Thermochemistry Chapter 6 PDF

Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  The Nature and Types of Energy Energy is the capacity to do work or to produce heat Radiant energy comes from the sun and is earth’s energy primary...

Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  The Nature and Types of Energy Energy is the capacity to do work or to produce heat Radiant energy comes from the sun and is earth’s energy primary source Thermal energy is the energy associated with the random motion of atoms and molecules 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 the atom Potential energy is the energy available by virtue of an object’s position Energy forms are interconvertible. When one form disappears, one other form of equal magnitude must appear (Law of conservation of energy)  Energy Changes in Chemical Reactions Almost all chemical reactions absorb or produce (release) energy, generally in the form of heat. Thermal energy is the energy associated with the random motion of atoms and molecules. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy 900C 400C greater thermal energy Useful Definitions for Energy Changes Analysis Thermochemistry: the study of heat change in chemical reactions. The system: the specific part of the universe that is of interest in the study. The surroundings: are the rest of the universe outside the system Universe = system + surroundings  Three Types of Systems Totally Closed insulated flask container open closed isolated Exchange: mass & energy energy nothing  Exothermic and Endothermic Processes Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2(g) 2H2O (l) + energy H2O(g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO(s) 2Hg (l) + O2 (g) H2O(l) + energy → H2O(g)  Thermodynamics Energy (heat) Movement or Transfer Thermodynamics is the field of chemistry that studies energy and its interconversions that accompany physical and chemical processes. Thermodynamics lets us predict whether a process will occur. Thermodynamics studies changes in the state of a system (defined by state properties or state functions).  State Functions State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Energy, pressure, volume, temperature are state functions Change in a state function depends only on initial and final states DE = Efinal - Einitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.  The First Law of Thermodynamics Energy can be converted from one form to another, but cannot be created or destroyed. (based on the law of conservation of energy) E universe is constant or ΔEuniverse = 0 ΔEsystem + ΔEsurroundings = 0 ΔEsystem = -ΔEsurroundings Example Consider the following reaction: C3H8 + 5O2 → 3CO2 + 4H2O Exothermic chemical reaction Chemical energy lost by = Energy gained by the Combustion surroundings System Surroundings  The First Law of Thermodynamics Another form of the first law for DEsystem DE = q + w DE is the change in internal energy of a system (internal Energy, E, is the sum of kinetic energy and potential energy). q is the heat exchange between the system and the surroundings w is the work done on (or by) the system Gain by the system (+) sign; Loss by the system (-) sign  Work Done by the System Gas in a cylinder fitted with a weightless, frictionless piston at a certain T, P, and V. The gas expands against a constant opposing external atmospheric pressure P. The work done by the gas on the surroundings is: w = - P ∆V The minus sign takes care of the convention for w. W depends on P. If P is zero (if the gas is expanding against a vacuum), the work done must be zero  For gas expansion For gas compression, w = Fd Work done by the work done on the system system w = -P DV DV > 0 DV < 0 F 3 P x V = 2x d = Fd = w -PDV < 0 -PDV > 0 d wsys < 0 wsys > 0 1L·atm = 101.3 J Work is not a state function! Dw = wfinal - winitial initial final  A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P DV (a) DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L atm = 0 joules (b) DV = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 L atm w = -14.1 L atm x101.3 J = -1430 J 1L atm Note: Work is not a state function as proved!  Enthalpy of Chemical Reactions Consider two situations: DE = q + w DE = q - PDV 1. At constant volume: DV = 0, no p-V work will result DE = qv 2. At constant pressure: DE = qp + w ΔE = qp- PΔV or qp = ΔE + PΔV Enthalpy (H) is a thermodynamic function of a system H = E + PV (by definition) If the pressure is held constant, then DH = DE + DPV (Since E, P, and V are state functions, H is also a state function) Similar The change in enthalpy is given by DH = DE +P DV to qp above Therefore, at constant pressure: ΔH = qp Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0  Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ/mol  Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ/mol  Thermochemical Equations  The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ/mol  If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ/mol  If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ  The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ/mol H2O (l) H2O (g) DH = 44.0 kJ/mol How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ x 1 mol P4 x 3013 kJ= 6470 kJ 266 g P4 123.9 g P4 1 mol P4  Calorimetry The measurement of heat changes in physical and chemical processes using a calorimeter. Substances respond differently to being heated. The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. (J/goC) The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = m x s x Dt , q = CDt Dt = tfinal - tinitial  How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x 0.444 J/g 0C x –890C = -34,000 J The heat capacity (C) is an extensive property. The specific heat (s) is an intensive property.  Constant-Volume Calorimetry A constant volume calorimeter is used to measure the heat of combustion of a substance A known mass of a compound is placed in a steel container called a constant- volume bomb calorimeter, which is filled with O2 at about 30 atm of pressure. The closed bomb is immersed in a known amount of water. The sample is ignited electrically, and the heat produced by the combustion reaction can be calculated by recording the rise in T of water. The heat given off by the sample is absorbed by the water and bomb. The special design of the calorimeter enable us to assume that no heat (or mass) is lost to the surroundings during the time it takes to make the measurement  Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn Because no heat enters or leaves the system throughout the process, qsys = 0 qrxn = - (qwater + qbomb) qwater = msDt qbomb = CbombDt Reaction at Constant V DH = qrxn DH ~ qrxn No heat enters or leaves!  Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J  Constant-Pressure Calorimetry The constant-pressure calorimeter is a simpler device than the constant-volume calorimeter is used to determine the heat changes for non combustion reactions. A crude constant-pressure calorimeter can be constructed from two Styrofoam coffee cups. This device measures the heat effects of a variety of reactions, such as acid-base neutralization, as well as the heat of solution and heat of dilution. Because the pressure is constant, the heat change for the process (qrxn) is equal to the enthalpy change (H). As in the case of a constant-volume calorimeter, the calorimeter is treated as an isolated system  Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P: DH = qrxn qrxn DH( 1mol )  No heat enters or leaves! n  Standard Enthalpy of Formation Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0f ) as a reference point for all enthalpy expressions. Standard enthalpy of formation (DH0f): is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero. DH0f (O2) = 0 DH0f (C, graphite) = 0 DH0f (O3) = 142 kJ/mol DH0f (C, diamond) = 1.90 kJ/mol  Standard Enthalpy of Reaction, ΔH°rx 0 ) is the enthalpy of The standard enthalpy of reaction (DHrxn a reaction carried out at 1 atm. aA + bB cC + dD DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ] DH0rxn = S nDH0f (products)- S mDHf0 (reactants) Example: Calculate ΔHºrxn of: C(graphite) + O2 (g) → CO2(g) ΔHºrxn = ΔHºf (CO2,g) - [ΔHºf (C ,gaphite)+ ΔHºf (O2 ,g)] ΔHºrxn = -393.5 – 0 = -393.5 kJ/mol  Standard Enthalpy of Reaction, ΔH°rx (Indirect Method) Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)  Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0rxn = -393.5 kJ S(rhombic) + O2 (g) SO2 (g) DH0rxn = -296.1 kJ CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DHrxn 0 = -1072 kJ 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. C(graphite) + O2 (g) CO2 (g) DH0rxn = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0rxn = -296.1x2 kJ + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DHrxn 0 = +1072 kJ C(graphite) + 2S(rhombic) CS2 (l) DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ 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) DHrxn 0 = SnDH0f (products)- SmDHf0 (reactants) DHrxn 0 = [12DH0f (CO2) +6DH0f (H2O)] - [2DH0f (C6H6)] DHrxn 0 = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ = - 2973 kJ/mol C6H6 2 mol  Heat of Solution The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?  The Solution Process for NaCl Two steps are involved: 2. Hydration of the 1. Separation of one formed gaseous ions: mole of a solid ionic Heat of hydration compound into (ΔHhydr+) gaseous ions: lattice energy (U). DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol  Heat of Dilution Is the heat change associated with the dilution process If a certain solution process is endothermic and the solution is diluted more heat will be absorbed by he same solution from the surroundings. If the solution process is exothermic, more heat will be liberated if additional solvent is added to dilute the solution. Caution! Concentrated H2SO4 has a high exothermic heat of dilution. Never attempt to dilute the concentrated acid by adding water to it! always add acid to water