ThermodynamicsGCSE backgroundThermodynamics involves the study of the energy changes in chemical reactions.Energy cannot be created or destroyed but can be converted from one form to another.This basic fact of science is called the first law of thermodynamics. When most chemical reactions take place, energy changes occur:

In chemistry, heat energy is usually measured at constant temperature and pressure. In such cases, heat energy is more properly called enthalpy (H). The amount of enthalpy that is transformed during a reaction can be shown using a H value (in science, is used to show a change). It is these enthalpy changes which chemists can measure

exothermic reactions have negative H values, endothermic reactions have positive H values.

e.g. CH4(s) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJmol-1

exothermic negative

CaCO3(s) CaO(s) + CO2(g) H = +572 kJmol-1

endothermic positive

Common exothermic reactions include oxidation, for example the combustion of fuels and respiration. In respiration, plants & animals oxidize sugars such as glucose to provide them with the energy they need:

C6H12O6 + 6O2 → 6CO2 + 6H2O H = -2801 kJmol-1

Common endothermic reactions include the thermal decomposition of carbonates and photosynthesis, where plants absorb light energy to build up sugars from carbon dioxide and water (the reverse of respiration!):

6CO2 + 6H2O → C6H12O6 + 6O2 H = +2801 kJmol-1

Chemical energy is stored in bonds. In order to break any chemical bond or intermolecular force, energy must be used. This also means that energy is released when any bond is formed.

In exothermic reactions, chemical energy locked up in the bonds of the reactants is converted into heat energy. The heat energy released makes the surroundings feel hotter.

In endothermic reactions, heat energy is absorbed and converted into chemical energy. Heat energy is taken from the surroundings making them feel cooler.

Bond breaking is always an endothermic process,Bond formation is always an exothermic process

The Kelvin temperature scaleIn everyday life, non-scientists usually use the Celcius scale to measure temperature. However, scientists often prefer to use the Kelvin scale because it is based on a fundamental idea – Zero Kelvin or Absolute Zero (0 k) is the temperature at which all particles would stop moving. This temperature is -273C.To convert a temperature expressed in Celcius to Kelvin, simply add 273.

e.g. 50C is equal to 50 + 273K or 323 K

Standard Enthalpy Changes of Reaction, Hro

When a chemical reaction occurs, the size of the enthalpy change depends on the surroundings of the experiment. If identical experiments were conducted in London and at the top of Mount Everest, the two experiments would give different H values due to the different temperatures and pressures. For this reason, chemists around the world have agreed a set of Standard Conditions so that “fair test” comparisons can be made.

The chosen standards are* A pressure of 1 atmosphere* A temperature of 298 K (25C)* A concentration of 1.0 moldm-3

When measured under standard conditions, an enthalpy change is given using Ho

Similarly, the standard state of a substance gives its physical state and chemical formula when under standard conditions.

Examples: sodium has standard state Na(s)chlorine Cl2(g)carbon C(s, graphite)

It is especially important to include state symbols in equations when studying thermodynamics because a change in state always involves an enthalpy change.

e.g. Na(s) Na(g) H = +109 kJmol-1

(endothermic as energy is needed to overcome the metallic bonds to make gaseous sodium.)

All forms of energy are measured in Joules (J), with larger amounts in kilojoules kJ. The size of the energy change in a chemical reaction obviously depends on the quantity of reagents mixed together. This is why H is measured in kilojoules per mole shown in an equation (kJmol-1).

Basically, it is always best to write a chemical equation next to any H value to show exactly which reaction is involved.

e.g. C(s) + O2(g) CO2(g) H = -393 kJmol-1

BUT2C(s) + 2O2 2CO2(g) H = 2 x -393 = -786 kJmol-1

i.e. if twice the amount of carbon burns then twice the amount of enthalpy is released!

The standard enthalpy change of reaction Hro is the enthalpy change that

accompanies a reaction in the molar quantities expressed in a chemical equation under standard conditions (298 K, 1 Atm).

Once H for a chemical equation is known, the actual amount of enthalpy change for a given quantity of reagents can be calculated using

e.g. calculate the enthalpy released when 32g of methane is burned.CH4(g) + 2O2(g) CO2(g) + 2H2OH = -890 kJmol-1

moles of methane burned = mass = 32 = 2 rmm 16

actual enthalpy released = H x moles = –890 x 2 = -1780 kJ

(Note: the actual enthalpy change Q is measured in kJ, NOT kJmol-1)

Enthalpy level diagrams and thermodynamic stabilityThese diagrams offer an excellent way to illustrate enthalpy changes. By showing the “internal enthalpy” of the reactants and products on a scale, it is easy to see what type of enthalpy change will occur during the reaction.

e.g.1. the exothermic enthalpy level diagram for the combustion of methane

In this example, the reactants have high “internal enthalpy” whilst the products have much less. This means that the reaction must release this extra enthalpy. The enthalpy difference between the reactants and products must therefore be equal to H. Exothermic reactions always have H arrows pointing downwards.e.g.2. the endothermic enthalpy level diagram for the decomposition of calcium

carbonate

actual enthalpy change (Q) = H x moles

CH4(g) + 2O2(g) reactants

Reaction coordinate

Enthalpy

CO2(g) + 2H2O(g) products

H = -890 kJmol-1

CaCO3(g) reactants

Enthalpy

H = +572 kJmol-1

CaO(s) + CO2(g) products

Reaction coordinate

Here, the products have more internal enthalpy than the reactants and so enthalpy must be added in order for the reaction to occur. The enthalpy difference between reactants and products is again H. Endothermic reactions always have H arrows pointing upwards.From these diagrams it is simple to see that the enthalpy change in any chemical reaction is simply the difference between the levels of internal enthalpy in the reactants and products. Mathematically, this can be written:

H = Hproducts - Hreactants

Enthalpy level diagrams are also useful to explain the idea of thermodynamic stability.Compare the enthalpy level diagram below with the diagram showing the gravitational potential energy of two rocks on a hillside:

Obviously, the higher rock is less stable than the rock which has lower gravitational potential energy and is much more likely to move. Similarly, the reactants A and B have a higher level of internal enthalpy than the products. This means that reaction is likely to occur – the reactants are more thermodynamically unstable than the products and will undergo an exothermic reaction. The reverse reaction is unlikely to happen as enthalpy is needed to “push the products up” – they are thermodynamically stable.

Note: just because a reaction is thermodynamically unstable, it does not mean that it will definitely occur as there are other factors to consider. It does make it more likely.

Specific forms of enthalpy change

1. The standard enthalpy change of combustion, Hco

Combustion reactions basically involve burning a substance in oxygen. Combustion reactions are always exothermic and so Hc

o values are always negative. Enthalpy changes of combustion are especially useful when comparing the advantages of different fuels.

Chemical equations can be written which fit the definition of Hco. However, it is common to see

fractions used to balance such equations so that only 1 mole of substance appears.

PotentialEnergy

A + Breactants

Reaction coordinate

Enthalpy

C + Dproducts

H

The standard enthalpy change of combustion is the enthalpy released when 1 mole of a substance is completely burned in excess oxygen under standard conditions (298 K, 1 Atm).

e.g. for ethane (C2H6) the equation representing Hco is

C2H6(g) + 3.5O2(g) 2CO2(g) + 3H2O(l) Hco = -1561 kJmol-1

but NOT

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) because 2 moles of ethane react in the equation.

2. The standard enthalpy change of formation, Hfo

Formation reactions involve making compounds from elements.

By definition, the enthalpy of formation of any element is zero. Again, chemical equations can be written which fit the definition of Hf

o.

e.g.1. for ethene (C2H4),2C(s,graphite) + 2H2(g) C2H4(g) Hf

o = +33 kJmol-1

e.g.2. for water,H2(g) + ½ O2(g) H2O(l) Hf

o = -286 kJmol-1

The definition includes “enthalpy change,” because Hfo values can be positive or negative

depending on the compound. Indeed, the thermodynamic stability of a compound is related to its Hf

o value. Consider the enthalpy level diagrams showing the enthalpy of formation of ethene (C2H4) and ethane (C2H6):

As ethane has a negative (exothermic) enthalpy of formation, this means there must be very little “internal enthalpy” in the compound. Ethane is more thermodynamically stable than its elements. However, as ethene has a positive (endothermic) enthalpy of formation, it contains high “internal enthalpy” and is more thermodynamically unstable than its elements. Ethene is much more likely to react exothermically than ethane.

The standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (298K, 1 Atm).

2C(s) + 2H2(g) reactants

Reaction coordinate

E C2H4(g) products

Hfo = + 33 kJmol-1

2C(s) + 3H2(g) reactants

Reaction coordinate

E

C2H6(g) products

Hfo = - 85 kJmol-1

Measurement of enthalpy changes in the laboratory

1. Measuring an enthalpy change of combustion, Hc

Apparatus

A simple calorimeter

Method1. Weigh an empty beaker, fill with water and then re-weigh using a top-pan balance. Hence calculate the mass of water in the beaker2. Weigh the fuel burner with fuel.3. Set up the apparatus, allow the water to equilibrate and take its initial temperature.4. Light the fuel and allow to burn for about 5 minutes, stirring the water occasionally.5. Record the maximum temperature of the water.6. Re-weigh the fuel burner and calculate the mass of fuel which has been burned.

Treatment of results The following calculation assumes that all of the enthalpy released by the combustion is used to heat up the water. The amount of heat energy required to raise the temperature of water is called its specific heat capacity (c), and is equal to 4.18 Joules per Kelvin per gram. This means that the enthalpy entering the water in the experiment and hence the enthalpy released by the combustion (Q) is given by

The moles of fuel burned in the experiment is then calculated using mass / rmm.

Finally, the enthalpy released per mole of fuel (Hc) is found in Joules per mole:

The final answer is then divided by 1000 to change the joules into kilojoules and a negative sign added to show the exothermicity of the combustion.

thermometer

insulation + lid

known mass of water (m)in beaker

tripod and gauze

fuel under test in burner

Q = m x c x T where m = mass of water in beaker in g, ( in joules) c = 4.18 JK-1g-1

T = max temp – initial temp

Hc = Q moles

Example calculationA calorimeter was used in an experiment to find Hc for methanol. The table shows the readings taken:

Initial mass of methanol + burner 24.14gFinal mass of methanol + burner 23.50gInitial temp of water 24CMaximum water temp 32CMass of water 200g

Use the data to find Hc for methanol.Temp change T = 32 – 24 = 8KEnthalpy released Q = mcT

= 200 x 4.18 x 8 = 6688 J

mass methanol burned = 24.14 - 23.50 = 0.64gmoles methanol burned = mass / rmm = 0.64 / 32 = 0.02

So H = Q = 6688 = 334400 Jmol-1 so Hc = -334 kJmol-1

moles 0.02

Experimental errorsThese simple calorimeter experiments give very poor results for 2 main reasons:a) Not all of the enthalpy released by the combustion enters the water – much heat is lost to the surroundings and some used to heat up the actual apparatus.b) Unless there is a plentiful supply of oxygen, some of the fuel will form carbon (soot) or carbon monoxide rather than carbon dioxide. This incomplete combustion means less enthalpy is released.c) Some of the fuel may evaporate from the burner before it is combusted.

These errors result in a low value for T and hence a low value for Hc. The data book value for Hc methanol is –726 kJmol-1. This is more than double the value obtained in the experiment above.

Specialist equipment such as the bomb calorimeter can be used to get much more accurate results.

2. Measuring enthalpy changes in solution Apparatus

Method1. The required volume of one reactant solution is placed in a plastic cup and the apparatus set up as shown.2. The solution is allowed to equilibrate and the temperature recorded.

thermometer

insulation and lid

reaction mixture in plastic cup/beaker

3. The other reagent is added, the lid replaced and the mixture stirred thoroughly.4. The maximum / minimum temperature on the thermometer is noted.

Treatment of resultsThree assumptions are made in this experiment:a) No heat is lost from/gained by the reaction mixture to the surroundingsb) The specific heat capacity of the reaction mixture is 4.18 JK-1g-1 (as for pure water)c) The density of the reaction mixture is 1gcm-3 (as for pure water). This allows solution volumes to be measured instead of masses (e.g. 50 cm3 = 50 g).

First, the enthalpy released/gained by the reaction is found using

Q = m x c x T where T is the change in temperature,m is the total mass of the reaction mixture (often found by adding two volumes together)c is 4.18 JK-1g-1

Then the moles of the limiting reagent in the reaction is found using mass/rmm or vol x conc / 1000 etc. Again, H is found by

H = Q moles

Finally, this is divided by 1000 to get units of kJmol-1 and the relevant sign added.

Example calculations1. A large excess of iron filings(5g) were added to 50cm3 of copper(II) sulphate solution (conc. 1.0 mol dm-3). The temperature increased from 24C to 28C. Calculate H for the reaction.

CuSO4 + Fe FeSO4 + Cu

Q = mcT = 50 x 4.18 x 4 = 836J (m = 50 because 50cm3 = 50g)

moles CuSO4 = 50 x 1.0 / 1000 = 0.05

H = Q = 836 = 16720 Jmol-1 so H = -16.7 kJmol-1

moles 0.05

[Notice that the mass of the iron filings is ignored as their heat capacity is not known and they absorb little heat in comparison to the 50cm3 of solution.]

2. A 25cm3 sample of sodium hydroxide (conc. 1.1 moldm-3) was treated with 25cm3 of hydrochloric acid (conc. 1.0 moldm-3). The temperature increased from 25C to 31C. Calculate the enthalpy change of neutralization, HN.

HCl + NaOH NaCl + H2O

Q = mcT = 50 x 4.18 x 6 = 1254 J (total mass m = 25 + 25 = 50)

moles of H2O formed = moles of limiting reagent (= HCl as its conc. is lower)= 25 x 1.0 / 1000 = 0.025

H = Q = 1254 = 50160 Jmol-1 so HN = -50.2 kJmol-1

moles H2O 0.025Hess’s law and the theoretical calculation of enthalpy changes

For some reactions it is impossible to measure H changes in the laboratory. For example, the enthalpy change of formation for methane:

C(s, graphite) + 2H2(g) CH4(g) H = ?

When 1 mole of carbon is mixed with 2 moles of hydrogen, the reaction produces a whole range of compounds (C2H6, C3H8 etc) and not just 1 mole of methane. In such cases, Hess’ law can be applied to find H.

Consider the reaction between ethyne (C2H2) and hydrogen gas to form ethane:

C2H2 + 2H2 C2H6

The reaction can be considered to take place all at once (as shown) or by a different 2-step route:

step1: C2H2 + H2 C2H4 then,step2: C2H4 + H2 C2H6

These two routes can be shown together in a Hess Cycle.

Energy cannot be created or destroyed. Therefore, if the starting materials and products both have certain fixed amounts of internal enthalpy then H must be equal for both routes.This means mathematically, Hroute A = Hroute B

orHoverall = Hstep1 + Hstep

Hess’ law can thus be similarly applied in many ways in order to find H values theoretically. However, it is vital to understand the definitions of Hc and Hf and to be able to write their accompanying chemical equations.

Using enthalpy of combustion values with Hess law If the Hc values for all of the reactants and products of a chemical reaction are known, a Hess cycle can be drawn and H for the reaction calculated. The Hess cycle will have two arrows pointing downwards (usually to carbon dioxide and water) to fit the definitions of Hc. The arrows can also carry the balanced amount of oxygen needed for the combustions.

C2H2 + 2H2 C2H4

step1 step2

C2H4 + H2

Route A

Route B

Hess’ Law: The enthalpy change for a chemical reaction is independent of the route taken as long as the initial and final states are the same

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

Hc of C2H4 +3.5O2 +3.5O2 Hc of C2H6

+ Hc of H2

2CO2(g) + 3H2O(l)

Note: 2 moles of CO2 and 3 moles of H2O are needed to balance the numbers of carbon and hydrogen atoms in the reactants and products

H = ?

Hess law states that Hroute A = Hroute B but in this case route B needs special consideration. To get from reactants to products, route B involves the combustion of C2H4 and H2 followed by the “reverse combustion” of C2H6 (i.e. the arrow is pointing the wrong way !). Mathematically, this means

H = [Hc of C2H4 + Hc of H2] – [Hc of C2H6]

Example questions1. Using the Hc data, draw a Hess cycle and calculate H for the reaction:

2C(s) + 3H2(g) C2H6(g)

Substance C(s) H2(g) C2H6(g)Hc / kJmol-1 -394 -286 - 1561

First draw the Hess cycle under the equation with arrows pointing downwards to balanced numbers of CO2 and H2O. Label the arrows H1 and H2:

Then calculate the values of H1 and H2:

H1 = (2 x -394) + (3 x -286) = -1646 (2 moles of C and 3 moles of H2 are burned)H2 = -1561

Finally apply Hess’ law and substitute in the values using brackets to avoid sign errors.

H = H1 - H2 = (-1646) – (-1561) = -85 kJmol-1

2. Using the data, calculate Hf for ethanol, C2H5OH.

Substance C(s) H2(g) C2H5OH(l)Hc / kJmol-1 -394 -286 -1400

In this question, first the equation representing Hf for ethanol must be written and then the cycle written beneath:

H1 = (2 x -394) + (3 x -286) = -1646(the O2 can be ignored as it doesn’t combust!)

H2 = -1400

By Hess’ law,

H = H1 - H2 = (-1646) – (-1400) = -246 kJmol-1

H1 +3O2 +3O2 H2

2CO2 + 3H2O

2C(s) + 3H2(g) + ½ O2(g) C2H5OH(l) Hf

(the negative sign accounts for the “reverse” combustion.)

H1 H2

2C(s) + 3H2(g) C2H6(g)

2CO2(g) + 3H2O(l)

H

Using enthalpy of formation values with Hess lawUnknown H values can be calculated from the Hf values of the reactants and products. A Hess cycle is drawn under the equation with two arrows pointing upwards from the required amount of elements in their standard states. This arrangement follows from the definitions of Hf.The calculations then follow the same method as above.

Example question Use the Hf values to find H for the reaction:

Hf C2H5OH(s) = -278 kJmol-1

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) Hf CO2(g) = -394 kJmol-1

Hf H2O(g) = -286 kJmol-1

The Hess cycle is:

H1 = Hf C2H5OH(s) = -278 kJmol-1 [Hf for O2 = 0 by definition]

H2 = (2 x Hf CO2) + (3 x Hf H2O) = (2 x –394) + (3 x –286) = -1646 kJmol-1

By Hess’ law, H = -H1 + H2 [the H1 arrow points the wrong way]

= -(-278) + (-1646) = -1368 kJmol-1

Calculating H using average bond enthalpies.In order to break any chemical bond, energy must be added. This automatically means that energy must be released when any bond is formed:

The stronger the bond, the more energy needed to break it. A good measure of the strength of a covalent bond is therefore given by its average bond enthalpy:

2C(s) + 3H2(g) + 3.5O2(g)

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

H1 H2

H

Bond formation is always exothermicBond breaking is always endothermic

The average bond enthalpy is the average energy needed to break 1 mole of bonds in the gas phase.

The average appears in the definition because identical covalent bonds tend to have slightly different strengths depending on the rest of the molecule in which they are found.

e.g. A (C-H) bond in CH4 is slightly different strength to a (C-H) bond in CHCl3.

Average bond enthalpies are always positive by definition (“bond breaking”) and the larger the value, the stronger the bond. The table shows some important examples:

Bond Average bond enthalpy / kJmol-1

Bond Average bond enthalpy / kJmol-1

C-C +346 C-Cl +338C=C +612 C-Br +276C-H +412 H-Cl +432C-O +360 H-Br +366C=O +743 Cl-Cl +242O-H +463 O=O +496

As most chemical reactions basically involves the breaking of “old” bonds followed by the making of “new” ones, average bond enthalpies can be used to calculate H values.As breaking is endothermic and making exothermic,

However, as the values used are only averages, they may not be especially accurate for the compounds in the reaction. The calculations using Hess’ law and Hc or Hf give much more accurate results because they use the actual values for the compounds concerned.

Example questionUse the average bond enthalpy data in the table to calculate H for the reaction:

CH4 + 2O2 CO2 + 2H2O

The bonds broken in the reactants include four (C-H) bonds and two (O=O) bonds

so bonds broken = (4 x 412) + (2 x 496) = 2640 kJmol-1

The bonds made in the products include two (C=O) bonds and four (O-H) bonds

so bonds made = (2 x 743) + (4 x 463) = 3338 kJmol-1

H = bonds broken – bonds made = 2640 – 3338 = -698 kJmol-1

H = bonds broken – bonds made