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Chapter 6 Thermochemistry

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1 Chapter 6 Thermochemistry
© 2012 Pearson Education, Inc.

2 © 2012 Pearson Education, Inc.
Energy Energy is the ability to do work or transfer heat. Energy used to cause an object that has mass to move is called work. w = F  d where w is work, F is the force, and d is the distance over which the force is exerted. Energy used to cause the temperature of an object to rise is called heat.(Heat flows from warmer objects to cooler objects) © 2012 Pearson Education, Inc.

3 Potential and Kinetic Energy
Potential energy is energy an object possesses by virtue of its position or chemical composition. Kinetic energy is energy an object possesses by virtue of its motion: © 2012 Pearson Education, Inc.

4 © 2012 Pearson Education, Inc.
Units of Energy The SI unit of energy is the joule (J): An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = J 1 J = 1  kg m2 s2 © 2012 Pearson Education, Inc.

5 Definitions: System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston). © 2012 Pearson Education, Inc.

6 First Law of Thermodynamics
Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. © 2012 Pearson Education, Inc.

7 © 2012 Pearson Education, Inc.
Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal − Einitial © 2012 Pearson Education, Inc.

8 DE = Efinal - Einitial = Eproducts - Ereactants
Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. DE = Efinal - Einitial = Eproducts - Ereactants

9 Changes in Internal Energy
If E < 0, Efinal < Einitial Therefore, the system released energy to the surroundings. This energy change is called exergonic. The reaction is exothermic If E > 0, Efinal > Einitial Therefore, the system absorbed energy from the surroundings. This energy change is called endergonic. The reaction is endothermic © 2012 Pearson Education, Inc.

10 Two forms of energy transfer
Heat (q): Energy transferred between a system and its surroundings as a result of a difference in their temperatures only. Work (w): Energy transferred when a force acts on an object. Total change in a system’s internal energy ∆E = q + w

11 © 2012 Pearson Education, Inc.
E, q, w, and Their Signs © 2012 Pearson Education, Inc.

12 Determining the Change in Internal Energy of a System
PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal. PLAN: Define system and surroundings, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = J w = J DE = q + w = -325 J + (-451 J) = -776 J -776J 103J kJ -0.776kJ 4.18kJ kcal = kJ = kcal

13 The Meaning of Enthalpy
Enthalpy is a thermodynamic function of a system, equivalent to the sum of the internal energy of the system plus the product of its volume multiplied by the pressure exerted on it by its surroundings. H = E + PV where H is enthalpy (greek: to warm) we cannot know the exact enthalpy of the reactants and products, we then measure H DH = DE + PDV qp = DE + PDV = DH at constant pressure

14 DHrxn = Hfinal - Hinitial = Hproducts - Hreactants
Heat of Reaction DHrxn = Hfinal - Hinitial = Hproducts - Hreactants Exothermic process Hfinal < Hinitial so ∆H < 0 Heat is released Decrease in enthalpy Endothermic process Hfinal > Hinitial so ∆H > 0 Heat is absorbed Increase in enthalpy

15 Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g) CH4 + 2O2 H2O(g) Hfinal Enthalpy, H Hinitial Enthalpy, H heat out heat in DH < 0 DH > 0 CO2 + 2H2O H2O(l) Hfinal Hinitial A Exothermic process B Endothermic process

16 Sample Problem Drawing Enthalpy Diagrams and Determining the Sign of DH PROBLEM: In each of the following cases, determine the sign of DH, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. (a) H2(g) + 1/2O2(g) H2O(l) kJ (b) 40.7kJ + H2O(l) H2O(g) PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction SOLUTION: (a) The reaction is exothermic. (b) The reaction is endothermic. H2(g) + 1/2O2(g) (reactants) (products) H2O(g) EXOTHERMIC DH = kJ ENDOTHERMIC DH = +40.7kJ H2O(l) (products) (reactants) H2O(l)

17 © 2012 Pearson Education, Inc.
Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. © 2012 Pearson Education, Inc.

18 Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity. We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C). © 2012 Pearson Education, Inc.

19 Heat Capacity and Specific Heat
Specific heat, then, is Specific heat = heat transferred mass  temperature change Cs = q m  T © 2012 Pearson Education, Inc.

20 The Practice of Calorimetry At constant Pressure
Heat lost by the system(sample) is equal in magnitude but opposite in sign to the heat gained by the surroundings(H2O) -qsample = qH2O Substitute the equation q = Cs x mass x ∆T -(Cssample x masssample x ∆Tsample) = (CsH20 x massH2O x ∆TH2O) Cssample = CsH20 x massH2O x ∆TH2O masssample x ∆Tsample -

21 Sample Problem Finding the Quantity of Heat from Specific Heat Capacity PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 250C to 300.0C? The specific heat capacity (c) of Cu is J/g*K. PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x DT to find the answer. DT in 0C is the same as for K. SOLUTION: 0.387 J g*K q = x 125 g x (300-25)0C = 1.33x104 J

22 The Practice of Calorimetry At constant Volume
Called bomb calorimetry Compound reacts completely with O2 example of a combustion reaction Withstands high pressures q rxn = -Ccal x ∆T

23 Sample Problem Calculating the Heat of Combustion PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2(the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases C. Is the manufacturer’s claim correct? PLAN: - q sample = qcalorimeter SOLUTION: qcalorimeter = heat capacity x DT = kJ/K x K = kJ 40.24 kJ kcal 4.18 kJ = 9.63 kcal or Calories The manufacturer’s claim is true.

24 Molar Heat Capacity (Cm)
The quantity of heat required to change the temperature of 1 mole of a substance by 1 K Molar heat capacity (Cm) = q/(moles x ∆T) [ units of J/mol·K] Liquid H2O has a Cm of J/g·K So Cm of H2O(l) = J/g·K x g/1 mole = J/mol·K

25 Sample Problem Determining the Heat of a Reaction PROBLEM: You place 50.0 mL of M NaOH in a coffee-cup calorimeter at C and carefully add 25.0 mL of M HCl, also at C. After stirring, the final temperature is C. Calculate qsoln (in J) and DHrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and Cs = 4.18 J/g*K)

26 HCl is the limiting reactant.
Sample Problem Determining the Heat of a Reaction continued SOLUTION: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l) For NaOH 0.500 M x L = mol NaOH For HCl 0.500 M x L = mol HCl HCl is the limiting reactant. mol of H2O will form during the rxn. total volume after mixing = L L x 103 mL/L x 1.00 g/mL = 75.0 g of water q = mass x specific heat x DT = 75.0 g x 4.18 J/g*0C x ( ) = 693 J (-693 J/ mol H2O)(kJ/103 J) = kJ/ mol H2O formed

27 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 = kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) H2O (l) DH = 2 x 6.01 = 12.0 kJ 6.4

28 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 = kJ 1 mol P4 123.9 g P4 x -3013 kJ 1 mol P4 x 266 g P4 = kJ 6.4

29 Hess’s Law of Heat Summation
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. Many reactions cannot be carried out separately They are often occur as a series of reactions The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.

30 Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown DH PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) CO2(g) + 1/2N2(g) DH = ? Given the following information, calculate the unknown DH: Equation A: CO(g) + 1/2O2(g) CO2(g) DHA = kJ Equation B: N2(g) + O2(g) NO(g) DHB = kJ SOLUTION: Multiply Equation B by 1/2 and reverse it. CO(g) + 1/2O2(g) CO2(g) DHA = kJ NO(g) /2N2(g) + 1/2O2(g) DHB = kJ CO(g) + NO(g) CO2(g) + 1/2N2(g) DHrxn = kJ

31 Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure) page 184 or table 5.3 The standard enthalpy of formation of any element in its most stable form is zero. DH0 (C, graphite) = 0 f DH0 (O2) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f

32 ∆H0f Standard Enthalpies found on page 184 or Table 5.3.
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) - (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) ∆H0f Standard Enthalpies found on page 184 or Table 5.3.

33 © 2012 Pearson Education, Inc.
Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(− kJ) + 5(0 kJ)] = [(− kJ) + (− kJ)] – [(− kJ) + (0 kJ)] = (− kJ) – (− kJ) = − kJ © 2012 Pearson Education, Inc.

34 Groups for your challenge question


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