Energetics HL and SL

An exothermic reaction releases heat energy. An endothermic reaction takes in heat energy. During a chemical reaction bonds in the reactants are broken and new bonds are formed in the products. Energy has to be put in to break bonds and energy is released on forming new bonds. If the energy needed to break the bonds is greater than the energy released on forming new bonds then the reaction is ENDOTHERMIC. If the energy released on forming new bonds is greater than the energy needed to break bonds then the reaction is EXOTHERMIC. 5.1 Exothermic and Endothermic Reactions

Enthalpy Change,  H The enthalpy change occurring during a chemical reaction is the heat energy change at constant pressure. When a system releases heat energy to the surroundings, enthalpy is lost so  H is negative for exothermic reactions. When a system gains heat energy from the surroundings, the enthalpy increases so  H is positive for endothermic reactions.  H is measured in kJmol -1.

The heat energy changes during reactions can be shown on a reaction profile: reactants products HH enthalpy, H For an exothermic reaction

reactants products HH enthalpy, H For an endothermic reaction Type of reaction Heat energy change Sign of  H Relative enthalpies Bond energyTemperature change Exothermic Endothermic

The value for  H depends on a number of factors e.g. amount of reactants, states of the reactants, temperature. A standard enthalpy change occurs under standard conditions of 101 kPa and 298 K. The symbol  H Ϧ is used for a standard enthalpy change. Standard Enthalpy of Formation,  H Ϧ f (HL only) The enthalpy change when 1 mole of a compound is produced from its elements under standard conditions, all reactants and products being in their standard states. e.g. Na (s) + ½Cl 2(g)  NaCl (s)  H Ϧ f = kJmol -1 By definition  H Ϧ f for all elements in their standard states = 0 kJmol -1

Standard Enthalpy of Combustion,  H Ϧ c (HL only) The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, all reactants and products being in their standard states. e.g. CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l)  H Ϧ c = kJmol -1

5.2 Calculation of enthalpy changes The heat energy, q, required to change the temperature of a substance can be calculated using: q = mc  T Where m is the mass of the substance in kg c is the specific heat capacity of the substance in kJK -1 kg -1 (usually water is heated so value is 4.18 kJK -1 kg -1 )  T is the change in temperature in K

Sample Calculation 1 In an experiment 1.00 g of methanol was burned in air and the flame was used to heat 100 g of water, which rose in temperature from 293 K to 313 K. Write an equation for the combustion reaction and calculate  H. Heat energy absorbed by waterq = mc  T = 0.1 x 4.18 x 20 = 17.6 kJ Moles of methanol burned = mass / Mr = 1.00 / 32 = mol Enthalpy change  H = -(heat energy/moles) = / = 570 kJmol -1

Sample Calculation 2 In an insulated container 50 cm 3 of 2.0 M HCl at 293 K were added to 50 cm 3 of 2.0 M NaOH also at 293 K. After the reaction the temperature has risen to 307 K. Write an equation for the neutralisation reaction and calculate  H. Heat energy absorbed by waterq = mc  T = 0.1 x 4.18 x 14 = 5.85 kJ Moles of acid = vol (in dm 3 ) x conc = 50 x 2.0 / 1000 = 0.10 mol Enthalpy change  H = -(heat energy/moles) = / 0.10 = 59 kJmol -1

5.3 Hess’s Law (used when  H is hard to measure) The first law of thermodynamics: Energy can neither be created nor destroyed but it can be converted from one form to another. Hess’s Law: The enthalpy change of a reaction depends only on the initial and final states of the reaction and is independent of the route by which the reaction occurs.

To illustrate: Solid calcium carbonate can be converted to an aqueous solution of calcium chloride by: i)Adding dilute hydrochloric acid, or ii)By heating to make solid calcium oxide and then adding dilute hydrochloric acid. According to Hess’s Law:  Hi) =  Hii)

Using Hess’s Law: i)Calculating  H f for organic compounds This involves using standard enthalpy of combustion data (given in Q) e.g. calculate the enthalpy of formation of methane. First step is to write an equation for the formation of methane. C (s) + 2H 2(g)  CH 4(g) We then need to draw what is often referred to as an energy cycle. This provides us with an alternative route

C (s) + 2H 2(g)  CH 4(g) CO 2(g) + 2H 2 O (l) We know the enthalpy change from the reactants to CO 2(g) + 2H 2 O (l), labelled  H 1, this is the enthalpy of combustion of carbon and 2 x the enthalpy of combustion of hydrogen. What else could it be given as? We also know the enthalpy change from the product to CO 2(g) + 2H 2 O (l), labelled  H 2, this is the enthalpy change of combustion of methane. H1H1 H2H2

C (s) + 2H 2(g)  CH 4(g) CO 2(g) + 2H 2 O (l) H1H1 H2H2 Using Hess’s Law:  H =  H 1 –  H 2 Why –  H 2 ? = {  H c (carbon) + 2  H c (hydrogen)} – {  H c (methane)} = (2 x –285) – (-890) = -73 kJmol -1

ii) Calculating  H for all other reactions This involves using standard enthalpy of formation data (given in Q) e.g. calculate the enthalpy change for the reaction: Fe 2 O 3(s) + 3CO (g)  2Fe (s) + 3CO 2(g) Draw an energy cycle. Fe 2 O 3(s) + 3CO (g)  2Fe (s) + 3CO 2(g) 2Fe (s) + 3O 2(g) + 3C (s) H1H1 H2H2

Using Hess’s Law:  H = -  H 1 +  H 2 Why –  H 1 ? = -{  H f (Fe 2 O 3 ) + 3  H f (CO)} + {3  H f (CO 2 )} Why is there no value needed for Fe? = -{(-822) + (3 x –111)} + (3 x –394) = -27 kJmol -1 Fe 2 O 3(s) + 3CO (g)  2Fe (s) + 3CO 2(g) 2Fe (s) + 3O 2(g) + 3C (s) H1H1 H2H2

5.4 Bond Enthalpies Or better, bond dissociation enthalpies. Give a measure of bond strength Refers to the enthalpy change for the process: A – B (g)  A (g) + B (g) For molecules made up of several atoms, it is best to use the term average bond enthalpy. Average bond enthalpies can be used to calculate the enthalpy change for a reaction.  H =  (enthalpy of bonds broken) -  (enthalpy of bonds formed)

Using Average Bond Energies to Calculate Enthalpy Changes These only give approximate values for  H as the values used are only average bond energies, B(X – Y). The value of a bond energy changes depending upon the surrounding atoms. To illustrate: H 2 O (g)  H (g) + OH (g)  H  diss = +492 kJmol -1 CH 3 OH (g)  CH 3 O (g) + H (g)  H  diss = +437 kJmol -1 The average bond energy is defined as the mean of the enthalpy required to break a particular covalent bond in a range of molecules.

Example 1: Use bond enthalpy values to calculate the enthalpy change for the reaction: CH 4(g) + Cl 2(g)  CH 3 Cl (g) + HCl (g)  H =  (enthalpy of bonds broken) -  (enthalpy of bonds formed) = {E(C – H) + E(Cl – Cl)} – {E(C – Cl) + E(H – Cl)} = ( ) - ( ) = kJmol -1

Use bond energies to calculate  H  c (standard enthalpy of complete combustion) of propane. B(C – C) = 348 kJmol -1 B(C – H) = 413 kJmol -1 B(O = O) = 498 kJmol -1 B(C = O) = 743 kJmol -1 B(O – H) = 463 kJmol -1  H  c =  B(bonds broken) –  B(bonds formed) = 6490 – 8162 = kJmol -1 Explain why this value differs from the data book value of kJmol -1.