Chapter 5 Thermochemistry Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 5 Thermochemistry © 2009, Prentice-Hall, Inc.

Overview of Thermochemistry System vs. Surroundings Energy vs. Work Potential energy vs. Kinetic Energy Changes in internal energy  endothermic vs. exothermic changes © 2009, Prentice-Hall, Inc.

Energy and Chemical Reactions Energy changes are due to rearranging chemical bonds. Energy of chemical bonding = a form of chemical potential energy: breaking of existing bonds and forming of new bonds. Energy is always needed to break bonds. Energy release occurs when new bonds are formed. The balance between these two processes  overall reaction is endothermic or exothermic. © 2009, Prentice-Hall, Inc.

Units of Energy The SI unit of energy is the joule (J). kg m2 An older, non-SI unit is still in widespread use: the calorie (cal). 1 cal = 4.184 J 1 J = 1  kg m2 s2 © 2009, Prentice-Hall, Inc.

E, q, w, and Their Signs © 2009, Prentice-Hall, Inc.

Work w = F x d In chemical systems, only P/V work (contraction or expansion of a gas) is of significance, and only when there is an increase or a decrease in the amount of gas present. There must be a change in volume in order for the system to do work on the surroundings (or have work done on it by the surroundings). © 2009, Prentice-Hall, Inc.

State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. © 2009, Prentice-Hall, Inc.

State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached room temperature from either direction. © 2009, Prentice-Hall, Inc.

State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so, E depends only on Einitial and Efinal. © 2009, Prentice-Hall, Inc.

Enthalpy Enthalpy = the heat transferred between the system and the surroundings during a chemical reaction carried out under constant pressure = is the internal energy plus the product of pressure and volume: H = E + PV © 2009, Prentice-Hall, Inc.

Enthalpy When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) This can be written H = E + PV © 2009, Prentice-Hall, Inc.

Enthalpy Since E = q + w and w = -PV, we can substitute these into the enthalpy expression: H = E + PV H = (q+w) − w H = q So, at constant pressure, the change in enthalpy is the heat gained or lost. © 2009, Prentice-Hall, Inc.

Enthalpy Heat from surroundings  system endothermic, DH > 0 Heat from system  surroundings exothermic, DH < 0 Enthalpy is a state function – doesn’t matter how it got there. © 2009, Prentice-Hall, Inc.

Endothermicity and Exothermicity A process is endothermic when H is positive. © 2009, Prentice-Hall, Inc.

Endothermicity and Exothermicity A process is endothermic when H is positive. A process is exothermic when H is negative. © 2009, Prentice-Hall, Inc.

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). © 2009, Prentice-Hall, Inc.

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. © 2009, Prentice-Hall, Inc.

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic. © 2009, Prentice-Hall, Inc.

Enthalpy DH = change in enthalpy = the heat transferred between the system and surroundings during a chemical reaction carried out under constant pressure © 2009, Prentice-Hall, Inc.

Enthalpy Heat from surroundings  system endothermic, DH > 0 Heat from system  surroundings exothermic, DH < 0 Enthalpy is a state function – doesn’t matter how it got there. © 2009, Prentice-Hall, Inc.

Overview of Thermochemistry Our focus: Enthalpy is a state function. DH and stoichiometry Specific Heat q = c x m x DT (and related calculations, including calorimetry) Hess’s Law and its application, including enthalpies of formation Chapter 19: Entropy and Gibbs Free Energy (second semester) © 2009, Prentice-Hall, Inc.

Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants © 2009, Prentice-Hall, Inc.

Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. © 2009, Prentice-Hall, Inc.

The Truth about Enthalpy Enthalpy is an extensive property – the magnitude of enthalpy is directly proportional to the amount of reactant consumed. © 2009, Prentice-Hall, Inc.

The Truth about Enthalpy H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) DH = –890 kJ, But CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) DH = +890 kJ. © 2009, Prentice-Hall, Inc.

The Truth about Enthalpy H for a reaction depends on the state of the products and the state of the reactants. 2H2O(g)  2H2O(l) DH = –88 kJ © 2009, Prentice-Hall, Inc.

Sample Exercise 5.4 (p. 179) How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) DH = –890 kJ (-250 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.4 Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2 H2O2(l)  2 H2O(l) + O2(g) DH = -196 kJ Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure. (-14.4 kJ) © 2009, Prentice-Hall, Inc.

Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Heat Capacity and Specific Heat Molar heat capacity = the heat capacity of 1 mol of a substance Specific heat capacity (or simply specific heat) = the heat capacity of 1 g of a substance. © 2009, Prentice-Hall, Inc.

Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change c = q m  T © 2009, Prentice-Hall, Inc.

Another useful version of this equation: q = (specific heat) x (grams of substance) x DT = c x m x DT © 2009, Prentice-Hall, Inc.

Sample Problem 5.5 (p. 181) a) How much heat is needed to warm 250 g of water from 22oC to near its boiling point, 98oC? The specific heat of water is 4.18 J/g-K. (7.9 x 104 J) b) What is the molar heat capacity of water? (75.2 J/mol-K) © 2009, Prentice-Hall, Inc.

Practice Problem 5.5 a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0oC. (4.9 x 105 J) b) What temperature change would these rocks undergo if they emitted 450 kJ of heat? (11k = 11oC decrease) © 2009, Prentice-Hall, Inc.

Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. © 2009, Prentice-Hall, Inc.

Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: qsoln = m  c  T = -qrxn Note the reversal of the sign of q. © 2009, Prentice-Hall, Inc.

Constant Pressure Calorimetry For dilute aqueous solutions, the specific heat of the solution will be close to that of pure water. © 2009, Prentice-Hall, Inc.

Sample Exercise 5.6 (p.182) When a student mixes 50. mL of 1.0 M HCl and 50. mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0oC to 27.5oC. Calculate the enthalpy change for the reaction, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100. mL, that its density is 1.00 g/mL, and that its specific heat is 4.18 J/g-K. (-54 kJ/mol) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.6 When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30oC to 23.11oC. The temperature increase is caused by the following reaction: AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq) Calculate DH for this reaction, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g-oC. (-68 kJ/mol) © 2009, Prentice-Hall, Inc.

Bomb Calorimetry Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. © 2009, Prentice-Hall, Inc.

Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. For most reactions, the difference is very small. © 2009, Prentice-Hall, Inc.

Sample Exercise 5.7 (p. 184) Methylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. The combustion of methylhydrazine with oxygen produces N2(g), CO2(g), and H2O(l): 2 CH6N2(l) + 5 O2(g)  2 N2(g) + 2 CO2(g) + 6 H2O(l) When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00oC to 39.50oC. In a separate experiment the heat capacity of the bomb calorimeter is measured to be 7.794 kJ/oC. What is the heat of reaction for the combustion of a mole of CH6N2 in this calorimeter? (-1.30 x 103 kJ/mol CH6N2) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.7 A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/oC. The temperature increases from 23.10oC to 24.95oC. Calculate the heat of combustion of a) lactic acid per gram (-15.2 kJ/g) and b) per mole (-1370 kJ/mol) © 2009, Prentice-Hall, Inc.

Hess’s Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using published H values and the properties of enthalpy. © 2009, Prentice-Hall, Inc.

Hess’s Law Hess’s law states that “if a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” © 2009, Prentice-Hall, Inc.

Hess’s Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. © 2009, Prentice-Hall, Inc.

Hess’s Law Combustion of methane CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = –802 kJ 2H2O(g)  2H2O(l) DH = –88 kJ __________________________________________ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) DH = –890 kJ © 2009, Prentice-Hall, Inc.

Sample Exercise 5.8 (p. 186) The enthalpy of combustion of C to CO2 is -393.5 kJ/mol C, and the enthalpy of combustion of CO to CO2 is -283.0 kJ/mol CO: 1) C(s) + O2(g)  CO2(g) DH1 = -393.5 kJ 2) CO(g) + ½ O2(g)  CO2(g) DH2 = -283.0 kJ Using these data, calculate the enthalpy of combustion of C to CO: 3) C(s) + ½ O2(g)  CO(g) DH3 = ? (-110.5 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.8 HW Carbon occurs in two forms, graphite and diamond. The enthalpy of combustion of graphite is -393.4 kJ/mol and that of diamond is -395.4 kJ: C(graphite) + O2(g)  CO2(g) DH1 = -393.5 kJ C(diamond) + O2(g)  CO2(g) DH2 = -395.4 kJ Calculate DH for the conversion of graphite to diamond: C(graphite)  C(diamond) DH3 = ? (+1.9 kJ) © 2009, Prentice-Hall, Inc.

Sample Exercise 5.9 (p. 187) HW Calculate DH for the reaction 2 C(s) + H2(g)  C2H2(g) Given the following reactions and their respective enthalpy changes: C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l) DH = -1299.6 kJ C(s) + O2(g)  CO2(g) DH = -393.5 kJ H2(g) + ½ O2(g)  H2O(l) DH = -285.8 kJ (226.8 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.9 HW Calculate DH for the reaction NO(g) + O(g)  NO2(g) given the following reactions and their respective enthalpy changes: NO(g) + O3(g)  NO2(g) + O2(g) DH = -198.9 kJ O3(g)  3/2 O2(g) DH = -142.3 kJ O2(g)  2 O(g) DH = 495.0 kJ (-304.1 kJ) © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Standard Enthalpies of Formation Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). © 2009, Prentice-Hall, Inc.

Heats of Formation We can use the heats of formation to predict the DH of a particular reaction – another version of Hess’s Law © 2009, Prentice-Hall, Inc.

Sample Exercise 5.10 (p. 190) For which of the following reactions at 25oC would the enthalpy change represent a standard enthalpy of formation? For those where it does not, what changes would need to be made in the reaction conditions? a) 2 Na(s) + ½ O2(g)  Na2O(s) b) 2 K(l) + Cl2(g)  2 KCl(s) c) C6H12O6(s)  6 C(diamond) + 6 H2(g) + 3 O2(g) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.10 Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4). © 2009, Prentice-Hall, Inc.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) © 2009, Prentice-Hall, Inc.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) © 2009, Prentice-Hall, Inc.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) © 2009, Prentice-Hall, Inc.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) The sum of these equations is: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) © 2009, Prentice-Hall, Inc.

Calculation of H We can use Hess’s law in this way: H = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients. © 2009, Prentice-Hall, Inc.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ © 2009, Prentice-Hall, Inc.

Sample Exercise 5.11 (p. 192) a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l). (-3267 kJ) b) Compare the quantity of heat produced by combustion of 1.00 g propane (C3H8) to that produced by 1.00 g benzene. (C3H8(g): -50.3 kJ/g; C6H6(l): -41.8 kJ/g) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.11 HW Using the standard enthalpies of formation listed in Table 5.3 or Appendix C, calculate the enthalpy change for the combustion of 1 mol of ethanol: C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l) (-1367kJ) © 2009, Prentice-Hall, Inc.

Sample Exercise 5.12 (p. 192) The standard enthalpy change for the reaction CaCO3(s)  CaO(s) + CO2(g) is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) and CO2(g) given in Table 5.3, calculate the standard enthalpy of formation of CaCO3(s). (-1207.1 kJ/mol) © 2009, Prentice-Hall, Inc.

Practice Exercise 5.12 Given the following standard enthalpy of reaction, use the standard enthalpies of formation in Table 5.3/Appendix C to calculate the standard enthalpy of formation of CuO(s): CuO(s) + H2(g)  Cu(s) + H2O(l) DHo = -129.7 kJ (-156.1 kJ/mol) © 2009, Prentice-Hall, Inc.

Sample Integrative Exercise (p. 186) Trinitroglycerin, C3H5N3O9, (usually referred to simply as nitroglycerin) has been widely used as an explosive. Alfred Nobel used it to make dynamite in 1866. Rather surprisingly, it also is used as a medication, to relieve angina (chest pains resulting from partially blocked arteries to the heart), by dilating the blood vessels. The enthalpy of decomposition at 1 atmosphere pressure of trinitroglycerin to form nitrogen gas, carbon dioxide gas, liquid water, and oxygen gas at 25oC is -1541.4 kJ/mol. © 2009, Prentice-Hall, Inc.

Sample Integrative Exercise (p. 186) a) Write a balanced chemical equation for the decomposition of trinitroglycerin. b) Calculate the standard heat of formation of trinitroglycerin. c) A standard dose of trinitroglycerin for relief of angina is 0.60 mg. Assuming that the sample is eventually completely combusted in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released? © 2009, Prentice-Hall, Inc.

Sample Integrative Exercise (p. 186) d) One common form of trinitroglycerin melts at about 3oC. From this information and the formula for the substance, would you expect it to be a molecular or ionic compound? Explain. e) Describe the various conversions of forms of energy when trinitroglycerin is used as an explosive to break rockfaces in highway construction. © 2009, Prentice-Hall, Inc.