Thermochemistry Chapter 6

Energy is the capacity to do work. Radiant energy comes from the sun and is earth’s primary energy 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 Potential energy is the energy available by virtue of an object’s position Kinetic energy is the energy produced by a moving object.

Thermochemistry Changes of heat content and heat transfer Follow Law of Conservation of Energy Or, 1st Law of Thermodynamics Energy can neither be created nor destroyed One form of energy can be converted to another form of energy, but can not be destroyed. Examples?p.231 last paragraph

Examples Automobile Engine - Chemical � Kinetic Hydroelectric - Gravitational � Electrical Solar -radiant � Electrical Nuclear -Nuclear � Heat, Kinetic Battery -Chemical � Electrical Food -Chemical � Heat, Kinetic Photosynthesis -radiant � Chemical

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two objects that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy

System and Surrounding(s) System is the specific part of the universe of our interest Surrounding(s) = everything outside the system When both system and surrounding at same temperature ⇒ thermal equilibrium When not, -Heat transfer to surrounding = exothermic -Heat transfer to system = endothermic

closed open isolated

Heat transfers until thermal equilibrium is established. Thermodynamics Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions. Heat transfers until thermal equilibrium is established.

1st Law of Thermodynamics The 1st Law of Thermodyamics simply states that energy can be neither created nor destroyed (conservation of energy). Thus power generation processes and energy sources actually involve conversion of energy from one form to another, rather than creation of energy from nothing.

∆E = E(final) - E(initial) Energy Level Diagram for Heat Energy Transfer CO2 gas ∆E = E(final) - E(initial) = E(gas) - E(solid) CO2 solid

Heat Energy Transfer in Physical Change CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Two things have happened! Gas molecules have higher kinetic energy. Also, WORK is done by the system in pushing aside the atmosphere.

FIRST LAW OF THERMODYNAMICS heat energy transferred ∆E = q + w work done by the system energy change Energy is conserved!

Another form of the first law for DEsystem 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

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. DEsystem + DEsurroundings = 0, (∆E represents change in internal energy) or DEsystem = -DEsurroundings C3H8 + 5O2 3CO2 + 4H2O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings

Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved, it does not depend on its history, only on its initial and final states. energy , pressure, volume, temperature DE = Efinal - Einitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial

ENTHALPY ∆H = Hfinal - Hinitial Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆H where H = enthalpy and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = Hfinal - Hinitial

Enthalpy and the First Law of Thermodynamics DU = q + w At constant pressure (happens usually): q = DH (Enthalpy change) and w = -PDV DE = DH - PDV DH = DE + PDV

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) energy + H2O (s) H2O (l)

Schematic of Exothermic and Endothermic Processes

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

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

Thermochemical Equations 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

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 x s Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial

6.5 A 466-g sample of water is heated from 8.50°C to 74.60°C. Calculate the amount of heat absorbed (in kilojoules) by the water.

6.5 Strategy We know the quantity of water and the specific heat of water. With this information and the temperature rise, we can calculate the amount of heat absorbed (q). Solution Using Equation (6.12), we write Check The units g and °C cancel, and we are left with the desired unit kJ. Because heat is absorbed by the water from the surroundings, it has a positive sign.

q= ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J = 25 kJ 17. How many kJ of heat must be removed from 1000 g of water (specific heat = 4.184 J /g oC) to lower the temperature from 18.0°C to 12.0°C?   A. 2.5 x 10–2 kJ B. 1.4 kJ C. 4.2 kJ D. 25 kJ q= ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J = 25 kJ

Constant-Pressure Calorimetry qrxn = - (qwater + qcal) qwater = m x s x Dt qcal = Ccal x Dt (negligible) Reaction at Constant P DH = qrxn = qwater DHreaction = - qwater Reaction evolved or lost heat (general for coffe cup cal.meter) No heat enters or leaves!

Constant-Volume Calorimetry qrxn = - (qwater + qbomb) qwater = m x s x Dt qbomb = Cbomb x Dt Reaction at Constant V DH ~ rxn No heat enters or leaves!

6.6 A quantity of 1.435 g of naphthalene (C10H8), a pungent-smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28°C to 25.95°C. If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

6.6 Strategy Knowing the heat capacity and the temperature rise, how do we calculate the heat absorbed by the calorimeter? What is the heat generated by the combustion of 1.435 g of naphthalene? What is the conversion factor between grams and moles of naphthalene?

6.6 Solution The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. From Equation (6.16), assuming no heat is lost to the surroundings, we write Because qsys = qcal + qrxn = 0, qcal = −qrxn. The heat change of the reaction is − 57.66 kJ. This is the heat released by the combustion of 1.435 g of C10H8; therefore, we can write the conversion factor as

6.6 The molar mass of naphthalene is 128.2 g, so the heat of combustion of 1 mole of naphthalene is

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 Substance DHcombustion (kJ/g) Apple -2 Beef -8 Beer -1.5 Gasoline -34

DH0 f of any element in its stable form is 0. (Table 6.4) 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) - DH0 f of any element in its stable form is 0. (Table 6.4) (

The heat of combustion of fructose, C6H12O6, is –2812 kJ/mole The heat of combustion of fructose, C6H12O6, is –2812 kJ/mole. Using the following information, calculate H°f for fructose.   C6H12O6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (l) H = –2812 kJ H°f of CO2 = –393.5 kJ/mole H°f of H2O = –285.83 KJ/mole –210.3 kJ B. 210.3 kJ C. –1264 kJ D. 1264 kJ Ans: H = H°f of products - H°f of reactants H°f of elements is zero. -2812 kJ =[ 6 (-393.5kJ/mole) + 6 ( –285.83 KJ/mole)] - H°f for fructose H°f for fructose = [ 6 (-393.5kJ/mole) + 6 ( –285.83 KJ/mole)] + 2812kJ = -1264 kJ

Calculate the standard enthalpy change, Ho, for the following gas phase reaction using bond energy data.   Bond Dissociation Energies Ans: Bonds broken minus bonds formed:   ∆H = [ (1 mol)(614 kJ/mol) + (1 mol)(435 kJ/mol) ] – [ (1 mol)(348 kJ/mol) + (1 mol)(328 kJ/mol) + (1 mol)(413 kJ/mol) ] = 1049 kJ – 1098 kJ = –40 kJ

Hess’s Law 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

Hess’s Law For example, suppose you are given the following data: Could you use these data to obtain the enthalpy change for the following reaction? 3

Hess’s Law If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

C (graphite) + O2 (g) CO2 (g) Also look at CO formation alone, p.258 C (graphite) + O2 (g) CO2 (g) C (graphite) + 1/2O2 (g) → CO (g) CO (g) + 1/2O2 (g) CO2 (g) Addition of the following gives: C (graphite) + O2 (g) CO2 (g) CO2 (g) → CO (g) + 1/2O2 (g) C (graphite) + 1/2O2 (g) → CO (g) Calculate ∆H for CO frmation

6.9 Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements: The equations for each step and the corresponding enthalpy changes are

6.9 Strategy Our goal here is to calculate the enthalpy change for the formation of C2H2 from its elements C and H2. The reaction does not occur directly, however, so we must use an indirect route using the information given by Equations (a), (b), and (c). Solution Looking at the synthesis of C2H2, we need 2 moles of graphite as reactant. So we multiply Equation (a) by 2 to get Next, we need 1 mole of H2 as a reactant and this is provided by Equation (b). Last, we need 1 mole of C2H2 as a product.

6.9 Equation (c) has 2 moles of C2H2 as a reactant so we need to reverse the equation and divide it by 2: Adding Equations (d), (b), and (e) together, we get

6.9 Therefore, This value means that when 1 mole of C2H2 is synthesized from 2 moles of C(graphite) and 1 mole of H2, 226.6 kJ of heat are absorbed by the reacting system from the surroundings. Thus, this is an endothermic process.

18. From the following heats of reaction,   2 C (graphite) + H2 (g)  C2H2 (g) Ho = 227 kJ/mole……. (1) 6 C (graphite) + 3 H2 (g)  C6H6 (l) Ho = 49 kJ/mole ……..(2) Calculate Ho for the reaction 3 C2H2 (g)  C6H6 (l) 632 kJ/mole B. –632 kJ/mole C. –178 kJ/mole D. 178 kJ/mole Ans: Reverse eqn (1) and multiply by 3, 3 C2H2 (g)  3(2C) + 3 (H2) Ho = 3( -227 kJ/mole)……..(3) Blue are in common, so the resulting eqn is: 3 C2H2 (g)  C6H6 (l) Ho =(3)+(2) = -632 kj/mole

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

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

From the following heats of reaction,   2 C (graphite) + H2 (g)  C2H2 (g) Ho = 227 kJ/mole……. (1) 6 C (graphite) + 3 H2 (g)  C6H6 (l) Ho = 49 kJ/mole ……..(2) Calculate Ho for the reaction 3 C2H2 (g)  C6H6 (l) 632 kJ/mole B. –632 kJ/mole C. –178 kJ/mole D. 178 kJ/mole Ans: Reverse eqn (1) and multiply by 3, 3 C2H2 (g)  3(2C) + 3 (H2) Ho = 3( -227 kJ/mole)……..(3) Blue are in common, so the resulting eqn is: 3 C2H2 (g)  C6H6 (l) Ho =(3)+(2) = -632 kj/mole

The heat of combustion of fructose, C6H12O6, is –2812 kJ/mole The heat of combustion of fructose, C6H12O6, is –2812 kJ/mole. Using the following information, calculate H°f for fructose.   C6H12O6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (l) H = –2812 kJ H°f of CO2 = –393.5 kJ/mole H°f of H2O = –285.83 KJ/mole –210.3 kJ B. 210.3 kJ C. –1264 kJ D. 1264 kJ Ans: H = H°f of products - H°f of reactants H°f of elements is zero. -2812 kJ =[ 6 (-393.5kJ/mole) + 6 ( –285.83 KJ/mole)] - H°f for fructose H°f for fructose = [ 6 (-393.5kJ/mole) + 6 ( –285.83 KJ/mole)] + 2812kJ = -1264 kJ

Calculate the standard enthalpy change, Ho, for the following gas phase reaction using bond energy data.   Bond Dissociation Energies Ans: Bonds broken minus bonds formed:   ∆H = [ (1 mol)(614 kJ/mol) + (1 mol)(435 kJ/mol) ] – [ (1 mol)(348 kJ/mol) + (1 mol)(328 kJ/mol) + (1 mol)(413 kJ/mol) ] = 1049 kJ – 1098 kJ = –40 kJ

Home Work Problems worked out in class, Example problems in the chapter, practice exercise, Review of concepts, and additional problems given below

Problem # 1 (1)Hydrogen peroxide, decomposes as follows: H2O2 (l) → H2O (l) + ½ O2 (g); ∆H = -98.2 kJ How much heat is released when 1 g of H2O2 decomposes? What is ∆H when 1 g of O2 is formed by the decomposition of hydrogen peroxide?

Problem#2 In a coffee-cup calorimeter, the following acid-base reaction is carried out: H+ (aq) + OH- (aq) → H2O (l) When 0.010 mol of H+ reacts with 0.010 mol of OH- , the temperature of 110 g of water rises from 25.0 to 26.2 ⁰C. Calculate (a) Qwater, (b)∆H for the reaction , © ∆H if 1.00 mol of H+ reacts with 1.00 mol of OH-

Problem#3 A 1.000 g sample of rocket fuel hydrazine, N2 H4 , is burned in a bomb calorimeter containing 1200 g of water. The temperature rises from 24.62 to 28.16 ⁰C. Taking C for the bomb to be 840 J/ ⁰C , calculate Qreaction for the combustion of the one-gram sample. Qreaction for the combustion of the one mole of hydrazine in the bomb calorimeter.

Problem#4 How many kJ of heat must be removed from 1000 g of water (specific heat = 4.184 J /g oC) to lower the temperature from 18.0°C to 12.0°C?   A. 2.5 x 10–2 kJ B. 1.4 kJ C. 4.2 kJ D. 25 kJ

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