Doc Brown's Chemistry KS4 science GCSE/IGCSE/GCE-AS Revision Notes

Exothermic and Endothermic Energy Changes

Energetics Introduction -  energy transfers in physical state changes and chemical reactions

Why are there energy changes when a chemical reaction takes place? Do physical state changes involve energy changes? Exothermic

energy changes and endothermic energy changes in chemical reactions are described, and exothermic reactions and endothermic

reactions are discussed in terms of bond energies - including calculations of energy transfers for GCSE/IGCSE and basic stuff for GCE

Advanced Level AS students. Methods of obtained vales for energy changes in chemical reactions are described and how to do the

calculations based on calorimeter experiment results. Also discussed are activation energies, reaction profiles, catalysts.

Sub-index: 1. Heat changes in chemical/physical changes * 2. Reversible reactions and energy changes * 3. Activation energy and

reaction profiles * 4. Catalysts and activation energy * 5. Bond energy/enthalpy calculations * 6. Calorimeter methods of determining

energy changes * 7. Energy calculations from calorimeter results

Alphabetical keywords-phrases: Activation energy * Breaking a chemical bond * Bond energy explained * Bond energy calculations

(theoretical) *calculations (practical) * Catalyst action * Chemical bond * Endothermic reaction * Energy level diagrams * Exothermic

reaction * Experimental methods for determining energy changes * Reaction profiles * Making a chemical bond

  Advanced Level Energetics-Thermochemistry - Enthalpies of Reaction, Formation & Combustion

  Foundation tier-easier multiple choice QUIZ on exothermic/endothermic reactions etc.

  Higher tier-harder multiple choice QUIZ on exothermic/endothermic reactions etc.

1. Heat changes

1a. Heat Changes in Chemical Reactions

When chemical reactions occur, as well as the formation of the products - the chemical change, there is also a heat energy

change which can often be detected as a temperature change.

This means the products have a different energy content than the original reactants (see the reaction profile diagrams below).

If the products contain less energy than the reactants, heat is released or  given out to the surroundings and the change is

called exothermic. The temperature of the system will be observed to rise in an exothermic change.

o Examples:

the burning or combustion of hydrocarbon fuels (see Oil Products) e.g. petrol or candle wax.

the burning of magnesium, reaction of magnesium with acids, or the reaction of sodium with water (see Metal Reactivity

Series)

the neutralisation of acids and alkalis (see Acids, Bases and salts)

using hydrogen as a fuel in fuel cells (see Electrochemistry)

If the products contain more energy than the reactants, heat is taken in or absorbed from the surroundings and the change is

called endothermic. If the change can take place spontaneously, the temperature of the reacting system will fall but, as is more likely, the

reactants must be heated to speed up the reaction and provide the absorbed heat.

o Examples:

the thermal decomposition of limestone (see Industrial Chemistry)

the cracking of oil fractions (see Oil products)

There are brief descriptions of many examples of exothermic and endothermic reactions on the "Types of Reaction" page.

The difference between the energy levels of the reactants and products gives the overall energy change for the reaction (the

activation energies are NOT shown on the diagrams below, but see section 3.).

At a more advanced level the heat change is called the enthalpy change is denoted by delta H, ΔH.

o ΔH is negative (-ve) for exothermic reactions i.e. heat energy is given out and lost from the system to the surroundings

which warm up.

o ΔH is positive (+ve) for endothermic reactions i.e. heat energy is gained by the system and taken in from the

surroundings which cool down OR, as is more likely, the system is heated to provide the energy needed to effect the

change.

o See later on for the bond energy arguments .

 

1b. Heat changes in physical changes of state

Changes of physical state i.e. gas <==> liquid <==> solid are also accompanied by energy changes.

To melt a solid, or boil/evaporate a liquid, heat energy must be absorbed or taken in from the surroundings, so these are

endothermic energy changes (ΔH +ve). The system is heated to effect these changes.

To condense a gas, or freeze a solid, heat energy must be removed or given out to the surroundings, so these are exothermic

energy changes (ΔH -ve). The system is cooled to effect these changes.

PLEASE NOTE that much of section 1b. is for advanced level students NOT GCSE students.

A comparison of energy needed to melt or boil different types of substance

The heat energy change (ΔH, enthalpy change) involved in a state change can be expressed in kJ/mol of substance for a fair comparison.

o In the table below

o ΔHmelt is the energy needed to melt 1 mole of the substance (formula mass in g) and is known as the enthalpy of fusion.

o ΔHvap is the energy needed to vaporise by evaporation or boiling 1 mole of the substance (formula mass in g) and is known as

the enthalpy of vaporization.

The energy required to boil or evaporate a substance is usually much more than that required to melt the solid.

o This is because in a liquid the particles are still quite close together with attractive forces holding together the liquid, but in a gas

the particles of the structure must be completely separated with virtually no attraction between them.

The stronger the forces between the individual molecules, atoms or ions, the more energy is needed to melt or boil the substance.

o As this is shown by the varying energy requirements to melt or boil a substance.

For simple covalent molecules, the energy absorbed by the material is relatively small to melt or vaporise the substance and the bigger

the molecule the greater the inter-molecular forces.

o These forces are weak compared to those that hold the atoms together in the molecule itself BUT these forces do not hold the

molecules together in a liquid or solid.

For strongly bonded 3D networks e.g. (i) an ionically bonded lattice of ions, (ii) a covalently bonded lattice of atoms or (iii) a metal lattice of

ions and free outer electrons, the structures are much stronger in a continuous way throughout the structure and consequently much

greater energies are required to melt or vaporise the material.

Substance formula Type of bonding, structure and

attractive forces operating

Melting

point K

Enthalpy of Boiling point

K (Kelvin)

Enthalpy of

(Kelvin) = oC

+ 273fusion ΔHmelt = oC + 273 vaporisationΔHvap

methane CH4

small covalent molecule - very weak

intermolecular forces91K/-182oC 0.94kJ/mol 112K/-161oC 8.2kJ/mol

ethanol 

('alcohol')C2H5OH

larger covalent molecule than methane,

greater, but still weak intermolecular

forces

156K/-117oC 4.6kJ/mol 352K/79oC 43.5kJ/mol

sodium

chlorideNa+Cl-

ionic lattice, very strong 3D ionic

bonding due toattraction between (+)

and (-) ions

1074K/801oC 29kJ/mol 1740K/1467oC 171kJ/mol

iron Fe

strong 3D bonding by attraction of

metal ions (+) with free outer electrons

(-)

1808K/1535oC 15.4kJ/mol 3023K/2750oC 351kJ/mol

silicon

dioxide

(silica)

SiO2

giant covalent structure, strong

continuous 3D bond network1883K/1610oC 46.4kJ/mol 2503K/2230oC 439kJ/mol

See Gases, Liquids and Solids notes and Structure and bonding notes for more details on structure and physical properties.

2. Reversible Reactions and energy changes

If the direction of a reversible reaction is changed, the energy change is also reversed.

For example: the thermal decomposition of hydrated copper(II) sulphate.

On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of anhydrous copper(II) sulphate is

formed.

o This a thermal decomposition and is endothermic as heat is absorbed (taken in)

o The energy is needed to break down the crystal structure and drive off the water.

When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed.

o The reverse reaction is exothermic as heat is given out.

o i.e. on adding water to white anhydrous copper(II) sulphate the mixture heats up as the blue crystals reform.

blue hydrated copper(II) sulphate + heat     white anhydrous copper(II) sulphate + water

CuSO4.5H2O(s)     CuSO4(s) + 5H2O(g)

For more on reversible reactions and chemical equilibrium see GCSE notes OR Advanced Level Notes.

3. Activation Energy and Reaction Profiles

3a. The significance of activation energy

When gases or liquids are heated the particles gain kinetic energy and move faster increasing the chance of collision between

reactant molecules and therefore the increased chance of a fruitful collision (i.e. one resulting in product formation).

However! this is NOT the main reason for the increased reaction speed on increasing the temperature of reactant molecules because

most molecular collisions do not result in chemical change.

Before any change takes place on collision, the colliding molecules must have a minimum kinetic energy called the activation

energy.

Do not confuse activation energy with the overall energy change also shown in the diagrams below, that is the overall energy absorbed-

taken in by the system (endothermic) or given out to the surroundings (exothermic). 

It does not matter whether the reaction is an exothermic or an endothermic energy change (see the pair of reaction profile diagrams

below).

Higher temperature molecules in gases and liquids have a greater average kinetic energy and so a greater proportion of them

will then have the required activation energy to react on collision.

The increased chance of higher energy collisions greatly increases the speed of the reaction because it greatly increases the

chance of a fruitful collision forming the reaction products by bonds being broken in the reactants and new bonds formed in the

reaction products.

The activation energy 'hump' can be related to the process of bond breaking and making (see section 5.).

o Up the hump is endothermic, representing breaking bonds (energy absorbed, needed to pull atoms apart),

o down the other side of the hump is exothermic, representing bond formation (energy released, as atoms become electronically

more stable).

 

For advanced students only: How to derive activation energies from reaction rate data and the Arrhenius

equation is explained in section 5. of the Advanced Level Notes on Kinetics section 5.

3b. Further examples of reaction progress profiles

Reaction profile diagram Relative comments

Very endothermic reaction with a big activation energy.

Very exothermic reaction with a small activation energy.

Moderately endothermic reaction with a moderately high activation energy.

Moderately exothermic reaction with a moderately high activation energy.

A small activation energy reaction with no net energy change. This is theoretically possible if

the total energy absorbed by the reactants in bond breaking equals the energy released by

bonds forming in the products (see section 5.).

Energy level diagram for an exothermic chemical reaction without showing the activation

energy.

It could also be seen as quite exothermic with a highly unlikely zero activation energy, but

reactions between two ions of opposite charge usually has a very low activation energy.

Energy level diagram for an endothermic chemical reaction without showing the activation

energy.

It could also be seen as quite endothermic with a zero activation energy (highly unlikely,

probably impossible?).

4. Catalysts and Activation Energy

Catalysts increase the rate of a reaction by helping break chemical bonds in reactant molecules.

This effectively means the activation energy is reduced (see diagram 'humps' below).

Therefore at the same temperature, more reactant molecules have enough kinetic energy to react compared to the uncatalysed

situation and so the reaction speeds up with the greater chance of a 'fruitful' collision.

o Note that a catalyst does NOT change the energy of the molecules, it reduces the threshold kinetic energy needed for a

molecules to react.

Although a true catalyst does take part in the reaction, it does not get used up and can be reused with more reactants, it may change

chemically on a temporary basis but would be reformed as the reaction products also form.

However a solid catalyst might change physically permanently by becoming more finely divided, especially if the reaction is exothermic.

Also note from the diagram that although the activation energy is reduced, the overall exothermic or endothermic energy change is

the same for both the catalysed or uncatalysed reaction. The catalyst might help break the bonds BUT it cannot change the actual

bond energies.

 

5. Calculation of heat transfer using bond energies (bond enthalpies)

PLEASE NOTE that section 5. is for higher GCSE students and an introduction for advanced level students of how to do bond

enthalpy (bond dissociation energy) calculations.

Atoms in molecules are held together by chemical bonds which are the electrical attractive forces between the atoms.

The bond energy is the energy involved in making or breaking bonds and is usually quoted in kJ per mole of the particular bond involved.

To break a chemical bond requires the molecule to take in energy to pull atoms apart, which is an endothermic change. The atoms of the

bond vibrate more until they spring apart.

To make a chemical bond, the atoms must give out energy to become combined and electronically more stable in the molecule, this is an

exothermic change.

The energy to make or break a chemical bond is called the bond energy and is quoted in kJ/mol of bonds.

o Each bond has a typical value e.g. to break 1 mole of C-H bonds is on average about 413kJ,

o the C=O takes an average 743 kJ/mol in organic compounds and 803 kJ/mol in carbon dioxide, and note the stronger double

bond, so more energy is needed,

o and not surprisingly, a typical double bond needs more energy to break than a typical single bond.

During a chemical reaction, energy must be supplied to break chemical bonds in the molecules, this the endothermic 'upward' slope on

the reaction profile on diagrams above.

When the new molecules are formed, new bonds must be made in the process, this is the exothermic 'downward' slope on the reaction

profile on diagrams above.

If we know all the bond energies (enthalpies) f the molecules involved in a reaction, we can theoretically calculate what the net energy

change is for that reaction and determine whether the reaction is exothermic or endothermic.

o These arguments can then be used to explain why reactions can be exothermic or endothermic.

We do this by calculating the energy taken in to break the bonds in the reactant molecules. We then calculate the energy given out when

the new bonds are formed. The difference between these two gives us the net energy change.

o In a reaction energy must be supplied to break bonds (energy absorbed, taken in, endothermic).

o Energy is released when new bonds are formed (energy given out, releases, exothermic).

o If more energy is needed to break the original existing bonds of the reactant molecules, than is given out when the new

bonds are formed in the product molecules, the reaction is endothermic i.e. less energy is released to the surroundings than

is taken in to break the reactant molecule bonds.

o If less energy is needed to break the original existing bonds of the reactant molecules, than is given out when the new

bonds are formed in the product molecules, the reaction is exothermic i.e. more energy released to surroundings than is

taken in to break bonds of reactants.

o So the overall energy change for a reaction (ΔH) is the overall energy net change from the bond making and bond forming

processes. These ideas are illustrated in the calculations below.

Example 5.1 Hydrogen + Chlorine ==> Hydrogen Chloride

o The symbol equation is: H2(g) + Cl2(g) ==> 2HCl(g)

o but think of it as: H-H + Cl-Cl ==> H-Cl + H-Cl

o (where - represents the chemical bonds to be broken or formed)

o the bond energies in kJ/mol are: H-H 436; Cl-Cl 242; H-Cl 431

o Energy needed to break bonds = 436 + 242 = 678 kJ taken in

o Energy released on bond formation = 431 + 431 = 862 kJ given out

o The net difference between them = 862-678 = 184 kJ given out (92 kJ per mole of HCl formed)

o More energy is given out than taken in, so the reaction is exothermic.

Example 5.2 Hydrogen Bromide ==> Hydrogen + Bromine

o The symbol equation is: 2HBr(g) ==> H2(g) + Br2(g)

o but think of it as: H-Br + H-Br ==> H-H + Br-Br

o (where - represents the chemical bonds to be broken or formed)

o the bond energies in kJ/mol are: H-Br 366; H-H 436; Br-Br 193

o Energy needed to break bonds = 366 + 366 = 732 kJ taken in

o Energy released on bond formation = 436 + 193 = 629 kJ given out

o The net difference between them = 732-629 = 103 kJ taken in (51.5kJ per mole of HBr decomposed)

o More energy is taken in than given out, so the reaction is endothermic

Example 5.3 hydrogen + oxygen ==> water

o 2H2(g) + O2(g) ==> 2H2O(g)

o or 2 H-H  +  O=O ==> 2 H-O-H (where - or = represent the covalent bonds)

o bond energies in kJ/mol: H-H is 436, O=O is 496 and O-H is 463

o bonds broken and energy absorbed (taken in):

o (2 x H-H) + (1 x O=O) = (2 x 436) + (1 x 496) = 1368 kJ

o bonds made and energy released (given out):

o (4 x O-H) = (4 x 463) = 1852 kJ

o overall energy change is:

o 1852 - 1368 = 484 kJ given out (242 kJ per mole hydrogen burned or water formed)

o since more energy is given out than taken in, the reaction is exothermic.

o NOTE: Hydrogen gas can be used as fuel and a long-term possible alternative to fossil fuels (see methane combustion

below in example 5..

It burns with a pale blue flame in air reacting with oxygen to be oxidised to form water.

hydrogen + oxygen ==> water

2H2(g) + O2(g) ==> 2H2O(l) 

It is a non-polluting clean fuel since the only combustion product is water and so its use would not lead to all

environmental problems associated with burning fossil fuels.

It would be ideal if it could be manufactured cheaply by electrolysis of water e.g. using solar cells, otherwise

electrolysis is very expensive due to high cost of electricity.

Hydrogen can be used to power fuel cells.

Example 5.4 nitrogen + hydrogen ==> ammonia

o N2(g) + 3H2(g) ==> 2NH3(g)

o or N N + 3 H-H ==> 2 

o bond energies in kJ/mol: N N is 944, H-H is 436 and N-H is 388

o bonds broken and energy absorbed (taken in):

(1 x N N) + (3 x H-H) = (1 x 944) + (3 x 436) = 2252 kJ

o bonds made and energy released (given out):

2 x (3 x N-H) = 2 x 3 x 388 = 2328 kJ

o overall energy change is: 

2328 - 2252 = 76 kJ given out (38 kJ per mole of ammonia formed)

o since more energy is given out than taken in, the reaction is exothermic.

Example 5.5 methane + oxygen ==> carbon dioxide + water

o CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g)

o or             

o or   using displayed formulae

o bond energies in kJ/mol:

C-H single bond is 412, O=O double bond is 496, C=O double bond is 803 (in carbon dioxide), H-O single bond is

463

o bonds broken and heat energy absorbed from surroundings, endothermic change

(4 x C-H) + 2 x (1 x O=O) = (4 x 412) + 2 x (1 x 496) = 1648 + 992 = 2640 kJ taken in

o bonds formed and heat energy released and given out to surroundings, exothermic change

(2 x C=O) + 2 x (2 x O-H) = (2 x 803) + 2 x (2 x 463) = 1606 + 1852 = 3458 given out

o overall energy change is:

3338 - 2640 = 698 kJ/mol given out per mole methane burned,

since more energy is given out than taken in, the reaction is exothermic.

o At AS level this will be expressed as enthalpy of combustion = ΔHcomb = -818 kJmol-1 

o This shows that heats of combustion can be theoretically calculated.

o NOTE: This is the typical very exothermic combustion chemistry of burning fossil fuels but has many associated

environmental problems. (seeOil Note and Extra Organic Chemistry)

Example 5.6 analysing the bonds in more complex molecules

o  ethyl ethanoate

2 x C-C single covalent bonds

8 x C-H single covalent bonds

2 x C-O single covalent bonds

1 x C=O double covalent bond

o  ethanol

1 x C-C single covalent bond

5 x C-H single covalent bonds

1 x C-O single covalent bond

1 x O-H single covalent bond

o If you wanted to work out the theoretical enthalpy/heat of combustion of propane, you could base your calculation on the displayed

formula equation

o

o Endothermic bond breaking: 8 C-H bonds broken, 2 C-C bonds broken, 5 O=O bonds broken.

o Exothermic bond formation: 6 x C=O bonds made, 8 x O-H bonds made.

6. The experimental determination of energy changes using simple calorimeters

This method 6.1 is can be used for any non-combustion reaction that will happen spontaneously at room

temperature involving liquids or solid reacting with a liquid. The reactants are weighed in if solid and a known

volume of any liquid (usually water or aqueous solution). The mixture could be a salt and water (heat change

on dissolving) or an acid and an alkali solution (heat change of neutralisation). It doesn't matter whether the

change is exothermic (heat released or given out, temperature increases) or endothermic (heat absorbed or

taken in, temperature decreases). See calculations below.

This method 6.2 is specifically for determining the heat energy released (given out) for burning fuels. The

burner is weighed before and after combustion to get the mass of liquid fuel burned. The thermometer records

the temperature rise of the known mass of water (1g = 1cm3).

You can use this system to compare the heat output from burning various fuels. The bigger the temperature

rise, the more heat energy is released. See calculations below for expressing calorific values.

This is a very inaccurate method because of huge losses of heat e.g. radiation from the flame and calorimeter,

conduction through the copper calorimeter, convection from the flame gases passing by the calorimeter etc.

BUT, at least using the same burner and set-up, you can do a reasonable comparison of the heat output of

different fuels.

7. Calculations from the experimental calorimeter results

PLEASE NOTE that section 7. is for higher GCSE students and an introduction for advanced level students of how to do energy

change (enthalpy change) calculations from experimental data.

The calculation method described below applies to both experimental methods 6.1 and 6.2 described above.

You need to know the following:

o the mass of material reacting in the calorimeter (or their concentrations and volume),

o the mass of water in the calorimeter,

o the temperature change (always a rise for method 6.2 combustion),

o the specific heat capacity of water, (shorthand is SHCwater), and this is 4.2J/goC (for advanced 4.2J g-1 K-1),

this means it means the addition of 4.2 J of heat energy to raise the temperature of 1g of water by 1oC.

Example 7.1 typical of method 6.1

o 5g of ammonium nitrate (NH4NO3) was dissolved in 50cm3 of water (50g) and the temperature fell from 22oC to 14oC.

o Temperature change = 22 - 14 = 8oC (endothermic, temperature fall, heat energy absorbed)

o Heat absorbed by the water = mass of water x SHCwater x temperature

= 50 x 4.2 x 8 = 1680 J (for 5g)

heat energy absorbed on dissolving = 1680 / 5 = 336 J/g of NH4NO3 

o this energy change can be also expressed on a molar basis.

Relative atomic masses Ar: N = 14, H = 1, O = 16

Mr(NH4NO3) = 14 + (1 x 4) + 14 + (3 x 16) = 80, so 1 mole = 80g

Heat absorbed by dissolving 1 mole of NH4NO3 = 80 x 336 = 26880 J/mole

At AS level this will be expressed as enthalpy of solution = ΔHsolution = +26.88 kJmol-1 

The data book value is +26 kJmol-1

Example 7.2 typical of method 6.2

o 100 cm3 of water (100g) was measured into the calorimeter.

o The spirit burner contained the fuel ethanol C2H5OH ('alcohol') and weighed 18.62g at the start.

o After burning it weighed 17.14g and the temperature of the water rose from 18 to 89oC.

o The temperature rise = 89 - 18 = 71oC (exothermic, heat energy given out).

o Mass of fuel burned = 18.62-17.14 = 1.48g.

o Heat absorbed by the water = mass of water x SHCwater x temperature

= 100 x 4.2 x 71 = 29820 J (for 1.48g)

heat energy released per g = energy supplied in J / mass of fuel burned in g

heat energy released on burning = 29820 / 1.48 = 20149 J/g of C2H5OH

o this energy change can be also expressed on a molar basis.

Relative atomic masses Ar: C = 12, H = 1, O = 16

Mr(C2H5OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g

Heat released (given out) by 1 mole of C2H5OH = 46 x 20149 = 926854 J/mole or 927 kJ/mol (3 sf)

At AS level this will be expressed as the ...

Enthalpy of combustion of ethanol = ΔHcombustion (ethanol) = -927 kJmol-1

This means 926.9 kJ of heat energy is released on burning 46g of ethanol ('alcohol').

The data book value for the heat of combustion of ethanol is -1367 kJmol-1, showing lots of heat loss in the experiment!

It is possible to get more accurate values by calibrating the calorimeter with a substance whose energy release on

combustion is known