Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Copyright © Cengage Learning. All rights reserved Energy Capacity to do work or to produce heat. Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed. The total energy content of the universe is constant. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Energy Potential energy – energy due to position or composition. Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Initial Position In the initial position, ball A has a higher potential energy than ball B. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Final Position After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Energy Heat involves the transfer of energy between two objects due to a temperature difference. Work – force acting over a distance. Energy is a state function; work and heat are not State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Copyright © Cengage Learning. All rights reserved

Energy is the capacity to do work 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 Electrical energy is the energy associated with the flow of electrons Potential energy is the energy available by virtue of an object’s position 6.1

Energy Changes in Chemical Reactions 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 400C 900C greater thermal energy 6.2

Thermochemistry is the study of heat change in chemical reactions. 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 6.2

2H2 (g) + O2 (g) 2H2O (l) + energy 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 + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g) 6.2

CONCEPT CHECK! Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Exo Endo Exothermic (heat energy leaves your hand and moves to the ice) Endothermic (heat energy flows into the ice) Endothermic (heat energy flows into the water to boil it) Exothermic (heat energy leaves to condense the water from a gas to a liquid) Endothermic (heat energy flows into the ice cream to melt it) Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Is the freezing of water an endothermic or exothermic process? Explain. Exothermic process because you must remove energy in order to slow the molecules down to form a solid. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! For each of the following, define a system and its surroundings and give the direction of energy transfer. Methane is burning in a Bunsen burner in a laboratory. Water drops, sitting on your skin after swimming, evaporate. System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic) System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic) Copyright © Cengage Learning. All rights reserved

Internal Energy Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system. To change the internal energy of a system: ΔE = q + w q represents heat w represents work Copyright © Cengage Learning. All rights reserved

Work vs Energy Flow To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Internal Energy Thermodynamic quantities consist of two parts: Number gives the magnitude of the change. Sign indicates the direction of the flow. Copyright © Cengage Learning. All rights reserved

Internal Energy Sign reflects the system’s point of view. Endothermic Process: q is positive Exothermic Process: q is negative Copyright © Cengage Learning. All rights reserved

Internal Energy Sign reflects the system’s point of view. System does work on surroundings: w is negative Surroundings do work on the system: w is positive Copyright © Cengage Learning. All rights reserved

Work Work = P × A × Δh = PΔV P is pressure. A is area. Δh is the piston moving a distance. ΔV is the change in volume.

Work For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = –PΔV To convert between L·atm and Joules, use 1 L·atm = 101.3 J. Copyright © Cengage Learning. All rights reserved

Thermodynamics DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDV when a gas expands against a constant external pressure 6.7

Enthalpy and the First Law of Thermodynamics DE = q + w At constant pressure, q = DH and w = -PDV DE = DH - PDV DH = DE + PDV 6.7

They perform the same amount of work. EXERCISE! Which of the following performs more work? a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. They both perform the same amount of work. w = -PΔV a) w = -(2 atm)(4.0-1.0) = -6 L·atm b) w = -(3 atm)(3.0-1.0) = -6 L·atm Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Determine the sign of ΔE for each of the following with the listed conditions: a) An endothermic process that performs work. |work| > |heat| |work| < |heat| b) Work is done on a gas and the process is exothermic. Δ E = negative Δ E = positive a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – ΔE is negative; second condition – ΔE is positive b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – ΔE is positive; second condition – ΔE is negative Copyright © Cengage Learning. All rights reserved

Calorimetry Science of measuring heat Specific heat capacity: The energy required to raise the temperature of one gram of a substance by one degree Celsius. Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved

Calorimetry If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution. Copyright © Cengage Learning. All rights reserved

A Coffee–Cup Calorimeter Made of Two Styrofoam Cups Copyright © Cengage Learning. All rights reserved

Calorimetry Energy released (heat) = s × m × ΔT s = specific heat capacity (J/°C·g) m = mass of solution (g) ΔT = change in temperature (°C) Copyright © Cengage Learning. All rights reserved

Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = msDt qbomb = CbombDt Reaction at Constant V DH = qrxn DH ~ qrxn No heat enters or leaves! 6.4

Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.4

Heat (q) absorbed or released: 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 = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.4

Dt = tfinal – tinitial = 50C – 940C = -890C 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 6.4

CONCEPT CHECK! A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C The correct answer is b). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A 100.0 g sample of water at 90.°C is added to a 500.0 g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 23°C The correct answer is c). The final temperature of the water is 23°C. - (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 23°C Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C The correct answer is c). The final temperature of the water is 18°C. - (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 18°C Copyright © Cengage Learning. All rights reserved

Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.7

DH = H (products) – H (reactants) 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 6.3

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 6.3

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 6.3

Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ 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 6.3

Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.3

6.4

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Copyright © Cengage Learning. All rights reserved

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3. N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ ΔH1 = ΔH2 + ΔH3 = 68 kJ Copyright © Cengage Learning. All rights reserved

The Principle of Hess’s Law Copyright © Cengage Learning. All rights reserved

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Characteristics of Enthalpy Changes If a reaction is reversed, the sign of ΔH is also reversed. The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer. Copyright © Cengage Learning. All rights reserved

Example Consider the following data: Calculate ΔH for the reaction Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products. Multiply reactions to give the correct numbers of reactants and products. Copyright © Cengage Learning. All rights reserved

Example Reverse the two reactions: Desired reaction: Copyright © Cengage Learning. All rights reserved

Example Multiply reactions to give the correct numbers of reactants and products: 4( ) 4( ) 3( ) 3( ) Desired reaction:

Example Final reactions: Desired reaction: ΔH = +1268 kJ Copyright © Cengage Learning. All rights reserved

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 (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f 6.5

Conventional Definitions of Standard States For a Compound For a gas, pressure is exactly 1 atm. For a solution, concentration is exactly 1 M. Pure substance (liquid or solid) For an Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C. Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy: Enthalpy Calculations When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes. When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer. Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy: Enthalpy Calculations The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: ΔH°rxn = ΣnpΔHf°(products) - ΣnrΔHf°(reactants) 4. Elements in their standard states are not included in the ΔHreaction calculations because ΔHf° for an element in its standard state is zero. Copyright © Cengage Learning. All rights reserved

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: EXERCISE! Calculate ΔH° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: ΔHf° (kJ/mol) Na(s) 0 H2O(l) –286 NaOH(aq) –470 H2(g) 0 ΔH° = –368 kJ [2(–470) + 0] – [0 + 2(–286)] = –368 kJ ΔH = –368 kJ Copyright © Cengage Learning. All rights reserved

Write the thermochemical equation for the enthalpy of formation of ethanol, C2H5OH (l) C (graphite) + 3 H2 (g) + O2 (g) C2H5OH(l) Write the thermochemical equation for the enthalpy of formation of aluminum oxide, Al2O3 (s) 2 Al (s) + O2 (g) Al2O3 (s) DHf = -1669.8 kJ/mol

6.5

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - 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.) 6.5

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ rxn 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. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.5

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) 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) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C6H6 6.5

DHsoln = Hsoln - Hcomponents 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? 6.6

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.6