The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.law of conservation of energy

 Heat is the energy transferred between objects that are at different temperatures.  The amount of heat transferred depends on the amount of the substance. ◦ Though energy has many different forms, all energy is measured in units called joules (J).

 Temperature is a measure of “hotness” of a substance and represent the average kinetic energy of the particles in a substance.  It does not depend on the amount of the substance. Do both beakers contain the same amount of heat?

 All chemical reactions are accompanied by some form of energy change  ExothermicEnergy is given out  Endothermic Energy is absorbed  Activity : observing exothermic and endothermic reactions

 Enthalpy (H) is the heat content that is stored in a chemical system.  We cannot measure enthalpy directly, only the change in enthalpy ∆ H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure, measured in kilojoules per mole (kJmol -1 ).. ∆ H = H (products) – H (reactants)  Standard enthalpy changes, Δ H Ɵ are measured under standard conditions: pressure of 1 atmosphere (1.013 x 10 5 Pa), temperature of 25 0 C (298K) and concentration of 1 moldm -1.

 The particles are dissolved in the solution  The heat is exchanged from the particles into the solution  The heat exchange is measured using the thermometer  There are limitations in using a simple calorimeter which need to be known.

 Loss of heat to the surroundings (exothermic reaction); absorption of heat from the surroundings (endothermic reaction). This can be reduced by insulating the calorimeter.  Using incorrect specific heat capacity in the calculation of heat change. If copper can is used, the s.h.c. of copper must be accounted for.

Enthalpy  For exothermic reactions, the reactants have more energy than the products, and the enthalpy change, ∆ H = H (products) - H (reactants)  ∆ H is negative since H (products) < H (reactants)  There is an enthalpy decrease and heat is released to the surroundings

 Self-heating cans ◦ CaO (s) + H₂O (l)  Ca(OH)₂ (aq)  Combustion reactions ◦ CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l)  neutralization (acid + base) ◦ NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l)  Respiration ◦ C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)

 For endothermic reactions, the reactants have less energy than the products, and the enthalpy change, ∆ H = H (products) - H (reactants)  ∆ H is positive since H (products) < H (reactants)  There is an enthalpy increase and heat is absorbed from the surroundings Enthalpy

 Self-cooling beer can ◦ H ₂O (l)  H₂O (g)  Thermal decomposition  CaCO₃ (s)  CaO (s) + CO ₂ (g)  Photosynthesis  6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)

 Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin.  Uint : Jg -1 0 C -1 Calculating heat absorbed and released – q = c × m × ΔT – q = heat absorbed or released – c = specific heat capacity of substance – m = mass of substance in grams – ΔT = change in temperature in Celsius

How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg -1 0 C -1 ) from 50 0 C to 70 0 C?

 If 1g of methanol is burned to heat 100g of water, raising its temperature by 42K, calculate the enthalpy change for the combustion of methanol.

When 3 g of sodium carbonate are added to 50 cm 3 of 1.0 M HCl, the temperature rises from 22.0 °C to 28.5°C. Calculate the enthalpy change for the reaction. Assume that the density of the solution is 1 gcm -3 and the specific heat capacity is 4.18J Jg -1 0 C -1.

100.0 cm 3 of mol dm -3 copper II sulphate solution is placed in a styrofoam cup g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction. First step Make sure you understand the graph. Extrapolate to determine the change in temperature. The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used? Example

Determine the limiting reactant Calculate Q Calculate the enthalpy for the reaction cm 3 of mol dm -3 copper II sulphate solution is placed in a styrofoam cup g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

 50 cm 3 of moldm -1 silver nitrate solution was put in a calorimeter and 0.200g of zinc powder is added. The temperature of th solution rose to C. Deduce which reagent was in excess and calculate the the enthalpy change for the reaction (per mole o Zn that reacts). Assume that the density of the solution is 1 gcm -3 and the specific heat capacity is 4.18J Jg -1 0 C -1. Ignore the heat capacity of metals and dissolved ions.

 The standard enthalpy change of combustion for a substance is the heat released when 1 mole of a pure substance is completely burnt in excess oxygen under standard conditions.  Example, CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) ΔH Ɵ c =-698 kJmol -1

The following measurements are taken:  Mass of cold water (g)  Temperature rise of the water ( 0 C)  The loss of mass of the fuel (g) We know that it takes 4.18J of energy to raise the temperature of 1g of water by 1 0 C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg -1 K -1. Hence, energy transferred can be calculated using: Energy transfer = mcΔT (joules)  If one mole of the fuel has a mass of M grams, then:  Enthalpy transfer = m x 4.18 x T x M/y  where y is mass loss of fuel.

To determine the enthalpy of combustion for ethanol (see reaction), a calorimeter setup (below) was used. The burner was lit and allowed to heat the water for 60 s. The change in mass of the burner was g and the temperature increase was measured to be 9.90 o C. What is the big assumption made with this type of experiment? Example

First step – calculate heat evolved using calorimetry Last step – determine ΔH for the reaction The burner was lit and allowed to heat the water for 60 s. The change in mass of the burner was g and the temperature increase was measured to be 9.90 o C.

Given that: Vol of water = 100 cm 3 Temp rise = C Mass of methanol burned = 0.75g Specific heat capacity of water = 4.18 Jg -10 C -1 Calculate the molar enthalpy change of the combustion of methanol.

 The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H + is completely neutralised by an alkali under standard conditions.  Example, NaOH(g) + HCl(g)  NaCl(g) + H 2 O(l) ΔH Ɵ =-57 kJmol -1 The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.

 Reaction between strong acid and strong base involves H + (aq) + OH - (aq)  H 2 O(l) ΔH Ɵ =-57 kJmol -1 For sulfuric acid, the enthalpy of neutralisation equation is ½ H 2 SO 4 (aq) + KOH(aq)  ½K 2 SO 4 (aq) + H 2 O(l) ΔH Ɵ =-57 kJmol -1

For neutralisation between a weak acid, a weak base or both, the enthalpy of neutraisation will be smaller than -57 kJmol -1 (less exothermic) CH 3 COOH(aq) + NaOH(aq)  CH 3 COONa(aq) + H 2 O(l) ΔH Ɵ =-55.2 kJmol -1  Some of the energy released is used to ionise the acid.

50.0 cm 3 of a 1.00 mol dm -3 HCl solution is mixed with 25.0 cm 3 of 2.00 mol dm - 3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7 o C. After stirring and accounting for heat loss, the highest temperature reached was 33.5 o C. Calculate the enthalpy change for this reaction. NaOHHCl both 26.7 o o  33.5 o Example

After writing a balanced equation, the molar quantities and limiting reactant needs to be determined. Note that in this example there is exactly the right amount of each reactant. If one reactant is present in excess, the heat evolved will associated with the mole amount of limiting reactant cm 3 of a 1.00 mol dm -3 HCl solution is mixed with 25.0 cm 3 of 2.00 mol dm - 3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7 o C. After stirring and accounting for heat loss, the highest temperature reached was 33.5 o C. Calculate the enthalpy change for this reaction.

Next step – determine how much heat was released. There are some assumptions in this calculation -Density of reaction mixture (to determine mass) -SHC of reaction mixture (to calculate Q) 50.0 cm 3 of a 1.00 mol dm -3 HCl solution is mixed with 25.0 cm 3 of 2.00 mol dm - 3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7 o C. After stirring and accounting for heat loss, the highest temperature reached was 33.5 o C. Calculate the enthalpy change for this reaction.

Final step – calculate ΔH for the reaction 50.0 cm 3 of a 1.00 mol dm -3 HCl solution is mixed with 25.0 cm 3 of 2.00 mol dm - 3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7 o C. After stirring and accounting for heat loss, the highest temperature reached was 33.5 o C. Calculate the enthalpy change for this reaction.

50.00 cm 3 of 1.0 moldm -1 hydrochloric acid was added to cm 3 of 1.0 moldm -1 sodium hydroxide solution. The temperatre rose by C. Calculate the enthalpy change f neutralisation for the reaction. Assume that the density of the solution is 1.00gcm -1 and the specific heat capacity of the solutio is 4.18 Jg -10 C -1.

States that  If a reaction consists of a number of steps, the overall enthalpy change is equal to the sum of enthalpy of individual steps.  the overall enthalpy change in a reaction is constant, not dependent on the pathway take.

Calculate the enthalpy change for the combustion of carbon to form carbon monoxide. Route 1:C(s) + O 2 (g)  CO 2 (g) ΔH Ɵ 1 =-394 kJmol -1 Route 2:C(s) + ½O 2 (g)  CO(g) ΔH Ɵ 2 = ? CO(s) + ½O 2 (g)  CO 2 (g) ΔH Ɵ 1 =-283 kJmol -1 C(s) + O 2 (g) CO 2 (g) CO(s) + ½O 2 (g) ΔHƟ1ΔHƟ1 ΔHƟ3ΔHƟ3 ΔHƟ2ΔHƟ2 Route 1 Route 2

Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide. NaOH(aq) NaOH(s) NaCl(s) + H 2 O(l) + HCl(aq) 1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔH Ɵ 1 =-43kJmol NaOH(aq) + HCl (aq)  NaCl(aq) + H 2 O(l) ΔH Ɵ 1 =-57kJmol NaOH(s) + HCl (aq)  NaCl(aq) + H 2 O(l). ΔH2ΔH2 ΔH1ΔH1 Indirect path Direct path + HCl(aq) + H 2 O(l)

Calculate the enthalpy change for the thermal decomposition of calcium carbonate. CaCO 3 (s)  CaO(s) + CO 2 (g) CaCO 3 (s) +2HCl(aq)  CaCl 2 (aq) + H 2 O(l) ΔH Ɵ 1 =-17 kJmol -1 CaO(s) +2HCl(aq)  CaCl 2 (aq) + H 2 O(l) ΔH Ɵ 1 =-195kJmol -1 CaCO 3 (s) CaO(s) +CO 2 (g) CaCl 2 (aq) + H 2 O(l) +CO 2 (g) ΔHΔH -195kJmol -1 Direct path Indirect path + 2HCl(aq) -17 kJmol -1

Calculate the enthalpy of hydration of anhydrous copper(II)sulfate change. CuSO 4 (s) +5H 2 O(l)  CuSO 4. 5H 2 O (s) CuSO 4 (s) +5H 2 O(l) CuSO 4. 5H 2 O (s) Cu 2+ (aq) + SO 4 2- (aq) ΔHΔH ΔH2ΔH2 ΔH1ΔH1 Direct pathway Indirect pathway

From the following data at 25 0 C and 1 atmosphere pressure: Eqn 1: 2CO 2 (g)  2CO(g) + O 2 (g) ΔH Ɵ =566 kJmol -1 Eqn 2: 3CO(g) + O 3 (g)  3CO 2 (g) ΔH Ɵ =-992 kJmol -1 Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O 2 (g)  O 3 (g)

Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions. C (s,graphite) + O 2 (g)  CO 2 (g) ΔH Ɵ =-393 kJmol -1 C (s, diamond) + O 2 (g)  CO 2 (g) ΔH Ɵ =-395 kJmol -1

 Enthalpy changes can also be calculated directly from bond enthalpies.  The bond enthalpy is the amount of energy required to break one mole of a specified covalent bond in the gaseous state.  For diatomic molecule the bond enthalpy is defined as the enthalpy change for the process X-Y(g) X(g) + Y(g) [gaseous state]  Average bond enthalpies are enthalpies calculated from a range of compounds,eg C-H bond enthalpy is based on the ave bond energies in CH 4, alkanes and other hydrocarbons.

Bond Ave bond enthalpy, ΔH Ɵ (Kjmol -1 ) Bond length (nm) H-H C-C C-H O-H N-H N-N C=C O=O C Ξ C NΞNNΞN 944

When a hydrocarbon e.g. methane (CH 4 ) burns, CH 4 + O 2  CO 2 + H 2 O What happens?

C H H H H + O O O O C H H H H O O O O C H H H H O O OO ENERGY Enthalpy Level (KJ) Progress of Reaction 2 O=O Bond Breaking Bond Forming 4 C-H 4 H-O2 C=O CH 4 + 2O 2  CO 2 + 2H 2 O

C H H H H + O O O O C H H H H O O OO Why is this an exothermic reaction (produces heat)?

Bond Ave Bond Enthalpy (kJ/mol) C-H412 H-O463 O=O496 C=O743 CH 4 + 2O 2  CO 2 + 2H 2 O Energy absorbed when bonds are broken = (4 x C-H + 2 x O=O) Energy given out when bonds are formed = ( 2 x C=O + 4 x H-O) = 4 x x 496 = 2640 kJ/mol C H H H H + O O O O C H H H H O O OO Break  Form = 2 x x 464 = 3338 kJ/mol

Energy absorbed when bonds are broken (a) = 2640 kJ/mol Energy released when bonds are formed (b) = 3338 kJ/mol Enthalpy change, ΔH = ∑(energy absorbed to break bonds) - ∑(energy released to form bonds) = a + (-b) = 2640 – 3338 = -698 kJ/mol Why is this an exothermic reaction (produces heat)?

What can be said about the hydrogenation reaction of ethene? H H C=C (g) + H-H (g)  H-C-C-H (g) H H HH HH

What can be said about the combustion of hydrazine in oxygen? H H N-N (g) + O=O (g)  N ΞN (g) + 2 O (g) H H HH

The combustion of both C and CO to form CO 2 can be measured easily but the combustion of C to CO cannot. This can be represented by the energy cycle. ΔH x = -393 – (-283) = kJmol -1 ΔHxΔHx ½O 2 (g) -393kJmol -1 C(s)+ ½O 2 (g) CO(g) CO 2 (g) -283kJmol -1 ½O 2 (g)

Calculate the standard enthalpy change when one mole of methane is formed from its elements in their standard states. The standard enthalpies of combustion of carbo, hydrogen and methane are - 393, -286 and -890 kJmol -1 resepctively.