Chapter 5: thermochemistry By Keyana Porter Period 2 AP Chemistry

What is thermochemistry? Thermodynamics: the study of energy and its transformations Thermochemistry is one aspect of thermodynamics The relationship between chemical reactions and energy changes Transformation of energy (heat) during chemical reactions

5.1 Kinetic & Potential Energy Energy is the capacity to do work or the transfer heat Objects possess energy in 2 ways Kinetic energy: due to motion of object Potential/Stored energy: result of its composition or its position relative to another object Kinetic energy (E k ) of an object depends on its mass (m) and speed (v): E k = ½ mv 2 Kinetic energy increases as the speed of an object and its mass increases Thermal energy: energy due to the substance’s temperature; associated with the kinetic energy of the molecules

5.1 Kinetic & Potential Energy Potential energy is a result of attraction & repulsion Ex: an electron has potential energy when it is near a proton due to the attraction (electrostatic forces) Chemical energy: due to the stored energy in the atoms of the substance

5.1 Units of Energy joule (J): SI unit for energy 1 kJ= 1000 J calorie (cal): non-SI unit for energy; amount of energy needed to raise the temperature of 1 g of water by 1 o C 1 cal = J (exactly) Calorie (nutrition unit) = 1000 cal = 1 kcal A mass of 2 kg moving at a speed of 1 m/s = kinetic energy of 1 J E k = ½ mv 2 = ½ (2 kg)(1 m/s) 2 = 1 kg-m 2 /s 2 = 1 J

5.1 System & Surroundings System (chemicals): portion that is singled out of the study Surroundings (container and environment including you): everything else besides the system Closed system: can exchange energy, in the form of heat & work, but not matter with the surroundings

5.1 Transferring Energy: Work & Heat Energy is transferred in 2 ways: Cause the motion of an object against a force Cause a temperature change Force (F): any kind of push or pull exerted on an object Ex: gravity Work (w): energy used to cause an object to move against force Work equals the product of the force and the distance (d) the object is moved: w = F x d heat: the energy transferred from a hotter object to a colder one Combustion reactions release chemical energy stored in the form of heat

5.2 Internal Energy The First Law of Thermodynamics: Energy is conserved; it is neither created nor destroyed Internal energy (E): sum of ALL the kinetic and potential energy of all the components of the system The change in internal energy = the difference between E final – E initial Δ E = E final – E initial We can determine the value of Δ E even if we don’t know the specific values of E final and E initial All energy quantities have 3 parts: A number, a unit, and a sign (exothermic versus endothermic)

5.2 Relating Δ E to Heat & Work A chemical or physical change on a system, the change in its internal energy is given by the heat (q) added to or given off from the system: Δ E = q + w both the heat added to and the work done on the system increases its internal energy

Sign Conventions Used and the Relationship Among q, w, and Δ E Sign Convention for q: q > 0: Heat is transferred from the surroundings to the system (endothermic) q < 0: Heat is transferred from the system to the surroundings (exothermic) Sign convention for w: w > 0: Work is done by the surroundings on the system w < 0: Work is done by the system on the surroundings Sign of Δ E = q + w q > 0 and w > 0: Δ E > 0 q > 0 and w < 0: the sign of Δ E depends on the magnitudes of q and w q 0: the sign of Δ E depends on the magnitudes of q and w q < 0 and w < 0: Δ E < 0

5.2 Endothermic & Exothermic Processes Endothermic: system absorbs heat Ex: melting of ice Exothermic: system loses heat and the heat flows into the surroundings Ex: freezing of ice The internal energy is an example of a state function Value of any state function depends only on the state or condition of the system (temperature, pressure, location), not how it came to be in that particular state Δ E = q + w but, q and w are not state functions

5.3 Enthalpy Enthalpy (H): state function; the heat absorbed or released under constant pressure The change in enthalpy equals the heat ( q P ) gained or lost by the system when the process occurs under constant pressure: Δ H = H final – H initial = q P only under the condition of constant pressure is the heat that is transferred equal to the change in the enthalpy The sign on Δ H indicated the direction of heat transfer + value of Δ H means it is endothermic - value of Δ H means it is exothermic

5.4 Enthalpies of Reaction Enthalpy of reaction ( Δ H rxn ): the enthalpy change that accompanies a reaction The enthalpy change for a chemical reaction is given by the enthalpy of the products minus the reactants: Δ H = H(products) – H(reactants) Thermochemical equations: balanced chemical equations that show the associated enthalpy change The magnitude of Δ H is directly proportional to the amount or reactant consumed in the process

5.4 Enthalpies of Reaction The enthalpy change for the reaction is equal in magnitude but opposite in sign to Δ H for the reverse reaction CO 2 (g) + 2H 2 O(l)  CH 4 (g) + 2O 2 (g) Δ H = 890 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH 1 = kJ Enthalpy ΔH 2 = 890 kJ Reversing a reaction changes the sign but not the magnitude of the enthalpy change: ΔH 2 = - ΔH 1

5.4 Enthalpies of Reaction The enthalpy change for a reaction depends on the state of the reactants and products Ex: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l) ΔH = -890 kJ If the product was H 2 O (g) instead of H 2 O (l), the ΔH would be kJ instead of kJ

5.5 Calorimetry/Heat Capacity & Specific Heat Calorimetry: the measure of heat flow Calorimeter: measures heat flow Heat capacity: the amount of heat required to raise its temperature by 1 K The greater the heat capacity of a body, the greater the heat required to produce a given rise in temperature Molar heat capacity: the heat capacity of 1 mol of a substance Specific heat: the heat capacity of 1 g of a substance; measured by temperature change ( ΔT) that a known mass (m) of the substance undergoes when it gains or loses a specific quantity of heat (q):

5.5 Calorimetry/Heat Capacity & Specific Heat specific heat = quantity of heat transferred (grams of substance) x (temperature change) = q m x ΔT Practice Exercise (B&L page 160) Calculate the quantity of heat absorbed by 50 kg of rocks if their temperature increases by 12.0 O C. (Assume the specific heat of the rocks is.82 J/g-K.)

5.5 Calorimetry/Heat Capacity & Specific Heat Solving the problem q = (specific heat) x (grams of substance) x ΔT = (0.82 J/g-K)(50,000 g)(285 K) = 4.9 x 10 5 J

5.5 Constant-Pressure Calorimetry The heat gained by the solution (q soln ) is equal in magnitude and opposite in sign from the heat of the reaction q soln = (specific heat of solution) x (grams of solution) x Δ T = - q rxn For dilute aqueous solutions, the specific heat of the solution is approx. the same as water (4.18 J/g-K) Practice Exercise (B&L page 161) When 50 mL of.100M AgNO3 and 50.0 mL of.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from o C to o C. The temperature increase is caused by this reaction: AgNO 3 + HCl  AgCl + HNO 3 Calculate Δ H for this reaction, assuming that the combined solution has a mass of 100 g and a specific heat of 4.18 J/g- o C

5.5 Constant-Pressure Calorimetry Solving the problem q rxn = -(specific heat of solution) x (grams of solution) x ΔT = - (4.18 J/g- o C)(100 g)(0.8 K) = - 68,000 J/mol

5.5 Bomb Calorimetry (Constant-Volume Calorimetry) Bomb calorimeter: used to study combustion reactions Heat is released when combustion occurs, absorbed by the calorimeter contents, raising the temperature of the water (measured before and after the reaction) To calculate the heat of combustion from the measured temperature increase in the bomb calorimeter, you must know the heat capacity of the calorimeter (C cal ) q rxn = - C cal x Δ T

5.6 Hess’s Law Hess’s Law: if a reaction is carried out in a series of steps, Δ H for the reaction will be equal to the sum of the enthalpy changes for the individual steps CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) ΔH = -802 kJ (ADD) 2H 2 O (g)  2H 2 O (l) ΔH = -88 kJ CH 4 (g) + 2O 2 (g) + 2H 2 O (g)  CO 2 (g) + 2H 2 O (g) + 2H 2 O (l) ΔH = -890 kJ

5.6 Hess’s Law Practice Exercise 5.8 (B&L page 165) Calculate ΔH for the reaction: 2C (s) + H 2  C 2 H 2 (g) Given the following reactions and their respective enthalpy changes: C 2 H 2 (g) + 5 / 2 O 2  2CO 2 (g) + H 2 O (l) ΔH = kJ C (s) + O 2 (g)  CO 2 (g) ΔH = kJ H 2 (g) + ½ O 2 (g)  H 2 O (l) ΔH = kJ

5.6 Hess’s Law Solving the problem 2CO 2 (g) + H 2 O (l)  C 2 H 2 (g) + 5 / 2 O 2 ΔH = kJ 2C (s) + 2O 2 (g)  2CO 2 (g) ΔH = kJ (2) kJ H 2 (g) + ½ O 2 (g)  H 2 O (l) ΔH = kJ 2C (s) + H 2  C 2 H 2 (g) ΔH = kJ if the reaction is reversed, the sign of ΔH changes if reaction is multiplied, so is ΔH

Enthalpy Diagram The quantity of heat generated by combustion of 1 mol CH4 is independent of whether the reaction takes place in one or more steps: ΔH 1 = ΔH 2 + ΔH 3

5.7 Enthalpies of Formation Enthalpies of vaporization: ΔH for converting liquids to gases Enthalpies of fusion: ΔH for melting solids Enthalpies of combustion: ΔH for combusting a substance in oxygen Enthalpy of formation (ΔH f ): enthalpy change where the substance has been formed from its elements Standard enthalpy (ΔH o ): enthalpy change when all reactants and products are at 1 atm pressure and specific temperature (298 K) Standard enthalpy of formation (ΔH o f ): the enthalpy change for the reaction that forms 1 mol of the compound from its elements, with all substances in their standard states ΔH o f = 0 for any element in its purest form at 295 K and 1 atm pressure

5.7 Enthalpies of Formation The standard enthalpy change for any reaction can be calculated from the summations of the reactants and products in the reaction ΔH o rxn = n ΔH o f (products) - m ΔH o f (reactants) Practice Problem 5.9 (B&L page 169) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C 6 H 6 (l), to CO 2 (g) and H 2 O (l).

5.7 Enthalpies of Formation Solving the problem C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O (l) Δ H o rxn = [6 Δ H o f (CO 2 ) + 3 Δ H o f (H 2 O)] – [ Δ H o f (C 6 H 6 ) + 15/2 Δ H o f (O 2 )] = [6( kJ) + 3( kJ)] – [(49.0 kJ) + 15/2 (0 kJ)] = (-2361 – – 49.0) kJ = kJ

Extra Equations Force = mass x 9.8 m/s 2 Internal energy… Δ E = E final – E initial Entropy… Δ S = S final – S initial Enthalpy… Δ H = H final – H initial Gibbs Free Energy… Δ G = G final – G initial Δ S = Δ H / T