N (g)

+ H C N C

CC

O

H

H

H

C

H

HH

H

N

C C

C

OH

H

H

HH

H

8) CH2=CH2 (g) + H2O (g) ⎯⎯⎯→ CH3-CH2-OH (g)

H

H H

H

+ H O

H

C CH

H

H

H

H

OH

9) CH3Cl (g) + H2O (g) ⎯⎯⎯→ CH3 OH (g) + HCl (g)

C ClH

H

H

CH

H

H

C OH

H

H

H

+ H O

H+ H Cl

10) Atomisation of (CH3)2C=O (g)

C C

C

OH

H

H

HH

H



ENTHALPY PROFILE DIAGRAMS AGAIN sections (d) and (e) Once we have the idea that not all the bonds in chemical substances need to be broken, the C≡N bond in question 7 for example, we start to realise that Enthalpy Profile Diagrams can be drawn slightly differently. So we do not have to break all the bonds in a substance to get it to react, in fact breaking all the bonds would not be a good thing to do. Exothermic case The energy given out in a reaction, enthalpy change ∆H, is still the difference between the enthalpy at the start of the reaction minus the enthalpy when the reaction is completed. For example hydrogen gas reacting with oxygen gas.

However all reactions still need energy to start them, in the case of hydrogen gas reacting with oxygen gas this is often a spark or match. This energy is called the activation energy of the reaction.

∗ Activation energy is the minimum energy needed to start and maintain a chemical reaction.

For an exothermic reaction it would be shown as in the diagram below.

Draw an enthalpy profile diagram for an endothermic reaction and label the energy changes.



ENTHALPY CHANGES section (f) SOME DEFINITIONS must learn these The Enthalpy change of a reaction is the amount of heat exchanged with the surroundings at constant pressure. Surroundings could be the room the reaction occurs in, for example with a combustion reaction. However, the “surroundings” are often the water in which the reaction occurs if the substances are in solution, for example an acid reacting with a base.

Enthalpy changes are always in units of kJ mol-1. The Standard Enthalpy Change for a reaction, symbol ∆H0, refers to the amounts of reactants shown in a chemical equation. For comparison all reactions should take place under standard conditions. These conditions are as follows: i) a pressure of 100 kPa = 1 bar or 101 kPa = 1 atmosphere, ii) a temperature of 298K or 25°C, iii) the substances must be in their normal physical states under these conditions, the

standard state, for example gas, liquid or solid (allotropic form). iv) any solutions involved must have a concentration of 1 mol dm-3 . The Standard Enthalpy Change of Combustion of a compound, symbol ∆H0

c, is the enthalpy change that takes place when one mole of the compound in its standard state undergoes complete combustion in excess oxygen under the standard conditions of 298K and 100 kPa. This enthalpy change is always exothermic, so ∆H is ……………….…….. The Standard Enthalpy Change of Formation of a compound, symbol ∆H0

f , is the enthalpy change that takes place when one mole of the compound is formed from its elements in their standard states under the standard conditions of 298K and 100 kPa. This enthalpy change is usually exothermic but can be endothermic. The Standard Enthalpy Change of Atomisation of an element, symbol ∆H0

atom, is the enthalpy change that takes place when one mole of gaseous atoms is made from the element in its standard state under the standard conditions of 298K and 100 kPa. Can also apply to a substance. This enthalpy change is always endothermic so ∆H is ………………….…..



ENTHALPY CHANGES section (g) Sometimes we can measure these changes in enthalpy rather easily. For example by measuring the amount of energy, remember energy is in kJ, given out by a measured mass of an element when it is burned in oxygen we can calculate the enthalpy change in kJ mol-1. A student burned 1.00 g of Mg in oxygen and found that 24.77kJ of heat was evolved. Calculate ∆H0

f [MgO(s)] Moles of Mg = 1.00/24.3 = 0.04115

∆H0 f [MgO(s)] = energy/moles

= -24.77/0.04115

= -601.9 kJ mol-1 exothermic as heat evolved

What is ∆H0

c [Mg(s)] ………………………………………………………. You will see calculations of this type shortly. Of course the big question is how do we measure the energy involved and that we will address next. Any ideas what we could actually measure to find the energy change? ………………………………………………………………………………………………………………………



ENTHALPY CHANGES section (g)

Experimental calculation of an enthalpy change from a rise in temperature. When an object is heated its ................................. increases. This rise in temperature depends on the ................ ......................... of the object. Now we have the relationship that:

∆T = ∆Q/C Where ∆T is the ......................................... ∆Q is the ......................................... C is the ......................................... Note: The heat capacity of substance is the quantity of heat needed to raise its temperature by 1 K. Heat capacity per unit mass is the ........................ ................ , of the substance this is given the symbol c, where c = C/m , m being the mass of the substance. Specific heat capacities have the units of J g-1 K-1 or J kg-1 K-1.

So ∆Q = ∆T . C which after substitution for C = mc , gives:

Heat change = ∆Q = m c ∆T in units of Joules This gives a method to calculate the heat absorbed in a substance. In most of the chemical reactions you will meet, the substance being heated or cooled will be water, or at least an aqueous solution, so c = 4.18 J g-1 K-1 and the mass m will be the mass of water used in grams. Remember that 1 cm3 of water weighs 1g. To convert the heat change, ∆Q, into an Enthalpy change, ∆H , we first change Joules into kJ by dividing by 1000. Then we need to work out the number of moles involved of the substance which is not in excess in the reaction from its mass and MR value. This is usually not the water or the aqueous solution used.

Enthalpy change = ∆H = ∆Q / moles in units of kJ mol-1.

Remember to check the power point on these calculations



ENTHALPY CHANGE FOR THE REACTION BETWEEN COPPER (II) IONS AND ZINC

Name: …………………………………………………………….. Using simple apparatus, enthalpy changes can be calculated by reacting known amounts of two substances together and measuring the temperature change produced. The degree of accuracy is dependant on various factors and these will be considered.

HAZARD WARNING: Copper and zinc salts are toxic, clear any spillages immediately and avoid skin and eye contact. Safety glasses must be worn. (1) Weigh an empty weighing boat and then weigh accurately into this weighing boat

approximately 2g of zinc powder. Record this mass to 2 d.p. in the Table below (2) Using a measuring cylinder, place 100 cm3 of 0.20M copper (II) sulphate solution

into the insulated cup and measure the temperature of the solution accurately, to 0.5oC.

(3) Measure the temperature of the solution accurately, to 0.5oC and do this every 30

seconds for 2.5 minutes. Record in the table below. (4) After 3 minutes add the zinc powder to the copper (II) sulphate solution and mix

thoroughly.

(5) Record the temperature every 30 seconds in the table until the temperature has either stabilised or decreased for about 2 minutes.

(6) After emptying the weighing boat, reweigh it. Record the mass to 2 d.p. in the

Table below and calculate the mass of zinc powder used. (7) Plot a graph of temperature against time and then calculate the temperature

change to the nearest 0.5 K from this graph. (8) Write a chemical equation for the reaction between zinc and copper (II) sulphate. (9) From your results and the fact that 4.2 J are required to raise the temperature of 1

cm3 of water by 1oC, calculate the enthalpy change for this reaction.

Mass of weighing boat and zinc powder / g

Mass of weighing boat /g

Mass of zinc powder used /g



Temperature record

Time /mins Temperature/°C Time /mins Temperature/°C 0.0 5.5 0.5 6.0 1.0 6.5 1.5 7.0 2.0 7.5 2.5 8.0 3.0 Add Zn powder 8.5 3.5 9.0 4.0 9.5 4.5 10.0 5.0

Use the above figures to plot a cooling curve graph of temperature against time With your data from graph complete the table below Initial temperature of copper sulphate solution / °C Maximum corrected temperature after adding zinc / °C

Temperature change /K

Chemical equation for the reaction …………………………………………………………………………

Heat change = ∆Q = m c ∆T in units of Joules m = …………………, c = …………………………., ∆T = …………………………………… ∆Q = Converting to kJ ∆Q =…………….kJ Moles of zinc powder = Moles of copper sulphate solution = So ……………………. is not in excess and this value will be used

Enthalpy change = ∆H = ∆Q / moles in units of kJ mol-1.

∆H = …………….. / …………… =



ERRORS AND IMPROVEMENT TO THE DESIGN (1) Give one safety precaution you took in this experiment.

…………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………..

(2) Can the following pieces of equipment used in the experiment be improved upon

and they can give an alternative piece of equipment and say why it has been changed?

Balance for weighing……………………………………………………………………………….………… …………………………………………………………………………………………………………………….... ……………………………………………………………………………………………………………………....

Measuring cylinder ..……………………………………………………………………………….………… …………………………………………………………………………………………………………………….... ……………………………………………………………………………………………………………………....

Thermometer ……….……………………………………………………………………………….………… …………………………………………………………………………………………………………………….... ……………………………………………………………………………………………………………………....

(3) What is the largest error in this type of experiment?

…………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………..

…………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………..



HESS’S LAW AND HESS’S CYCLES section (k) Hess’s Law

∗ In any chemical reaction the total enthalpy change is independent of the route taken, providing the initial and final states are the same.

This Law is often used to calculate enthalpy changes for reactions which either cannot be done directly as the activation energy is too high

Example: C + H2 (g) ⎯→ CH4 (g)

or for reactions which cannot be done at all

Example: C + ½O2 (g) ⎯→ CO (g)

There are two main cycles you will meet this year. 1) Enthalpy changes of combustion giving an enthalpy change of formation

2) Enthalpy changes of formation giving an enthalpy change of reaction

3) We can even use a Hess's cycle for bond breaking and making calculations



HESS’S CYCLES section (k) Part 1) Enthalpy changes of combustion to calculate an enthalpy change of formation We cannot do the following reaction which is a formation of CO from its elements. C (gr) + ½O2 (g) ⎯⎯⎯⎯→ CO (g) As although we can burn carbon we tend to get either complete combustion C (gr) + O2 (g) ⎯⎯→ CO2 (g) incomplete combustion xC (gr) + yO2 (g) ⎯→ CO2 (g) + CO (g) + C (gr) Note: the complete combustion reaction is both an enthalpy change of combustion of

carbon and an enthalpy change of …………………… of ……………………

Hess discovered that if we started with C and made CO2 (a combustion reaction) it involves the same amount of enthalpy as going from C ⎯→ CO ⎯→CO2

That is the combustion of carbon to give carbon dioxide = formation of CO from its elements added to the combustion of CO to give carbon dioxide again. This is written more easily as

∆HOc [C (gr)] = ∆HO

f [CO (gr)] + ∆HOc [CO (g)]

Hess’s Cycles are even easier to show this ∆HO

f [CO (g)] C (gr) + ½O2 (g) ⎯⎯⎯⎯⎯⎯→ CO (g)

½O2 (g) ½O2 (g) ∆HO

c [C (gr)] ∆HOc [CO (g)]

CO2 (g) Just follow the arrows to get the above equation. If from experiments we have two of the quantities we can calculate the third, so knowing ∆HO

c [C (gr)] = -393.5 kJ mol-1 ∆HOc [CO (g)] = -283.0 kJ mol-1

Then ∆HO

f [CO (gr)] = ∆HOc [C (gr)] - ∆HO

c [CO (g)] = -393.5 - (-283.0) = -110.5 kJ mol-1



HESS’S CYCLES section (k) Using Hess’s cycles calculate the following enthalpy changes of formation Marks will be given for the correct drawing of the cycle as well as the calculation. 1. Enthalpy change of formation of CH4 , use the combustion values given on page 17 and ∆HO

c [CH4 (g)] = -890.3 kJ mol-1, ∆HOc [H2 (g)] = -285.8 kJ mol-1

2. Enthalpy change of formation of H2S. You will need enthalpy change for combustion of H2S = -562.1 kJ mol-1

enthalpy change for combustion of S = -296.9 kJ mol-1

3. Enthalpy change of formation of CH3OH You will need enthalpy change combustion of CH3OH = -715.0 kJ mol-1



CALCULATIONS USING STANDARD ENTHALPY CHANGES OF COMBUSTION sections (f) and (k) Answer these questions on file paper 1. State Hess's Law of constant heat summation.

2. Define precisely the terms:

a) standard enthalpy change of formation

b) standard enthalpy change of combustion

3. Calculate the standard enthalpy change of formation of sulphur dioxide if on burning in

excess oxygen under standard conditions 1.00g of sulphur evolves 9.28 kJ of heat. 4. At 1 atmosphere pressure and 298K in the presence of excess oxygen burning 1.00g of

carbon gives out 32.8 kJ of heat. Calculate the standard enthalpy change of combustion of the carbon.

5. Draw a Hess’s cycle and calculate the standard enthalpy change of formation of ETHYNE

(C2H2) given the following standard enthalpy changes of combustion: ethyne (g) = -1300 kJ mol-1; carbon (graphite) = -394 kJ mol-1; hydrogen (g) = -286 kJ mol-1

6. Calculate the standard enthalpy change of formation of propan-1-ol (CH3CH2CH2OH )

given the following standard enthalpy changes of combustion : propan-1-ol (l) = -2010 kJ mol-1; carbon (graphite) = -394 kJ mol-1; hydrogen (g) = -286 kJ mol-1



THE REACTION BETWEEN CITRIC ACID AND SODIUM HYDROGENCARBONATE HAZARD WARNING: Avoid skin and eye contact, clear any spillages immediately. (1) Weigh accurately into a weighing boat approximately 12.5g of sodium hydrogencarbonate. Record the mass to 0.01g in a table of your design. (2) Using a measuring cylinder, place 50 cm3 of 1.0M citric acid solution into the insulated cup and measure the temperature of the solution accurately, to 0.5oC. (3) Add the sodium hydrogencarbonate to the citric acid solution a spatula full at a time, waiting between additions for the effervesence to subside. (4) When the additions have been made mix thoroughly. Record the maximum or minimum temperature and work out the temperature change to the nearest 0.5oC. (5) Rewigh the empty weighing boat to determine the amount of sodium hydrogencarbonate added. (6) The chemical equation for the reaction between sodium hydrogencarbonate and citric acid solution is: C3H4(OH)3(COOH)3 + 3NaHCO3 ⎯→ C3H4(OH)3(COONa)3 + 3CO2 + 3H2O (7) From your results and the fact that 4.2 J are required to raise the temperature of 1 cm3 of water by 1oC, calculate the enthalpy change for this reaction.



HESS’S CYCLES section (k) Part 2) Enthalpy changes of formation to calculate an enthalpy change of reaction Suppose we now consider this reaction NH3 (g) + HCl (g) ⎯⎯⎯⎯→ NH4Cl (s) Find we can use same ideas to calculate the enthalpy change for this reaction which is called ∆HR, and refers to a specific chemical equation. If we use enthalpy changes of formation of ammonia, HCl and ammonium chloride we can construct a Hess’s cycle NH3 (g) + HCl (g) ⎯∆HR ⎯⎯⎯→ NH4Cl (s) ∆Hf [NH3 (g)] ∆Hf [HCl (g)] ∆Hf [NH4Cl (s)] -46.2 -92.3 -315 ½N2 (g) + 2H2 (g) + ½Cl2 (g) Again the total enthalpy changes must be the same so following the arrows gives ∆HR + ∆Hf [NH3 (g)] + ∆Hf [HCl (g)] = ∆Hf [NH4Cl (s)] Rearranging gives ∆HR = ∆Hf [NH4Cl (s)] - { ∆Hf [NH3 (g)] + ∆Hf [HCl (g)] } putting in values, see cycle above ∆HR = -315 - { -46.2 + -92.3 } = - 315 -(-138.5) = -176.5 kJ mol-1

Remember this is for the reaction in the equation so if we made 2 moles of NH4Cl we

would give out ……………………… kJ



CALCULATIONS USING STANDARD ENTHALPY CHANGES OF

FORMATION sections (f) and (k)

Answer these questions on file paper 1. Calculate the standard enthalpy change of the reaction between ethene, H2C=CH2 ,

and hydrogen gas given the following standard enthalpy changes of formation: ethene = +52 kJ mol-1 ethane = -85 kJ mol-1

H2C=CH2 + H2 ⎯⎯→ H3C-CH3 2. Write an equation for the reaction of lead (II) oxide and carbon monoxide gas.

Calculate the standard enthalpy change for the reaction of lead (II) oxide and carbon monoxide gas given that the standard enthalpy changes of formation of lead (II) oxide, carbon monoxide and carbon dioxide are -219, -111 and -394 kJ mol-1 respectively.

3. Calculate the standard enthalpy change for the reaction of iron (III) oxide and

aluminium powder. The standard enthalpy changes of formation of iron (III) oxide and aluminium oxide are -822 kJ mol-1 and -1669 kJ mol-1. State whether the reaction is exothermic or endothermic.

4. Calculate the standard enthalpy change for the reaction of sulphur dioxide gas with

hydrogen sulphide gas giving solid sulphur and water. The standard enthalpy change of combustion of sulphur is -297 kJ mol-1 and the standard enthalpy changes of formation of hydrogen sulphide and water are -20.2 kJ mol-1 and -286 kJ mol-1 respectively.

5. Given that the standard enthalpy change of combustion of rhombic sulphur is -294.9

kJ mol-1 and the standard enthalpy change of combustion of monoclinic sulphur is -297.2 kJ mol-1. Calculate the standard enthalpy change for the conversion of the allotropic forms of sulphur from monoclinic to rhombic.

6. In the reaction of hydrogen chloride gas (∆Ho

f = -92.3 kJ mol-1) and ethene ( ∆Hof

= +52.3 kJ mol-1), the standard enthalpy change is -65 kJ mol-1. Calculate the standard enthalpy change of formation of one mole of the product, CH3CH2Cl (g).

H2C=CH2 + HCl ⎯⎯→ CH3CH2Cl



HESS’S CYCLES section (k) Part 3) Calculations using bond energies and enthalpy changes of formation ∆Hf [CH4 (g)] -75 kJ mol-1

C (gr) + 2H2 (g) ⎯⎯⎯⎯→ CH4 (g) ∆Hat [C(g)] ∆Hat [1/2H2(g)] 4 x BE(C-H) +715 4 x (+218) ??? C(g) + 4H (g) ∆Hf [CH4 (g)] + (4 x BE(C-H)) = ∆Hat [C(g)] + ( 4 x ∆Hat [1/2H2(g)]) 4 x BE(C-H) = ∆Hat [C(g)] + ( 4 x ∆Hat [1/2H2(g)]) - ∆Hf [CH4 (g)] = 715 + (4 x 218) - (-75) = 1662 BE(C-H) = 1662/4 = 415.5 kJ mol-1

Problem: Given the above data and that the enthalpy change of formation of ethane is ∆Hf [C2H6 (g)] = -85 kJ mol- calculate the bond energy of a carbon carbon bond, BE(C-C)



ENTHALPY CHANGES OF COMBUSTION section (f)

From experiments to find the enthalpy change of combustion of various alcohols the following results have been obtained.

----------------------------------------------------------------------------------------------------- ALCOHOL FORMULA ∆Hoc / kJ mol-1

----------------------------------------------------------------------------------------------------- methanol CH3OH - 726 ----------------------------------------------------------------------------------------------------- ethanol CH3CH2OH -1367 ----------------------------------------------------------------------------------------------------- propan-1-ol CH3CH2CH2OH -2021 ----------------------------------------------------------------------------------------------------- butan-1-ol CH3CH2CH2CH2OH -2676 ----------------------------------------------------------------------------------------------------- pentan-1-ol CH3CH2CH2CH2CH2OH ?????? ----------------------------------------------------------------------------------------------------- hexan-1-ol CH3CH2CH2CH2CH2CH2OH -3984 ----------------------------------------------------------------------------------------------------- heptan-1-ol CH3CH2CH2CH2CH2CH2CH2OH -4638 ----------------------------------------------------------------------------------------------------- octan-1-ol CH3CH2CH2CH2CH2CH2CH2CH2OH -5294 -------------------------------------------------------------------------------------------------------------- 1. Plot a graph of the enthalpy change of combustion against the number of carbon atoms in each alcohol.

Note: axes should be 0 to -6500 kJ mol –1 ( vertical) versus 1 to 10 carbon atoms (horizontal).

2. Estimate the value for ∆Hoc of pentan-1-ol from your graph ……………………….. 3. Predict using the graph the values for alcohols with 9 carbon atoms, nonan-1-ol and 10 carbon atoms, decan-1-ol.

nonan-1-ol ………………………………… decan-1-ol ……………………………. 4. Using the values in the table above determine how the extra CH2 group contributes to

the enthalpy change of combustion………………………………………………… 5. If you use this factor can you calculate and confirm the values obtained from the graph?

nonan-1-ol ………………………………… decan-1-ol …………………………….


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