8 Energetics I. Enthalpy* change Enthalpy (H) is the heat content of a system.

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1 * Inner warmth 8 Energetics I Enthalpy* change Enthalpy (H) is the heat content of a system. In a reaction this may increase or decrease and produce an enthalpy change (ΔH). ΔH = H2 - H1 where H1 = enthalpy of reactants and H2 = enthalpy of products

2 The enthalpy change is affected by pressure and temperature so standard conditions are used for measurements. These conditions are 1 atmosphere pressure (1atm = 100kPa) and a temperature of 25ºC or 298K. Eg. 2H2(g) + O2(g) 2H2O(l) standard enthalpy of reaction ΔHº 298 = kjmol -1 This enthalpy change is expressed per mole of one of the reactants or products. In the above case, the figure given refers to ΔHº per mole of oxygen.

3 Exothermic and endothermic reactions Definition: ΔHº R is the standard enthalpy change for a given reaction (under standard conditions of 298K and 1 atm). An exothermic reaction is one in which heat is given out. It results in a temperature increase of the surroundings. ΔH is given a negative sign to indicate an exothermic reaction, e.g. 2H2(g) + O2(g) 2H2O(l) ΔHº 298 = kjmol -1 An endothermic reaction is one in which heat is taken in. It results in a temperature decrease. ΔH is given a positive sign to indicate an endothermic reaction, e.g. 2Ag2O(s) 4Ag(s) + O2(g) ; ΔHº 298 = +61 kjmol -1

4 Enthalpies of formation, combustion and neutralisation Definition: The standard enthalpy change of formation (ΔHºf) is the enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states (under standard conditions of 298K and 1 atm). Eg. C(s) + 2S(s) CS2(l) ΔHºf = + 88 kjmol -1 The positive sign means that for each mole of CS2 formed, 88kJ of energy are absorbed. NB. The enthalpy of formation of an element is 0 kjmol -1

5 Definition: The standard enthalpy change of combustion (ΔHºc) is the enthalpy change when 1 mole of a substance is burnt completely in oxygen (under standard conditions of 298K and at 1 atm.; reactants and products in their standard states). C3H6 (g) + 41/2 O2(g) 3CO2 (g) + 3H2O (l) ΔHºc = kj per mole The negative sign means that each mole of propene burnt releases kj of energy. NB(1). Although the conditions during these reactions will change, the standard enthalpy change refers to the reactants and products in their standard states, which sometimes must be taken into account. Will the enthalpy change in the following reaction be more or less exothermic than the above? C3H6 (g) + 41/2 O2(g) ΔH 3CO2 (g) + 3H2O (g)

6 NB(2). The standard enthalpy of combustion will depend on the form of the reacting element, eg. carbon and sulphur. E.g. S (rhombic, solid) + O2(g) SO 2 (g) ΔHº= kj mol -1 S (monoclinic, solid) + O2(g) SO 2 (g) ΔHº = kJ mol -1 C (graphite, solid) + O 2 (g) CO 2 (g) ΔHº = -393 kj mol -1 Using diamond however would result in a different value for the reaction. Diamond is 2 kj per mole less thermodynamically stable than graphite. ΔHº f of CO2 refers to the graphite reaction. *As well as being the enthalpy of combustion of carbon in its standard state (graphite), this is also the enthalpy of formation of carbon dioxide (see definition earlier).

7 Definition: The standard enthalpy of neutralisation (ΔHºneut) is the enthalpy change when 1 mol of water is formed by the neutralisation of hydrogen ions by hydroxide ions (under standard conditions). H + (aq) + OH - (aq) H2O(l) ΔHºneut = -57 kjmol -1 e.g. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Removing spectator ions H + (aq) + OH - (aq) H 2 O(l) ΔH = -57kJ mol -1 The neutralisation of any strong acid by any strong alkali always produces the same final equation and the same value for the change in enthalpy due to the fact that ionisation is almost complete.

8 Revision: Specific heat capacity (Physics GCSE) Energy transferred (Q) = mass (m) x Specific head capacity (c) X temperature change ( T) In Chemistry, the units are usually: energy transfer, ΔH in J mass, m in g specific heat capacity, c in J g -1 ºC -1 temperature change, ΔT in ºC (or K) We also may use moles instead of mass, so you must be comfortable with the conversion.

9 Example: Finding the enthalpy of combustion of ethanol What measurements would you need to calculate ΔHºc? Mass of water: 150g Initial temp: 20 ºC Final temp: 40 ºC Initial mass of burner: g Final mass of burner: g c for water: 4.2 J g -1 ºC -1 EXPT. Read & do Qs Sources of error: Heat loss to the surroundings from the calorimeter. Heat energy absorbed by the calorimeter. Incomplete combustion (CO and C formed).

10 Energy changes for reactions in solution can be determined in a similar way. Eg. Find the ΔHneut for the following reaction: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) EXPT.

11 Example: 50 cm 3 of 1M HCl(aq) are neutralised with 50cm 3 of 1.1M NaOH(aq). Why the XS? Stirring is important. Why? Results. Initial temperature: 18 ºC Final temperature: 24 ºC Assume density of solution to be 1gcm -3 ; c = 4.2 J g -1 ºC -1 Calculate the ΔHneut A good estimation of the true temperature change can be obtained by extrapolating the line formed by the temperature measurements after mixing and finding where it crosses a vertical line drawn at time of mixing. Read 234& do Qs

12 Enthalpy level diagrams and reaction profiles

13 A reaction profile shows how the enthalpy of a system changes during the reaction. It shows that heat needs to be put into a system before molecules can react or break up. For example hydrogen and iodine react in a reversible reaction to form hydrogen iodide. H2(g) + I2(g) 2HI(g) The reaction profile is shown below.

14 Reaction profile Note the difference between reaction profiles and enthalpy dia Enthalpy Diagram (or Hess Diagram) Read 234& do Qs

15 Hess's law Hess s Law states that the total enthalpy change for a chemical reaction is the same regardless of the route taken for the reaction (given the same conditions for reactants and products). It is also a consequence of a more general physical law- the Law of Conservation of Energy which states that energy can not be created nor destroyed. Route A This is called a Hess s Law Cyc Reactants ΔH1 Products ΔH2 Intermediate products Route B ΔH3 Enthalpy change route A (ΔH1) = Enthalpy change route B (ΔH2 + ΔH3) ΔH1 = ΔH2 + ΔH3

16 Hess s Law can be very useful for calculating the enthalpy change in reactions which cannot be determined by experiment. Eg. Calculate ΔHreaction for: Cgraphite(s) Cdiamond(s) Cgraphite(s) + O 2 (g) CO 2 (g) ΔH = -393 kj mol -1 Cdiamond(s) + O 2 (g) CO 2 (g) ΔH = -395 kj mol -1 Draw a reaction cycle and a Hess diagram for these reactions. Construct an equation, and calculate ΔHreaction

17 Alternative method: Switch the direction of the 2nd reaction (the diamond reaction) and add the two reactions together. That is, Cgraphite(s) + O 2 (g) CO 2 (g) ΔH1 = -393 kjmol -1 CO 2 (g) Cdiamond(s) + O 2 (g) ΔH2 = +395 kjmol -1 Cgraphite(s) Cdiamond(s) ΔHreact = ΔH1 + ΔH2 = +2 kj kjmol -1 CORE PRACTICAL 8: To determine the enthalpy change of a reaction using Hess s Law

18 There are 2 other ways to find the ΔH for a given reaction: Using ΔHº f Using bond enthalpies Remember: The standard enthalpies of formation of the elements are taken as zero. The number of moles of the species gives the multiplication factor for calculations. Always respect the sign of enthalpy changes, and be careful with the maths involved. Read & do Qs

19 1. Using enthalpy of formation Eqn?... Number of moles of each product Number of moles of each reactant

20 For example, find ΔH reaction for: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) ΔH = [ 3(CO 2 (g))+ 4(H 2 O(g)) ] - [ (C 3 H 8 (g))+ 5(O 2 (g))] O 2 (g) is the stable form of oxygen so ΔHº f (O 2 (g)) = 0. Thus we have ΔH = [3 ( kj/mole) + 4 ( kj/mole ) ] - [ ( kj/mole) ] = kj SS example, PQs 2.2 and 2.3

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22 2. Using bond enthalpies (Sum of bond enthalpies of reactants) (Sum of bond enthalpies of products) Eqn?... ΔHreaction = ΣΕ(reactants) Σ Ε(products)

23 Bond Enthalpy and mean bond enthalpy Definition: The standard molar enthalpy change of bond dissociation (ΔH d ) is the energy change when 1 mole of bonds is broken, the molecules and resulting fragments being in the gaseous state (standard conditions). Remember: Breaking bonds requires energy (Endothermic) while making bonds releases energy (Exothermic). Strictly speaking, a bond enthalpy refers to a specific bond in a specific molecule. Even if a molecule has 4 of the same bond (eg the C-H bonds in methane), different dissociation energies can occur: CH 4 (g) CH 3 (g) + H(g) ΔH d = +427 kj mol -1 CH 3 (g) CH 2 (g) + H(g) ΔH d = +371 kj mol -1

24 It is much more useful to know the average amount of energy needed to break a particular bond, and in the case of methane, the following reaction is more interesting in calculations. The enthalpy change for this reaction is kj mol -1, so the average bond enthalpy is / 4 = +412 kj mol -1. This is the figure given in data books: Bond Type Average bond enthalpy /kj mol -1 C-H +413 C-C +347 O=O +498 C=O +805 H-O +464

25 NB: Short bonds tend to be strong. Double bonds are stronger overall, and shorter.

26 The value for the enthalpy of combustion of propane from the data book is kj mol -1. Average bond enthalpies are used to calculate enthalpy changes in reactions, but the answer will be slightly different compared to a result obtained by other methods. Eg. The complete combustion of propane can be represented by the following equation: Or: Which bonds are broken? 8x C-H 2x C-C 5x O=O And then how many bonds have been formed! 6x C=O 8x H-O

27 Using the data from the previous table: ΔH r = [(8x413)+(2x347)+(5x498)] - [(6x805)+(8x464)] = kj mol -1 This answer is different to the Data Book Value. Why? This apparent error is due to the fact that: (a) the water has not cooled to its standard state (b) we use Average Bond Enthalpies (less significant here) Draw an enthalpy level diagram to illustrate this. More practice. What is the enthalpy of formation of water given: E (O-H) = +462 kjmol -1, E (H-H) = +436 kjmol -1, E (O=O) = 498 kjmol -1. MIND BENDER: Try to construct a hess diagram to show how enthalpies of combustion can be used to calculate ΔHreaction Read & do Qs