Chapter 6: Enthalpy changes

Learning outcomes: Energy change that happens with a chemical reaction Exothermic and endothermic reactions Bond energies Hess’s law Lab experiment yeah finally!

Energy change in a chemical reaction 2H2 + O2 2H2O + heat C3H8 + 5O2 3CO2 + 4H2O + heat

Energy change in a chemical reaction A chemical reaction will either “release” energy or “absorb” energy from its surrounding. The same: temperature of the surrounding will either increase or decrease. Release of energy = “exothermic”, for example “combustion” reactions Absorb of energy = “endothermic”, for example photosynthesis 6CO2 + 6H2O C6H12O6 + 6O2

Time for a break and practice Please make Q.1 on page 91

But why? Energy needed to break bonds – energy release when bonds form It needs energy to break bonds! The bonds broken and formed are different so the overall energy of a reaction is not 0. Energy is released when bonds form

Example: CH4 + 2 O2 CO2 + 2 H2O E E

The energy change: ΔH The “energy of a reaction” can be measured by measuring the temperature change ΔT We call the energy change of a reaction “enthalpy change” with symbol ΔH and unit kilojoules per mol (kJ/mol) We calculate ΔH = Hproducts - Hreactants Exothermic reactions have negative ΔH and endothermic have positive ΔH

Example: CH4 + 2O2 CO2 + 2H2O What is the value of ΔH (in kJ/mol)? Given information: Bond: Bond energy: C-H 435 kJ/mol O=O 498 kJ/mol O-H 464 kJ/mol

Example profile CH4 + 2O2 CO2 + 2H2O ΔH = Hproducts - Hreactants

Enthalpy change and profile

Enthalpy change and profile Draw the enthalpy profile for the endothermic reaction: CaCO3 CaO + CO2 ΔH = +572 kJ/mol

Time for a break and practice Please make Q.2 on page 91

Why energy change in chemical reactions? Beside kinetic energy, molecules also posses potential energy (just like all matter) which is stored in the bonds of the molecule In a chemical reaction, the bonds change (otherwise the molecules can’t change) It has been empirically demonstrated that heat is either released into the surrounding or absorbed by the reaction The energy gained by the surrounding MUST be equal to the energy lost by the reaction That means in an exothermic reaction, the potential energy stored in the chemical bonds is being converted to thermal energy

Why energy change in chemical reactions? In a exothermic reaction the potential energy of the products is (on average) lower than the potential energy of the products

Standard conditions: R.T.P. To compare enthalpy changes of different reactions we need to use the same conditions consistently: R.T.P. Pressure = 1atm = 105 pa, Temperature = 298 K R.T.P. in enthalpy changes is indicated by the symbol ΔH Example: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ΔH = -890.3 kJ/mol

Enthalpy change of reaction Enthalpy change of formation, ΔHf Enthalpy change of combustion, ΔHc Enthalpy change of neutralisation, ΔHn Enthalpy change of solution, ΔHsol Enthalpy change of atomisation, ΔHat Enthalpy change of hydration, ΔH? Usually just called: Enthalpy change of reaction ΔHr

Enthalpy change of reaction, ΔHr At R.T.P. for the formation of 1 mol of product: H2 (g) + ½ O2 (g) H2O (l) E = -286 kJ 2H2 (g) + O2 (g) 2H2O (l) E = -572 kJ

Standard enthalpy change of formation, ΔHf The standard enthalpy change of formation is the enthalpy change when 1 mol of compound is formed from its elements under standard conditions. The reactants must be in their standard states Symbol is ΔHf Example: ΔHf of Fe2O3 = -824.2 kJ/mol also written as ΔHf [Fe2O3(s)] = -824.2 kJ/mol 2Fe(s) + 3/2O2(g) Fe2O3 (s) ΔHf = -824.2 kJ/mol

Standard enthalpy change of combustion, ΔHc The standard enthalpy change of combustion is the enthalpy change when one mole of a substance is burnt in excess oxygen under standard conditions. The symbol is ΔHc ΔHc is always negative (combustion is always ….) Example ΔHc [CH4(g)] = -890.3 kJ/mol, this means if we burn 1 mol of CH4 (or 16 grams) in excess oxygen 890.3 kJ of heat is released. Let’s construct the equation in R.T.P. for the combustion of CH4

Time for a break and practice Please make Q.3 on page 93

Standard enthalpy change of neutralization, ΔHn The standard enthalpy change of neutralization is the enthalpy change when one mole of water is formed by the reaction of an acid with an alkali (base) under standard conditions H+ (aq) + OH- (aq) H2O (l) ΔHn= -57.1 kJ/mol

Standard enthalpy change of atomisation, ΔHat The standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from its elements under standard conditions For instance the standard enthalpy change of atomisation of hydrogen relates to the equation: ½ H2 (g) H (g) with ΔHat = +218 kJ/mol

Standard enthalpy changes X + Y Fe2O3 (s) with ΔHf = -824.2 kJ/mol Na Na with ΔHat C3H8 + 5O2 3CO2 + 4H2O with What are X and Y? What are the state symbols for the second reaction? What kind of enthalpy is shown by the 3rd reaction and what is the sign (+ or -) of the number and why?

Calculate and measure energy changes We can calculate the energy change of a chemical reaction with data from the data booklet or the book (like bond energies) We can use that data because some people take the effort to do experiments providing us that data, they measure the energy change

Measuring the energy change of a reaction We know that any reaction is either exothermic or endothermic For most reactions we can design some kind of experiment to measure that energy change of the chemical reaction What do we need for that? For example we want to measure the energy change of a neutralization reaction or of dissolving a salt Scale Clock Lamp Thermometer Isolation material Cup with medium (like water)

Measuring the enthalpy change We know that a reaction is either exothermic or endothermic With that knowledge we can use an “calorimeter” to experimentally find the enthalpy change of some reactions So what is the relation between temperature and enthalpy? The calorimeter

Calorimetry It takes 4.18 Joule of energy to increase the temperature of 1 gram of water by 1 °C This is called the “specific heat capacity” of water. Other liquids have different numbers….

What else? Let’s do a reaction in water that gives out energy, so we know the temperature of the water increases We know it takes 4.18 Joules of energy to raise 1 gram of water by 1°C Now we can measure the temperature change of the reaction What other information do we need to know before we can calculate the energy change?

The formula Q = - m c ΔT Q = energy transferred (unit: J) m = mass of water (unit: g) c = the specific heat capacity (4.18 J g-1 °C-1) ΔT = the change in temperature…

Time for an example 1. Calculate the energy transferred when the temperature of 100 cm3 water rises from 28°C to 60°C. Do not forget the unit in your answer! 2. When 10 grams of sodium chloride is dissolved in 50 ml of water the water temperature drops by 2°C. Calculate the energy that is absorbed by the system.

Enthalpy change of neutralisation We do an experiment using a calorimeter to determine the enthalpy change of the neutralisation reaction ΔHn . For that we mix 2 solutions of acid and alkali: First we put 50 cm3 of 1 mol/L hydrochloric acid in the calorimeter and record the temperature Next we add 50 cm3 of 1 mol/L sodium hydroxide We stir the solution and let the reaction take place while we measure the temperature change

Enthalpy change of neutralisation Results: The mass of the total amount of solution = 50 + 50 = 100 grams Starting Temperature was 21.3°C and the final temperature was 27.8°C Given that C = 4.18 J g-1 °C-1 We can now calculate the enthalpy change in kJ/mol

50 cm3 of 2.50 mol dm–3 nitric acid was placed in a polystyrene beaker of negligible heat capacity. Its temperature was recorded and then 50 cm3 of 2.50 mol dm–3 NaOH at the same temperature was quickly added, with stirring. The temperature rose by 17 °C. The resulting solution may be considered to have a specific heat capacity of 4.18 J g–1 K–1. What is an approximate value for the molar enthalpy change of neutralisation of nitric acid and sodium hydroxide from this experiment?

Experiment to find the enthalpy change of combustion reactions The fuel that combusts will produce a flame and heats the water in the cup fuel

Example combustion Results: Let’s burn/combust some 1-propanol to find the enthalpy change of combustion ΔHc for this fuel We use 100 grams of water which has an initial temperature of 20.2 °C We measure the mass of the 1-propanol + the burner = 86.27 g Results: We stop the burning and find that the mass of the 1-propanol + burner decreased to 86.06 g and the temperature raised to 30.9 °C

How much energy is transferred? Q = - m c ΔT Q = - 100 x 4.18 x 10.7 Q = - 4472.6 Joule How much energy is transferred per mole of fuel?

The ΔHc of ethanol You want to find the enthalpy change of combustion ΔHc for ethanol and use 200 grams of water with an initial temperature of 293.1 K. The burner (including ethanol) weighted 65.10 grams After burning some time the mass of the ethanol left + the burner = 64.27 g while an increase in temperature was observed to 321.2 K. Calculate the energy transferred from the combustion using c= 4.18 Jg-1K-1. Give your answer in kJ/mol of ethanol that combusts.

Time for a break and practice Please make Q.4-7 on page 96

First law of thermodynamics The first law of thermodynamics states that energy can not be created or destroyed We have seen that this also applies to the chemical reaction (potential energy changes to thermal energy) The total energy of the molecules and their surrounding must remain constant This means the total energy change of a reaction is fixed and it does not matter by which route the reaction takes place as long as the initial and final conditions are the same

Enthalpy cycles and Hess’s law A can react to B, directly or indirectly We call X and Y “intermediates” Especially useful if we can not find a value by experiment Requires good mathematics skills

The catalyst Catalysts can take part in the reaction, however they are always chemically unchanged at the end Example: 2H2O2 2H2O + O2 (overall reaction not showing catalyst) 2H2O2 + I- OI- + H2O H2O2 + OI- H2O + O2 + I-

“The enthalpy change is independent of the path taken” Hess’s law “The enthalpy change is independent of the path taken” The enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. DHr = DH1 + DH2 + DH3 If we go in the opposite direction of an arrow, we subtract the value of the enthalpy change. For instance for DH2 we write: DH2 = - DH1 + DHr - DH3

From formation to reaction Remember we discussed the ΔHf enthalpy change of formation C (s) + 2H2 (g) CH4 (g) with ΔHf = -74.8 kJ/mol ½ O2 (g) + H2 (g) H2O (l) with ΔHf = -285 kJ/mol C (s) + O2 (g) CO2 (g) with ΔHf = -393.5 kJ/mol With this information we can find the enthalpy change for the combustion for CH4: Construct the final equation Construct all sub equations Sum all sub equations

Time for a break and practice Please make Q.8 on page 98

Combustion & Formation

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The enthalpy change of hydration Difficult to experimentally measure We use the Standard enthalpy change of solution ΔHsol for the calculation with Hess’s law Example Na2S2O3(s) + 5H2O(l) Na2S2O3.5H2O(s)

Bond energies and enthalpy changes As discussed before breaking bonds require energy and hence is an endothermic process Opposite of that energy is released when bonds form and this is an exothermic process Depending on the bonds broken and formed the net result is positive or negative The data for bond energies can be found in the data booklet and is for the gaseous state The data is an average number that means there is differences between molecules

Bond energy Symbol is E, example E(C-H) is the average energy of C-H bond (imagine there is CH4, C2H6, C2H5O etc.) The energy of double bonds and triple bonds are higher than single bonds Let’s look again at the combustion of CH4 CH4 + O2 CO2 + H2O How many C-H bonds need to be broken? How many O=O bonds need to be broken? How many C=O bonds are formed? How many O-H bonds are formed?

Example: CH4 C-H: 4 x E[C-H] O=O: 2 x E[O=O] O=C: 2 x E[O=C] O-H: 4 x E[H-O] E[C-H]: 410 kJ/mol E[O=O]: 496 kJ/mol E[O=C]: 805 kJ/mol E[O-H]: 465kJ/mol

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Bond energies Useful in reversible reactions like: N2 + H2 NH3 Calculate the formation of ammonia, given that E[N=N] = 945 kJ/mol E[H-H] = 436 kJ/mol E[N-H] = 391 kJ/mol

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