Brad Collins Thermochemistry-Part 1 Chapter 7
Thermochemistry Thermodynamics: The study of energy Thermochemistry: The study of energy in chemical reactions Energy: The capacity to do work Work = force x distance Force = mass area 7.1
Systems and Surroundings The system is the specific part of the universe that is of interest in the study. SURROUNDINGS SYSTEM open closed isolated Exchange: mass & energy energy nothing 7.1
An exothermic process is any process that gives off heat transfers thermal energy from the system to the surroundings. 2H 2 (g) + O 2 (g) H 2 O (g) 2H 2 O (l) + energy H 2 O (l) + energy An endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) energy + H 2 O (s) 2Hg (l) + O 2 (g) H 2 O (l) 7.1
Exo vs. Endothermic An Illustration Exothermic Endothermic 7.1
Types of Energy Potential energy is the energy of position, or stored energy. Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in an atom. Elastic energy is the energy stored in a compressed gas or coiled spring (sometimes called mechanical energy) 7.1
Types of Energy Kinetic energy is the energy of motion Radiant (electromagnetic) radiation comes from the sun and is the earth's primary energy source Thermal energy is the energy from the random motion of atoms and molecules Sound energy is the transmitted vibrational energy of atoms and molecules in bulk. 7.1
First Law of Thermodynamics Energy can be converted from one form to another, but cannot be created or destroyed.
More Thermochemistry Definitions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature X= Thermal Energy 90 0 C 40 0 C greater thermal energy 7.1
Units of Heat Heat is measured in joules or calories A joule (J) is defined as the energy expended in applying a force of 1 newton over a distance of 1 meter 1 J = N x m 2 = Pa x m 3 = (kg x m 2 )/s 2 A calorie (cal) is defined as the energy required to raise 1 gram of water 1ºC. A kilocalorie (kcal, or Cal) is defined as the energy required to raise 1 kilogram of water 1ºC. Used for foods. 1 calorie = 4.184 joules 7.1
Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) heat of reaction = 2801 kj/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J 7.1
Specific Heat The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m s Heat (q) absorbed or released: q = m s Δt q = C Δt Δt = tfinal - tinitial 7.1
Practice Problem How much heat is required to raise the temperature of 12.5 g of copper, 57ºC? q = m s T specific heat (s) of copper = 0.365 J/g ºC 7.1
How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C Δt = t final t initial = 5 0 C 94 0 C = -89 0 C q = msδt = 869 g x 0.444 J/g 0 C x 89 0 C = -34,000 J 6.5
Calorimetry The measurement of heat (q) transferred between a system and its surroundings Uses a calorimeter Types of calorimeters: Constant volume Constant pressure 7.1
Constant-Volume Calorimetry Bomb Calorimeter Reaction at Constant V q sys = q water + q bomb + q rxn q sys = 0 (an isolated system) q rxn = (q water + q bomb ) q water = msδt q bomb = C bomb Δt C bomb is the heat capacity of the calorimeter No heat enters or leaves! 7.1
Constant-Pressure Calorimetry Reaction at constant pressure q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = msδt q cal = C cal Δt C cal is the heat capacity of the calorimeter No heat enters or leaves! 7.1
Practice Problem A lead pellet having a mass of 26.47 g was heated to a temperature of 89.98ºC and placed in a constant-pressure calorimeter containing 100.0 g of water at 22.50ºC. If the calorimeter has a heat capacity of approximately zero and the final temperature of the water plus lead pellet is 23.17ºC, what is the specific heat of the lead pellet? 7.1
Solution Heat lost by the pellet is gained by the water in the calorimeter. Because the calorimeter has a heat capacity of zero, no heat is lost to the calorimeter, so: q pellet + q water = 0 q pellet = q water Heat gained by the water is: T = T f (ºC) T i (ºC) q water = m s T, where T = 23.17ºC 22.50ºC q water = 100.0 g 4.184 J/g ºC 0.67ºC = 280.328 J Heat lost by the pellet is: q pellet = q water = 280.328 J Specific heat of lead is: T = 23.17ºC 89.98ºC s lead = q lead /m T = 280.328 J/(26.47 g ( 66.81ºC)) = 0.158 J/g ºC 7.1
Enthalpy (H) Enthalpy (H) is used to measure the heat flow into or out of a system for a process that occurs at constant pressure. Enthalpy change ( H) is the difference between the enthalpy under initial conditions and enthalpy under final conditions H = H final H initial H and H are both state functions Enthalpy is a molar quantity, meaning it is expressed in J/mol Under normal lab conditions H = q Negative values for H indicate exothermic processes. Positive values for H indicate endothermic processes. 7.2
Enthalpy Change ( H) ΔH = heat given off or absorbed during a reaction at constant pressure ΔH = H (products) H (reactants) H products < H reactants ΔH < 0 H products > H reactants ΔH > 0 7.2
Enthalpy Change Examples Is ΔH negative or positive? System absorbs heat Endothermic ΔH > 0 6.01 kj are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. H 2 O (s) H 2 O (l) ΔH = 6.01 kj 7.2
Enthalpy Change Examples Is ΔH negative or positive? System gives off heat Exothermic ΔH < 0 890.4 kj are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) ΔH = -890.4 kj 7.2
Thermochemical Equations Chemical equations that include the enthalpy change as part of the balanced chemical reaction The stoichiometric coefficients always refer to the number of moles of a substance H 2 O (s) H 2 O (l) ΔH = 6.01 kj If you reverse a reaction, the sign of ΔH changes H 2 O (l) H 2 O (s) ΔH = -6.01 kj If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n. 2H 2 O (s) 2H 2 O (l) ΔH = 2 x 6.01 = 12.0 kj 7.2
Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (s) H 2 O (l) ΔH = 6.01 kj H 2 O (l) H 2 O (g) ΔH = 44.0 kj How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) ΔH = -3013 kj 1 mol P 4 266 g P 4 x 123.9 g P 4 x 3013 kj 1 mol P 4 = 6470 kj 7.2
Enthalpy and Phase Changes 7.2
Enthalpy and Phase Changes As heat is added to a substance it warms, and undergoes phase changes. The heat required to warm a substance is given by the specific heat for the phase (s, l, g) of the substance times its mass (q=m s T) The heat required to melt a substance is the enthalpy of fusion ( H fus ) # H 2 O (s) H 2 O (l) ΔH fus = 6.01 kj The heat required to vaporize a substance is the enthalpy of vaporization ( H vap ) H 2 O (l) H 2 O (g) ΔH vap = 44.0 kj 7.2
Practice Problem How much heat is required to raise the temperature of 10.0 g of water from 10ºC to +50ºC? 7.2
Hess's Law Germaine Henri Hess (Russian chemist) Thermochemical equations are additive + C(s) + ½O2(g) > CO(g) CO(g) + ½O2(g) > CO2(g) C(s) + O2(g) > CO2(g) H = 110.5 kj H = 283.0 kj H = 393.5 kj Hess s Law works because H is a state function 7.3
State Functions State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy (E), enthalpy (H), pressure, volume, temperature ΔE = E final - E initial ΔH = H final - H initial ΔP = P final - P initial ΔV = V final - V initial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. ΔT = T final - T initial 7.3
Hess s Law Practice Calculate the enthalpy change of: C2H2(g) + 2Cl2(g) > C2H2Cl4(l) Using the following thermochemical equations: 2C(s) + H2(g) > C2H2(g) 2C(s) + 2Cl2(g) > C2H2Cl4(l) H = 227 kj H = 130 kj 7.3
Enthalpy of Formation Definition: The enthalpy change required to form one mole of a substance from each of its elements in their standard states. Standard state of an element is its state in pure form at 1 atm. Standard states of some common elements Metals: are monatomic solids,except Hg(l) Diatomic elements: are gases, except Br(l) and I(s) Carbon: graphite, written as C(graphite) Phosphorus: white phosphorus, P(white) Sulfur: rhombic sulfur, S(rhombic) Enthalpy of formation ( H f ) is determined experimentally C(graphite) + O 2 (g) > CO 2 (g) H f = 393.5 kj 7.4
7.4
Using Hf to Calculate Hrxn Use Hº * f of the reactants and products C6H12O6(s) + 6 O2(g) > 6 CO2(g) + 6H2O(l) Calculate using n Hºf{products} m Hºf{reactants} H 0 rxn = [6 Hºf{CO2(g)} + 6 Hºf{H2O(l)}] [ Hºf{C6H12O6(s)} + 6 Hºf{O2(g)}] From Table, p. 253 Hºf{CO2(g)} = 393.5 kj/mol Hºf{O2(g)} = 0 kj/mol Hºf{H2O(l)} = 241.8 kj/mol Hºf{C6H12O6(s)} = 1274.5 kj/mol H 0 rxn = [6 ( 393.5) + 6 ( 241.8)] [ 1274.5 + 6 (0)] = 2537.3 kj Try this one: CaCO3(s) > CaO(s) + CO2(g) *Indicates Hf was measured at 25ºC and 1 atm
6.5
Hºrxn Practice Problem Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kj/mol. 2C6H6(l) + 15O2(g) > 12CO2(g) + 6H2O(l) Hºrxn = n Hºf(products) m Hºf(reactants) Hºrxn = [12 Hºf(CO2) + 6 Hºf(H2O)] 2 Hºf(C6H6) Hºrxn = [12 ( 393.5 kj) + 6 ( 285.8 kj)] 2 (49.04 kj) 6535 kj 2 mol = 3267 kj/mol C6H6