Chapter 5 Thermochemistry (5.1-5.2)

Energy Every chemical change has an accompanying change of energy Combustion of fossil fuels Metabolism of foods Discharging of a battery Chemical Cold Pack

Kinetic and potential energy Physics Context Kinetic Energy is the energy of motion KE = ½ mv2 Potential Energy is stored energy; (GPE = gravitational PE) Chemistry Context There is chemical potential (internal) energy in the bonds between atoms in a compound. When the bonds are broken, energy is absorbed. When bonds are formed, energy is released. Temperature is the measure of relative chemical energy

Thermochemistry The study of energy and its transformations is THERMODYNAMICS Study of the relationship between heat, work and energy in a system In this class we will study the relationships between chemical reactions and energy changes involving heat – THERMOCHEMISTRY

SI unit for energy is the Joule Another unit of energy is the calorie Units of energy SI unit for energy is the Joule 1kg moving at 1 m/s = 1 J of energy Another unit of energy is the calorie 1 g water raised 1ºC 1 cal = 4.184 J

System and Surroundings In a chemical reaction: System – the objects or chemicals we are studying Surroundings – container and everything else beyond the chemicals Universe – total of the system and the surroundings Transferring Energy: Work and Heat Work (w) is energy used to push or pull an object against a force Heat (q) is energy transferred from a hotter object to a cooler one

5.2 First Law of Thermodynamics Definition The First Law of says that in any process, the total change in energy of the system, is equal to the sum of the heat absorbed, q, and the work, w, done on the system This law is often referred to as the Law of Conservation of Energy Internal Energy (E) The sum of all the kinetic and potential energy in the system We measure the change in internal energy (∆E) ∆E = Efinal –Einitial (+) ∆E = system gained energy from surroundings = endothermic (-) ∆E = system lost energy to surroundings = exothermic

Relating ∆E to heat and work Two ways to calculate work Work done against force of gravity: w = F x d Pressure-volume work (compression & expansion of gases) w=-P∆V Equation for the First Law: ∆E = q + w Sign Convention: Heat (q) is positive if energy enters the system; it is negative when energy leaves the system Work (w) is positive if work is done on the system; it is negative if work is done by the system. ∆E is positive when there is a net gain of energy by the system; it is negative when there is a net loss.

Sign conventions for q, w, and ∆E + - q Gains heat Loses heat W Work done on system Work done by system ∆E Net gain of energy by the system Net loss of energy by the system

Examples Calculate the ∆E for a gas in which the gas: Absorbs 35 J of heat and does 10 J of work Q is absorbed = +; w is released = - ∆E = w + q = -10 J + 35 J = 25 J Produces 15 J of heat and has 12 J of work done on it as it contracts q is released = -; w is absorbed = + ∆E = +12 J – 15 J = -3 J

Energy diagrams Terminology Exothermic – a reaction that releases energy, ∆E is negative Endothermic – a reaction that absorbs energy, ∆E is positive Activation energy, Ea – the minimum energy that must be possessed by a pair of molecules if collision is to result in reaction Activated complex – a species, formed by collision of energetic particles, that can react to form products Energy of reaction, ∆E – the difference between the energies of the products & reactants.

State Properties The state of a system refers to the temperature, pressure, and composition of that system 25.00 g of hydrogen gas at 1.00 atm at 25C State properties are quantites that depend only upon the “state” of a system, not upon the way it reached that state Ex: volume, temperature Heat and work are not state properties – they do depend on the path

Enthalpy Chem II 5.3-5.4

Enthalpy The chemical and physical changes that occur around us essentially occur at constant atmospheric pressure. Most commonly, the only kind of work produced by chemical and physical changes open to the atmosphere is the mechanical work associated with change in the volume of the system.

Enthalpy, H From a Greek word meaning “to warm” ∆H Commonly called “heat of reaction.” In most cases ∆E and ∆H are equal, except with gases. In these cases the pressure or volume of a system is changing: In these cases ∆H = ∆E + P∆V For most reactions the difference in ∆H and ∆E are so small because P ∆ V is small.

Enthalpy 1. Enthalpy is an extensive property→∆H is directly proportional to the amount of reactant consumed in the process. 2. The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to the ∆H for the reverse reaction. 3. The enthalpy change for a reaction depends on the state of the reactants and products. Important to note the states in your equations.

Enthalpy of Reaction ∆Hrxn = ∆Hproducts– ∆Hreactants Enthalpy of reaction–the enthalpy change that accompanies a reaction Sometimes called = Heat of reaction

Thermochemical equations Balanced chemical equations that show the associated enthalpy change 2 H2(g) + O2(g) → 2 H20 (g) ∆H = -483.6 kJ Would this be exothermic or endothermic? exothermic

Example NH4NO3(s) → NH4+(aq) + NO3-(aq) ∆H= +28.1 kJ Is this reaction exothermic or endothermic? endothermic If 2.00g of solid ammonium nitrate react, how much energy is needed? 2.00g NH4NO3 1 mole 28.1 kJ = 0.702 kJ 80.06 g 1 mol

You try it… H2(g) + Cl2(g) → 2 HCl(g) ∆H = -185kJ Is this reaction exothermic or endothermic? exothermic How much energy is created when 3.50 g of HCl are formed? 3.50g HCl 1 mol HCl -185 kJ = -8.88 kJ 36.46 g 2 mol HCl

And again… Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2H2O2 (l) →2H2O (l) + O2 (g) ∆H= -196 kJ How much energy is created when 5.00 grams of hydrogen peroxide decomposes at constant pressure? 5.00g H2O2 1 mole -196 kJ = 0.147 mole H2O2 34.02 g 2 mol H2O2

Calorimetry Chem II 5.5

Calorimetry Calorimeters are used to measure heat flow or the temperature change experienced by an object when it absorbs a certain amount of heat. Heat Capacity –the amount of heat required to raise its temperature by 1 K. The greater the heat capacity, the greater the heat required to produce a given increase in temperature. It is a unique value for all substances.

Molar Heat Capacity The heat capacity of one mole of a substance. Cmolar Specific heat capacity (specific heat) –the heat capacity of one gram of a substance q c = ------ or q = c x m x ∆T m x ∆T

Specific Heat Values Examples: N2(g) = 1.04 J/g-K Al(s) = 0.90 J/g-K Fe(s) = 0.45 J/g-K H2O (l) = 4.18 J/g-K What do you notice about the value of water compared to others? Why would this be important to us?

You Try It… How much heat is needed to warm 250 grams of water from 22 C to near its boiling point, 98 C? What is the molar heat capacity of water? q = cm∆T q = (4.18 J/g-K)(250 g)(371 – 295) = 79420 J = 79 kJ 250 g H2O 1 mole = 13.87 = 14 moles H20 18.02 g 79.4 kJ/13.87 mole = 5.726 = 5.7 kJ/mole H2O

Constant Pressure Calorimetry Uses a “coffee-cup calorimeter” Not sealed so reaction occurs at constant pressure –atmospheric In a simple calorimeter in which essentially all of the heat given off by a reaction is absorbed by the known amount of water, qsolution = -qreaction

NH4NO3(s) → NH4+(aq) + NO3-(aq) ∆H= + 29 kJ What is the qreaction when a 1.00g sample of ammonium nitrate is dissolved in water if the temperature of 40.0 g of water changed from 25.0⁰C to 22.9⁰C? Q = cm∆T Qreaction = -(4.184J/g⁰C) (41.0 g) (-2.1⁰C) = 360 J How much heat would be evolved if one mole of ammonium nitrate dissolved? (This is called the ∆H for the reaction.) Write the thermochemical equation for this process. 1 mol NH4NO3 80.06g 360 J = 29,000J 1 mol 1.00g NH4NO3(s) → NH4+(aq) + NO3-(aq) ∆H= + 29 kJ

If 25. 0 g of NH4NO3 are placed in 125 grams of water at 22 If 25.0 g of NH4NO3 are placed in 125 grams of water at 22.0⁰C, what is the final temperature of the water? 25g 1 mol 29,000J = 9000J = 9.0 x 103 J 80.06g 1 mol ∆T = qsolution = -9000J = -14⁰C cm (4.184J/gC)(150g) so Tfinal = 22.0 – 14 = 8⁰C

What is the heat of reaction (ΔH) for the neutralization of HCl if 100 What is the heat of reaction (ΔH) for the neutralization of HCl if 100.0 mL of 1.0 M NaOH is added to 100.0 mL of 1.0 M HCl with an initial temperature of 24.6⁰C and after the reaction is complete, the temperature is 31.3⁰C? Assume the densities of the solutions are 1.00 g/mL & the specific heat is 4.184 J/g ⁰C. q = c x m x ΔT q = 4.184 J/g ⁰C x 200.0 g x (31.3 ⁰C - 24.6 ⁰C) q = 5600 J HCl + NaOH --> H2O + NaCl 0.1000 L x 1.0 M = 0.1000 moles of HCl & NaOH ΔH = -5600 J / 0.1000 moles = -56,000 J/mol OR -56 kJ/mol of reaction

1. Theory - a device used to measure heat flow in which a reaction is Bomb Calorimeter 1. Theory - a device used to measure heat flow in which a reaction is carried out with a sealed metal container 2. Calculations: qreaction = - Ccalorimeter x ΔT a. The reaction between 1.00 g octane & oxygen during combustion causes the temp. of the calorimeter to rise from 25.00⁰C to 33.20⁰C. If Ccalorimeter = 837 J/ ⁰C, how much heat is produced in the reaction? -(837J/g⁰C)(33.20-25.00) = 6863.4 = -6860 J (per 1 g of octane) * The negative sign means that heat is released from the reaction.

b. How much energy is produced if one mole of octane was burned? Write the thermochemical equation for this process. 1.00 mol 114.22g -6860J 1kJ = -784 kJ 1 mol 1.00g 1000J C8H18(l) + 25 2 O2(g) → 8 CO2(g) + 9 H2O(g) ΔH = -784kJ

Hess’s Law Chem II 5.6

Hess’s Law If a reaction is carried out in a series of steps, ∆H for the reaction will equal the summ of the enthalpy changes for the individual steps. Enthalpy is a state function. It is independent of the path. We can add equations to come up with the desired final product, and add the ∆H Two rules If the reaction is reversed the sign of ∆H is changed If the reaction is multiplied or divided, so is ∆H

Using Hess’s Law to Calculate ΔH The following information is known: C(s) + O2(g) →CO2(g) ΔH1 = -393.5 kJ CO(g) + ½ O2(g) → CO2(g) ΔH2 = -283.0 kJ Using these data, calculate the enthalpy for: C(s) + ½ O2(g) → CO(g)

Answer The following information is known: C(s) + O2(g) →CO2(g) ΔH1 = -393.5 kJ CO(g) + ½ O2(g) → CO2(g) ΔH2 = -283.0 kJ Using these data, calculate the enthalpy for: C(s) + ½ O2(g) → CO(g) (Reverse Rxn2) ΔHrxn = -393.5 + (+283.0) = -110.5 kJ

Practice with Hess’s Law Calculate ∆H for the reaction 2C(s) + H2(g) →C2H2(g) Given the following chemical equations and their respective ∆H. C2H2(g)+ 5/2O2 → 2CO2(g) + H2O(l) ∆H = -1299.6 kJ C(s) + O2(g) → CO2(g) ∆ H = -393.5 kJ H2(g) + ½ O2(g) → H2O(l) ∆H = -285.8 kJ

Answer Reverse Rxn1 Double Rxn 2 ΔH1 = + 1299.6 ΔHrxn = + 226.8

You Try It… Calculate ΔH for the reaction NO(g) + O(g) →NO2(g) Given the following information: NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -198.9 kJ O3(g) → 3/2 O2(g) ΔH = -142.3 kJ O2(g) → 2O(g) ΔH = 495.8 kJ

Answer Reverse Rxn 2 Reverse Rxn 3 Cut Rxn 3 in half ΔH1 = -198.9 kJ

Remember… H is a state function, so for a particular set of reactants and products, ΔH is the same whether the reaction takes place in one step or in a series of steps.

Standard Enthalpies of Formation ∆Hf is the enthalpy change associated with the formation of a given compound ∆Hffor a number of compounds is found in Appendix C starting on pg. 1041 Standard Enthalpies of formation is the change of enthalpy for the reaction that forms 1 mol e of the compound from its elements in standard states. Ex: 2C(s) + 3H2(g) + ½ O2(g)→ C2H5OH(l) ∆ 𝐻𝑓 ⁰ = -277.7 kJ

Standard Enthalpy of Formation The standard enthalpy of formation of the most stable form (elemental) of any element is zero. Thus, the ∆Hf⁰ for most individual elements is 0. The exception would be the diatomic gases of “Dr. HOFBrINCl. H2, O2, F2, Br2, I2, N2, and Cl2 are zero and the individual elements of hydrogen, oxygen, fluorine, bromine, iodine, nitrogen and chlorine are not zero.