SGCh06 - Chapter Six THERMOCHEMISTRY • • • • •...

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Unformatted text preview: Chapter Six THERMOCHEMISTRY • • • • • • Energy Changes in Chemical Reactions Enthalpy Calorimetry Standard Enthalpies of Formation and Reaction Heats of Solution and Dilution The First Law of Thermodynamics ENERGY CHANGES IN CHEMICAL REACTIONS STUDY OBJECTIVES 1. 2. Define heat, work, and energy. Distinguish between exothermic and endothermic reactions. Thermochemistry. Chemical reactions occur with either the absorption or release of energy. When a system does work or releases heat, its energy decreases. In this chapter we are concerned with the heat absorbed or released in chemical reactions. This topic is called t hermochemistry . Energy can be assigned to one of two classes: kinetic or potential. K inetic energy is energy that an object has because of its motion. T hermal energy comes under this class because it is the energy associated with the random motion of atoms and molecules. According to the kinetic-molecular theory, all matter has thermal energy , because atoms and molecules are in constant motion. They vibrate, rotate, and movefrom one point to another. The more energy of motion the higher the temperature. P otential energy is energy that results from an object's position. Potential energy is stored energy, and can only be released when some process occurs. Chemical energy is stored within molecules as a consequence of attractions between electrons and atomic nuclei. When substances undergo chemical reactions, chemical energy is released. Endothermic and Exothermic Reactions. In the study of thermochemistry, certain terms are defined very carefully and are used in a precise way. A thermodynamic system is the part of the universe that is selected for study. It may be a beaker containing a mixture of reactants, a bacteria colony, or the entire earth. The boundary of the system must be carefully defined. The surroundings are the part of the universe that interacts with the system. Systems can interact with the surroundings by the exchange of heat, work, and matter. Heat is energy that is transferred because of a temperature difference between the system and the surroundings. Heat always flows from a hotter to a colder body. Work is energy lost or gained by a system by mechanical means, rather than by heat conduction. Systems do not "contain" work or heat. What they do contain is energy. The SI unit for amount of energy is the joule (J). Since both work and heat are energy in transition, they also have units of joules. A non-SI unit for heat is the calorie. One calorie (cal) is the amount of heat required to raise the temperature of 1 gram of water from 14.5°C to 15.5°C. Heat or energy in calories can be converted to joules: 1 cal = 4.184 J and 1 kcal = 4.184 kJ 1 05 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 06 / Thermochemistry Chemical reactions that evolve heat to the surroundings are called e xothermic reactions . Combustion reactions involving wood, coal, and hydrogen are examples of exothermic reactions. During an exothermic reaction the surroundings warm up. When heat is absorbed from the surroundings by a reaction system, the reaction is referred to as e ndothermic . The melting of ice and the boiling of water are endothermic processes. During an endothermic reaction the surroundings cool down. The heat evolved during a chemical reaction results from the decrease in chemical potential energy as reactants are converted to products. In an exothermic reaction the total chemical energy of the products is less than the total chemical energy of the reactants [Figure 6.3(a) in the text]. In an endothermic reaction the products have greater chemical energy than the reactants [Figure 6.3(b) in the text]. Here, the heat absorbed brings about an increase in the potential energy of the system. The difference in energy between the reactants and products is equal to the heat absorbed by the system. ENTHALPY STUDY OBJECTIVES 1. 2. Use the thermochemical equation to calculate the heat evolved or absorbed when a given amount of reactant is converted to products. Interpret a thermochemical equation in terms of sign conventions for ∆ H and amounts of reactants consumed. Enthalpy Changes. Most reactions that you carry out in the general chemistry laboratory occur at constant pressure because the atmospheric pressure does not change appreciably in three or four hours. The heat evolved or absorbed by a reaction carried out at a constant pressure is equal to a quantity called the enthalpy change of the system, also called the e nthalpy of reaction , ∆ H. ∆ H = heat transferred to or from a system at constant pressure The enthalpy of reaction is the difference between the enthalpies of the products and the enthalpies of the reactants. ∆ H = enthalpy of the system after the reaction – enthalpy of the system before the reaction ∆ H = H(products) – H(reactants) Changes in enthalpy can be positive or negative. For an endothermic reaction H(products) > H(reactants) and ∆ H is positive. The heat that is added is converted into potential energy of the system. For an exothermic reaction, H(reactants) > H(products) and ∆ H is negative. Enthalpy and chemical reactions: exothermic endothermic H(reactants) > H(products) H(products) > H(reactants) ∆H < 0 ∆H > 0 ∆ H is negative ∆ H is positive A t hermochemical equation is a balanced chemical equation with a term added to show the enthalpy change, ∆ H. Two examples are CaCO3 (s) → CaO(s) + CO2 (g) 2SO2 (g) + O 2 (g) → 2SO 3 (g) Back Forward Main Menu TOC ∆ H° = 178 kJ ∆ H° = –198 kJ Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 07 Summary of Thermochemistry Principles There are several points to keep in mind when interpreting thermochemical equations. 1. The enthalpy change in kilojoules applies to the reaction of the amount of substance shown in the balanced equation. As shown in the above equation, when 2 mol SO2 react with 1 mol O2 , 198 kJ of heat will be released. The value of ∆ H must always relate to the way the equation is balanced. If the equation was divided by 2 and written: 1 SO2 (g) + 2 O2 (g) → SO 3 (g) ∆ H° = –99 kJ then the enthalpy change would be divide by 2 also. Remember that the heat evolved or absorbed by a chemical reaction is an extensive property: that is, it depends on the amount of reactants. 2. If we write the reverse of a chemical reaction, the magnitude of ∆ H is the same, but its sign changes. The following reaction absorbs heat: CaCO3 (s) → CaO(s) + CO2 (g) ∆ H° = 178 kJ But the reverse reaction must release heat: CaO(s) + CO 2 (g) → CaCO 3 (s) 3. ∆ H° = –178 kJ We must always specify the physical states of all reactants and products, because they help determine the magnitude of the enthalpy change. In the combustion of methane for instance, water is a product. The enthalpy change is different when water is a gas than when water is a liquid. CH4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) CH4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ∆ H° = –802 kJ ∆ H° = –890 kJ _______________________________________________________________________________ EXAMPLE 6.1 Thermochemical Equations The thermochemical equation for the combustion of propane is: C 3 H8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O a. b. o ∆ Hrxn = –2220 kJ How many kJ of heat are released when 0.50 mole of propane reacts? How much heat is released when 88.2 g of propane reacts? •Method of Solution The heat released by an exothermic reaction is an extensive property. Therefore q depends on the amount of propane consumed. The equation indicates that 2220 kJ of heat is evolved per mole of propane burned. The conversion factor is: –2220 kJ 1 mol C 3 H8 • Calculation for (a) q = mol C 3 H8 × kJ mol C 3 H8 q = 0.50 mol C3 H8 × – 2220 kJ 1 mol C 3 H8 = –1110 kJ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 08 / Thermochemistry •Calculation for (b) Since we know the heat of reaction per mole, we should first convert the number of grams of C3 H8 to moles. 1 m ol C 3 H8 – 2220 kJ q = 88.2 g C3 H8 × × 44.1 g C 3 H8 1 mol C 3 H8 = –4440 kJ _______________________________________________________________________________ EXERCISES 1. Given the thermochemical equation: 1 SO2 (g) + 2 O 2 (g) → SO 3 (g) o ∆ Hrxn = – 99.0 kJ how much heat is evolved when 75 g SO2 undergoes combustion? 2. How many grams of SO 2 must be burned to yield 251 kJ of heat? Given: 1 SO2 (g) + 2 O 2 (g) → SO 3 (g) ∆ H° = –99.0 kJ 3. The reaction that occurs when a typical fat, glyceryl trioleate, is metabolized by the body is C 57 H104 O6 (s) + 80O 2 (g) → 57CO 2 (g) + 52H 2 O(l) o ∆ Hrxn = –3.35 × 10 4 kJ. How much heat is evolved when 1 gram of this fat is completely oxidized? 4. 3 Given 2Al(s) + 2 O 2 (g) → Al 2 O3 (s) 2Al2 O3 (s) → 4Al(s) + 3O2 (g) ∆ H° = –1670 kJ. What is ∆ H for the reaction? ∆ H° = ? CALORIMETRY STUDY OBJECTIVES 1. 2. Calculate the heat evolved or absorbed when an object or substance undergoes a known temperature change. Determine the enthalpy of reaction using data from both constant pressure and constant volume calorimeter experiments. Specific Heat. The device used to measure the amount of heat evolved or absorbed in a reaction is called a calorimeter. All calorimetry is based on the temperature rise observed in some medium, usually water. The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by 1.0°C. A related term is the h eat capacity (C), which is the amount of heat required to raise the temperature of a given mass of substance by one °C. The amount of heat flowing in or out of a substance is q = C ∆t Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 09 ∆ t is the temperature change given by tf – t i , where tf is the final temperature and ti is the initial temperature. q = C (tf – t i ) In terms of the specific heat (s) q = ms ∆ t where m is the mass in grams of the object absorbing the heat and C = ms. With this equation we can calculate the amount of heat absorbed by a given amount of water or by an object of given mass. See Example 6.2. Constant-Volume Calorimetry. Bomb calorimeters, as shown in Figure 6.6 in the textbook, are used to measure the heat evolved in combustion reactions of foods and fuels. The stainless steel "bomb" is loaded with a small amount of substance and O 2 at 30 atm of pressure, and is immersed in a known amount of water. The heat evolved during combustion is absorbed by the water and the calorimeter assembly (bomb and contents, can, thermometer, etc.). Measurements of ∆ t of the water and calorimeter are related to the heat evolved by the reaction q and to ∆ H. We saw previously that the enthalpy change of a system is equal to the heat evolved at constant pressure, q. ∆H = q Reactions in a "bomb" occur under constant volume conditions rather than constant pressure conditions, and so the heat evolved is not quite equal to ∆ H: ∆H ≅ q However, for most reactions the difference is small and for our purpose can be neglected. For instance, for the combustion of 1 mole of pentane, the difference is only 7 kJ out of 3500 kJ. Therefore, we can often use q ≅ ∆H without significant error. To measure q, the heat evolved by the reaction, first measure the heat absorbed by both the bomb calorimeter and the surrounding water. Let qrxn be the amount of heat from the reaction. q of reaction = –(q of bomb calorimeter + q of water) qrxn = –(qbomb + q water) Where qbomb = C bomb ∆ t and qwater = m waters ∆ t Here Cbomb is the heat capacity of the bomb and s is the specific heat of water. The heat capacity of both the bomb calorimeter and the surrounding water must be known. Substituting gives: qrxn = –[C bomb ∆ t + mwaters ∆ t] Examples 6.3 and 6.4 illustrate the use of a bomb calorimeter. Constant-Pressure Calorimeter. A constant-pressure calorimeter is open to the atmosphere, and the measurement of heat given off by a reaction is a direct measure of the enthalpy change. q = ∆H Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 10 / Thermochemistry Such a device is shown if Figure 6.7 (text). A reaction involving solutes occurs in the solution and liberates heat that is absorbed by the solution (which is mostly water), the stirrer, the thermometer, and the container walls. These later three parts are considered the calorimeter, and so q = –(qcalorimeter + q water) _______________________________________________________________________________ EXAMPLE 6.2 Specific Heat a. b. What is the specific heat of lead if the temperature of a 425 g block increases 2.31°C when it absorbs 492 J of heat? What is the heat capacity of the block of lead? •Method of Solution In terms of the specific heat (s), the amount of heat absorbed when an object of mass m is heated from T i to Tf is: q = ms (Tf – T i ) Substituting into the equation: 492 J = (425 g) s (2.31°C) Rearranging to solve for the specific heat s: s= 492 J = 0.501 J/g·°C (425 g) (2.31°C) The heat capacitry of the 425 g block of lead is the mass of the block times the specific heat: C = ms C = 425 g × 0.501 J/g·°C = 213 J/°C _______________________________________________________________________________ EXAMPLE 6.3 Determining The Calorimeter Constant The combustion of benzoic acid is often used as a standard source of heat for calibrating combustion bomb calorimeters. The heat of combustion of benzoic acid has been accurately determined to be 26.42 kJ/g. When 0.8000 g of benzoic acid was burned in a calorimeter containing 950 g of water, a temperature rise of 4.08°C was observed. What is the heat capacity of the bomb calorimeter (the calorimeter constant)? •Method of Solution The combustion of 0.8000 g of benzoic acid produces a known amount of heat. qrxn = 0.8000 g × – 26.42 kJ = –21.14 kJ = –21.14 × 10 3 J 1g And since ∆ t and the amount of water are known, Cbomb , the heat capacity of the calorimeter, can be calculated. The total heat of combustion is absorbed by the bomb calorimeter and water. qrxn = –(qbomb + q water) qrxn = –(C bomb ∆ t + ms ∆ t) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 11 The heat absorbed by the water was qwater = ms ∆ t, where m = 950 g water s = 4.184 J/g·°C ∆ t = 4.08°C Substitution yields: qwater = 950 g (4.184 J/g·°C)(4.08°C) = 16.20 × 10 3 J The bomb calorimeter must have absorbed the difference between the 21.14 × 10 3 J released, and the 16.20 × 10 3 J absorbed by the water. qrxn = –(qbomb + q water) qbomb = –qrxn – qwater = –(–21.14 × 10 3 J) – 16.20 × 10 3 J qbomb = 4.94 × 10 3 J The heat capacity of the calorimeter is C bomb = qbomb 4 .94 × 1 0 3 J = ∆t 4.08°C C bomb = 1210 J/°C _______________________________________________________________________________ EXAMPLE 6.4 Determining the Heat of Combustion The thermochemical equation for the combustion of pentane (C5 H12 ) is C 5 H12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) ∆ H° = –3509 kJ From the following information calculate the heat of combustion, ∆ H°, and compare your result to the value given in the above equation. 0.5521 g of C5 H12 was burned in the presence of excess O2 in a bomb calorimeter. The heat capacity of the calorimeter was 1.800 × 10 3 J/°C, and the temperature of the calorimeter and 1.000 × 10 3 g of water rose from 21.22°C to 25.70°C. •Method of Solution The heat evolved by the combustion reaction is absorbed by the water and the calorimeter assembly. qrxn = –[qbomb + q water] where qrxn = –(C bomb ∆ t + ms ∆ t) •Calculation ∆ t = 25.70°C – 21.22°C = 4.48°C Substituting: 1 .800 × 1 0 3 J × 4.48°C = 8.06 × 10 3 J °C 4 .184 J qwater = 1.000 × 10 3 g × × 4.48°C = 1.87 × 10 3 J g. °C qbomb = – Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 12 / Thermochemistry qrxn = –[8,060 J + 18,700 J] qrxn = –26,800 J = – 26.8 kJ (three significant figures) ∆ H° refers to the reaction of 1 mole of pentane. The 26,800 J was evolved by 0.5521 g C5 H12 . The molar mass of pentane is 72.15 g. ∆ H° = q 7 2.15 g C 5 H12 –26.8 kJ = × 0.5521 g 1 mol ∆ H° = –3500 kJ/mol = –3.50 × 10 3 kJ/mol •Comment To three significant figures the result compares well with the value given in the thermochemical equation. Small differences will arise because the heat evolved at constant volume is not quite the same as the heat evolved at constant pressure (which is equal to ∆ E). Only the heat evolved at constant pressure exactly equals ∆ H, the enthalpy change. _______________________________________________________________________________ EXERCISES 5. A piece of iron initially at a temperature of 25.2°C absorbs 16.9 kJ of heat. If its mass is 82 g, calculate the final temperature. Given the specific heat of iron is 0.444 J/g·°C. 6. How much heat is absorbed by 52 g of iron when its temperature is raised from 25°C to 275°C? The specific heat of iron is 0.444 J/g·°C. 7. A 25.0 g piece of aluminum at 4.0°C is dropped into a beaker of water. The temperature of water drops from 75.0°C to 55.0°C. What amount of heat did the aluminum absorb? The specific heat of Al is 0.902 J/g·°C. 8. Benzoic acid (C6 H5 CO2 H) was used to calibrate a bomb calorimeter. Its enthalpy of combustion is accurately known to be –3226.7 kJ/mol. When 1.0236 g of benzoic acid was burned in a bomb calorimeter, the temperature of the calorimeter and the 1.000 × 10 3 g of water surrounding it rose from 20.66°C to 24.47°C. What is the heat capacity of the calorimeter? 9. To determine the heat capacity of a bomb calorimeter, a student added 150 g of water at 50.0°C to the bomb. The bomb initially was at 20.0°C. The final temperature of the water and the bomb was 32.0°C. What is the heat capacity of the bomb in J/°C? 10. Octane, C8 H18 , a constituent of gasoline burns according to the equation: 1 C 8 H18 (l) + 12 2 O 2 (g) → 8CO 2 (g) + 9H 2 O(l) 0.1111 g of C8 H18 was burned in the presence of excess O2 in a bomb calorimeter. The heat capacity of the calorimeter was 1.726 × 10 3 J/°C. The temperature of the calorimeter and 1.200 × 10 3 g of water rose from 21.22°C to 23.05°C. Calculate the heat of combustion per gram of octane. 11. Magnesium metal reacts with hydrochloric acid according to the following equation. Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (g) When 0.425 g of magnesium was added to 150.0 mL of 1.00 M HCl in a coffee-cup calorimeter the temperature of the solution increased from 24.5°C to 35.3°C. Given that the heat capacity of the calorimeter is 125 J/°C and that the density of the HCl solution is 1.00 g/mL, calculate the heat released per mole of magnesium. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 13 STANDARD ENTHALPIES OF FORMATION AND REACTION STUDY OBJECTIVES 1. 2. 3. Define standard enthalpy of formation. o Calculate ∆ Hrxn for any reaction with the aid of a table of enthalpies of formation. Use Hess's law to calculate ∆ H° for a reaction that is the sum of several given chemical reactions. Formation Reactions. In a formation reaction 1 mole of a compound is formed from the elements in their stable forms. For methane the formation reaction is: C(s) + 2H 2 (g) → CH 4 (g) o The s tandard enthalpy of formation ∆ Hf is the heat released or absorbed when 1 mole of a compound is formed from the elements in their standard states at a pressure of 1 atm. The enthalpy of formation of methane is –74.7 kJ. This is the enthalpy change for the reaction of carbon as graphite and hydrogen gas at 1 atm pressure that forms methane gas at 1 atm pressure at 25°C. In some cases the enthalpies of formation can be measured directly by calorimetry. For example, both H2 O and P4 O10 can be formed in a calorimeter. 1 H2 (g) + 2 O2 (g) → H2 O(g) P 4 (white) + 5O2 (g) → P 4 O10 (s) o ∆ Hf = ∆ Hcomb o ∆ Hf = ∆ Hcomb The Indirect Method. In many cases, enthalpies of formation cannot be determined by a direct calorimetry experiment. For example, the formation reaction of hydrogen sulfide (H2 S) will not go cleanly in a calorimeter, and many other compounds besides H2 S are formed when sulfur and hydrogen react. In cases like o these, a calculational method can used to determine ∆ Hf . S(s) + H2 (g) → H2 S(g) o ∆ Hf = ? According to H ess's law of heat summation if a reaction is the sum of several reaction steps, then the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes of the intermediate steps. For example, let's say that we want to determine the enthalpy change for the above reaction. The reaction enthalpies of the following reactions involving sulfur and hydrogen are known from combustion bomb calorimetry: o S(s) + O2 (g) → SO 2 (g) ∆ Hrxn = –296.1 kJ H2 (g) + 2 O2 (g) → H2 O(l) 3 H2 S(g) + 2 O2 (g) → SO 2 (g) + H 2 O(l) ∆ Hrxn = –285.8 kJ 1 Back Forward Main Menu TOC o o ∆ Hrxn = –561.7 kJ Study Guide TOC Textbook Website MHHE Website 1 14 / Thermochemistry These equations can be arranged so that their sum is the formation reaction of hydrogen sulfide. S(s) + O2 (g) → SO 2 (g) The first one is unchanged: The second one is unchanged: The third one is reversed: 1 H2 (g) + 2 O 2 (g) → H2 O(l) 3 SO 2 (g) + H 2 O(l) → H2 S(g) + 2 O 2 (g) S(s) + H2 (g) → H2 S(g) Their sum is the overall equation: We must reverse the sign of the enthalpy change of the third reaction before summing the enthalpy changes. o S(s) + O2 (g) → SO 2 (g) ∆ Hrxn = –296.1 kJ 1 o H2 (g) + 2 O 2 (g) → H2 O(l) ∆ Hrxn = –285.8 kJ o 3 SO2 (g) + H 2 O(l) → H2 S(g) + 2 O2 (g) ∆ Hrxn = 561.7 kJ ____________________________________________________ o S(s) + H2 (g) → H2 S(g) ∆ Hf = –20.2 kJ Summation of the three reaction steps yields the net reaction, and summation of the ∆ H° values yields the enthalpy change oof the net reaction. Example 6.5 in the text provides another example of the indirect method. The value of ∆ Hf for a compound depends on the temperature, pressure, and physical state of the compound. Standard enthalpies of formation at 25°C are tabulated in Appendix 3 of the text. o Using Enthalpies of Formation. Rather than measuring a standard enthalpy change ∆ Hrxn for every chemical reaction, chemists have devised a way to use the enthalpy of formation values of compounds involved in a reaction. Enthalpies of formation can be used to calculate the standard enthalpy change for any reaction for o which ∆ Hf is known for all reactants and products. The relationship is o o o ∆ Hrxn = ∑ n ∆ Hf (products) – ∑ m ∆ Hf (reactants) where ∑ (sigma) means "the sum of" and n and m are the stoichiometric coefficients. So here you simply add o o together all the ∆ Hf of the products, and add together all the ∆ Hf of the reactants, and subtract the reactant sum from the product sum. Example 6.5 illustrates the use of this equation and shows how the stoichiometric coefficients are entered into the equation. _______________________________________________________________________________ EXAMPLE 6.5 Using Enthalpies of Formation o Using ∆ Hf values in Appendix 2 of the textbook calculate the standard enthalpy change for the incomplete combustion of ethane (C 2 H6 ). 5 C 2 H6 (g) + 2 O 2 (g) → 2CO(g) + 3H2 O(l) •Method of Solution The enthalpy change for this chemical reaction in terms of enthalpies of formation is: o o o ∆ Hrxn = ∑ n ∆ Hf (products) – ∑ m ∆ Hf (reactants) o o o o o ∆ Hrxn = [2 mol × ∆ Hf (CO) + 3 mol × ∆ Hf (H2 O) ] – [1 mol × ∆ Hf (C2 H6 ) – 5/2 mol × ∆ Hf (O2 ) ] To avoid cumbersome notation, the physical states of the reactants and products were omitted from this o equation. Be careful to obtain from Appendix 2 the appropriate value of ∆ Hf . Note that the coefficients from Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 15 o the chemical equation are equal to the number of moles of each substance and that all the terms involving ∆ Hf of a reactant are preceded by a negative sign. Next, look up and insert values. •Calculation o ∆ Hrxn = [2 mol(–110.5 kJ/mol) + 3 mol(–285.8 kJ/mol)] – [1 mol(–84.68 kJ) + 5/2(0)] = (–221.0 kJ) + (–857.4 kJ) – (–84.68 kJ) = –221.0 kJ – 857.4 kJ + 84.68 kJ = –993.7 kJ _______________________________________________________________________________ EXAMPLE 6.6 Hess's Law The standard enthalpy change for the combustion of 1 mole of ethanol is: C 2 H5 OH(l) + 3O2 (g) → 2CO 2 (g) + 3H 2 O(l) o ∆ Hrxn = –1367 kJ o What is ∆ Hrxn for the following reaction in which H2 O is formed as a gas, rather than as a liquid? C 2 H5 OH(l) + 3O2 (g) → 2CO 2 (g) + 3H 2 O(g) Given the heat of vaporization of water H2 O(l) → H2 O(g) ∆ Hvap = 44 kJ •Method of Solution We can imagine a two-step path for this reaction. In the first step, ethanol undergoes combustion to form liquid H2 O, followed by the second step in which H2 O(l) is vaporized. The sum of the two steps gives the desired overall reaction, and according to Hess's law the sum of the two ∆ H's gives the overall ∆ H. •Calculation o C 2 H5 OH(l) + 3O2 (g) → 2CO 2 (g) + 3H 2 O(l) ∆ Hrxn = –1367 kJ o 3H 2 O(l) → 3H2 O(g) ∆ Hrxn = 132 kJ ______________________________________________________ o C 2 H5 OH(l) + 3O2 (g) → 2CO 2 (g) + 3H 2 O(g) ∆ Hrxn = –1235 kJ _______________________________________________________________________________ EXERCISES 12. Using data from Appendix 2 in the text, calculate the standard enthalpy of formation of octane, C8 H18 , given the following combustion reaction. 2C 8 H18 (l) + 25O 2 (g) → 16 CO2 (g) + 18 H 2 O(l) o ∆ Hrxn = –11,020 kJ 13. From the following heats of combustion with fluorine, calculate the enthalpy of formation of CH4 . CH4 (g) + 4F2 (g) → C F 4 (g) + 4HF(g) C(graphite) + F 2 (g) → C F 4 (g) H2 (g) + F2 (g) → 2HF(g) Back Forward Main Menu TOC o ∆ Hrxn = –1942 kJ o ∆ Hrxn = –933 kJ o ∆ Hrxn = –542 kJ Study Guide TOC Textbook Website MHHE Website 1 16 / Thermochemistry 14. A reaction used for rocket engines is N2 H4 (l) + 2H 2 O2 (l) → N2 (g) + 4H 2 O(l) What is the enthalpy of reaction in kJ? Given the following enthalpies of formation: o ∆ Hf (N2 H4 ) = 95.1 kJ o ∆ Hf (H2 O2 ) = –187.8 kJ o ∆ Hf (H2 O) = –285.8 kJ HEATS OF SOLUTION AND DILUTION STUDY OBJECTIVES 1. 2. Define the heat of solution, and the heat of dilution. Describe the energy changes that occur during the solution process, and apply Hess's law to calculate the enthalpy of solution. Heat of Solution. The heat of solution of a solute is the enthalpy change associated with the dissolution of a certain amount of solute in a given amount of solvent. The solution process can be represented by a thermochemical equation. H2 O o HCl(g) → HCl(aq) ∆ Hsoln = –75.1 kJ Values of the heat of solution ∆ Hsoln of several solutes in H2 O are given in Table 6.1. Table 6.1 Heats of solution. ______________________________ o Solute ∆ Hsoln (kJ/mol) ______________________________ NaOH –44.5 Ca(OH)2 –11.7 KF +2.0 NaCl +4.0 NH4 NO3 +26.2 ______________________________ The energy changes that occur when ionic solids dissolve in water can be understood by imagining the solution process to take place in two steps. We realize that when a compound such as KF dissolves in water the K+ and F– ions must break away from the crystal lattice and enter the aqueous phase where they are surrounded by water molecules. First, we imagine a step in which the lattice is broken up and the ions enter the gas phase. KF(s) → K+(g) + F– (g) The energy change for this process is called the l attice energy , U. All ionic crystals have a characteristic lattice energy. This is the energy required to completely separate all the ions in 1 mole of solid into ions in the gas phase. For KF(s), U = 821 kJ. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 17 Second, the gaseous K+ and F– ions enter the aqueous phase, where they are hydrated. During hydration, K+ and F– ions are surrounded by H2 O molecules. Ions in water are attracted to the polar water molecules, and a quantity of energy is released that is called the h ydration energy , ∆ Hhydr . For K+ and F– ions, ∆ Hhydr = – 819 kJ. H2 O K+(g) + F– (g) → K +(aq) + F– (aq) According to Hess's law, the overall energy change for the two steps will be the heat of solution. K + (g) + F – (g) ion hydration ∆ Hhydr lattice energy U heat of solution ∆ Hsoln KF(s) K + (aq) + F – (aq) where ∆ Hsoln = U + ∆ Hhydr = 2 kJ. Heat of Dilution. Dilution of concentrated laboratory reagents such as hydrochloric and sulfuric acids is accompanied by sufficient evolution of heat to make the procedure hazardous. The h eat of dilution i s the heat released or absorbed when additional solvent is added to a solution. If heat is released during the preparation of a certain solution, heat will also be released if that solution is diluted. It follows that when heat is absorbed during a solution process, then heat is also absorbed during dilution. _______________________________________________________________________________ EXAMPLE 6.7 Heat of Hydration The heat of solution of NaBr is –1.0 kJ/mol. The lattice energy of NaBr is 735 kJ/mol. Determine the heat of hydration of NaBr and write the equation for the hydration reaction. •Method of Solution Recall that the heat of solution is the sum of the lattice energy (U) and the heat of hydration (∆ Hhydr ). ∆ Hsoln = U + ∆ Hhydr •Calculation Rearrranging so as to solve for the heat of hydration: ∆ Hhydr = ∆ Hsoln – U = –1.0 kJ/mol – 735 kJ/mol = –736 kJ/mol ∆ Hhydr refers to the heat liberated or absorbed when the gas phase ions are dissolved in water: The equation is: H2 O Na+(g) + Br– (g) → Na+(aq) + Br– (aq) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 18 / Thermochemistry •Comment For NaBr, U and ∆ Hhydr have essentially the same values, but are opposite in sign. Therefore, when NaBr dissolves in water, the energy U required to break up the crystal lattice is "paid back" by the hydration of the Na+ and Br– ions. Also, NaBr does not dissolve in nonpolar solvents. Molecules of nonpolar solvents can only interact weakly with ions in the crystal lattice, and so "solvation" is not sufficient to compensate for the lattice energy. _______________________________________________________________________________ EXERCISES 15. Given that the lattice energy of LiCl is 828 kJ/mol and the heat of hydration of Li+ and Cl– ions is –865 kJ/mol, calculate the heat of solution of LiCl(s). 16. Given that the heat of solution of LiF is +32 kJ/mol and that the lattice energy (U) for LiF is 1006 kJ/mol, calculate the heat of hydration (∆ Hhydr ) of Li + and F– ions. THE FIRST LAW OF THERMODYNAMICS STUDY OBJECTIVES 1. 2. 3. 4. Describe a state function and give several examples. Calculate the work done when a gas expands. Write the equation for the first law of thermodynamics and give the sign conventions for q and w. Relate ∆ E to ∆ H. State Functions. Internal energy (E), is an example of a s tate function. This means that the internal energy of the system depends only on the conditions (the state) of the system, and not on how it got that way. Changes in internal energy (∆ E) during a chemical change depend only on the state and identity of the products, and the state and identity of the reactants. Energy change (∆ E) is independent of the path of the reaction. Temperature, pressure, volume, and enthalpy are also state functions .On the other hand, heat is not a state function because the heat released or absorbed for a given change depends on the path. Heat is not a property of the system. Work of Expansion. Energy can be transferred from one system to another by means of either heat flow or by work. Calculations involving the heat liberated by chemical reactions were performed previously in this chapter. There are many kinds of work, but in this chapter we are mainly concerned with the work done by expanding gases. In the reaction: 2HgO(s) → 2Hg(l) + O 2 (g) the products occupy more volume than the reactants by virtue of the fact that one mole of O 2 gas is formed. The oxygen "expanding" from an open reaction vessel must do work in pushing back the atmosphere. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 19 The work done by expansion through a volume change (∆ V) i:s w = –P ∆ V where P is the pressure opposing the expansion and ∆ V is the volume change. In order for expansion to take place the pressure of the gas within the cylinder (internal pressure) must be greater than the external pressure. In this equation, P is the opposing pressure (or external pressure), not the internal pressure. The First Law. A system contains a certain quantity of energy (E) called its internal energy. The f irst law of thermodynamics is the statement that energy is conserved during any process. Energy can be converted into many forms but the total amount of energy is constant. A useful form of the first law is : ∆E = q + w where ∆ E is a change in internal energy of the system, w is the work done on the system, and q is the heat absorbed by the system. A change in a system's internal energy may also be written as: ∆ E = Ef – E i Here ∆ E is the change in a state function (E) and depends only on the initial and final states, Ei and Ef. The change in energy of a system depends only on how much heat flows into (+q), or out of (–q) the system; and on how much energy is gained when work is done on the system (+w), or expended when the system does work on the surroundings (–w). These sign conventions are summarized below. The Sign Conventions. Thermodynamic quantities always have two parts. The number gives the amount of change, and the sign indicates the direction of flow. Here we use the convention that the sign reflects the system's point of view, rather than the surrounding's point of view. The symbol q denotes the heat transferred to the system. When q is positive, it signifies that heat has been transferred to the system (endothermic), and that it contributes to an increase in the internal energy of the system. When q is negative, it signifies that heat has flowed out of the system (exothermic) with a corresponding decrease in internal energy. This heat is absorbed by the surroundings. The symbol w denotes the work done on the system by the surroundings. When w is positive, it signifies that work has been done on the system and has contributed to an increase in the internal energy. When the volume of a system is compressed by the surroundings, work is done on the system. When w is negative, it signifies that work has been done by the system on the surroundings. If the system does work on the surroundings, its internal energy must decrease. When the volume of a system expands, the system pushes back the surroundings and work is done by the system. These conventions are the result of taking the system's point of view. Energy entering the system is positive. Energy leaving the system is negative. Enthalpy. Two important conditions under which chemical reactions can be carried out are constant volume and constant pressure. The energy change is ∆E = q + w = q – P ∆V where w = –P ∆ V and is the work done on the system. Under conditions of constant volume, ∆ V = 0, and the change in energy of the system is equal to the heat absorbed or released. ∆ E = qv Where the subscript "v" refers to a constant volume process. Under conditions of constant pressure, when the system expands against the atmosphere: ∆ E = qp + w = q p – P ∆ V qp = ∆ E + P ∆ V Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 20 / Thermochemistry Where the subscript "p" refers to a constant pressure process. Since the enthalpy of reaction (∆ H) is the heat evolved in a reaction at constant pressure, then the enthalpy change ∆ H: qp = ∆ H Therefore ∆H = ∆E + P ∆V We can learn two things from this equation. First enthalpy must be a state function and second ∆ H ≈ ∆ E. This second observation results from the fact that P ∆ V is usually much less than ∆ E. ∆ H is equal to the difference between the initial and final enthalpies of the system: qp = ∆ H = H2 – H 1 Figure 6.1 shows that the enthalpy change (∆ H) for the combustion of sulfur dioxide with oxygen. 1 SO2 (g) + 2 O 2 (g) → SO 3 (g) ∆ H° = –99.1 kJ The heat released at constant pressure (qp ) is equal to the enthalpy change of the system. The heat released by the reaction comes from the decrease in enthalpy of the system. 1 Enthalpy SO 2 + O 22 ∆ H = – 99.1 kJ SO3 Figure 6.1. An enthalpy diagram showing the change in enthalpy of the system. Enthalpy lost by the system is transferred to the surroundings as heat. It will be helpful if you learn to sketch enthalpy diagrams for exothermic and endothermic reactions as shown in Figures 6.4 and 6.5 in the text. _______________________________________________________________________________ EXAMPLE 6.7 Work of Expansion A gas, initially at a pressure of 10.0 atm and having a volume of 5.0 L, is allowed to expand at constant temperature against a constant external pressure of 4.0 atm until the new volume is 12.5 L. Calculate the work done by the gas on the surroundings. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 21 •Method of Solution In this problem the system does work on the surroundings as it expands, and by convention (Table 6.5 in the text) the sign is negative: w = –P ∆ V = –P(V2 – V 1 ) where P is the pressure opposing the expansion, and ∆ V is the change in volume of the system. •Calculation w = –4.0 atm (12.5 L – 5.0 L) = –30 L. atm This quantity can be expressed in units of joules. w = –30 L . atm × 1 01.3 J = –3.0 × 10 3 J 1 L . atm ___________________________________________________________ EXAMPLE 6.8 Work Done on the System Calculate the work done on the system when 6.0 L of a gas is compressed to 1.0 L by a constant external pressure of 2.0 atm. •Method of Solution The work done is: w = –P ∆ V = –P(V2 – V 1 ) = –2.0 atm (1.0 L – 6.0 L) = + 10 L. atm w = 1.0 × 10 3 J •Comment Obtaining a positive value for work means, that in a compression, work is done on the system by the surroundings. The positive value for work means that the system gains energy. ______________________________________________________________ EXAMPLE 6.9 The First Law of Thermodynamics A gas is allowed to expand at constant temperature from a volume of 10.0 L to 20.0 L against an external pressure of 1.0 atm. If the gas also absorbs 250 J of heat from the surroundings, what are the values of q, w, and ∆ E? •Method of Solution The work done by the system is w = –P ∆ V = –P(V2 – V 1 ) = –1.0 atm (20.0 L – 10.0 L) = – 10 L. atm w = –1.0 × 10 3 J The amount of heat absorbed was 250 J: q = 250 J Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 22 / Thermochemistry The first law of thermodynamics is ∆E = q + w = 250 J – 1000 J ∆ E = –750 J •Comment In this example, the system did more work than the energy absorbed as heat, therefore the internal energy E decreased. _______________________________________________________________________________ EXERCISES 17. Calculate the work done on a gas when 22.4 L of the gas is compressed to 2.24 L under a constant external pressure of 10.0 atm. 18. A system does 975 J of work on its surroundings while at the same time it absorbs 625 J of heat. What is the change in energy ∆ E for the system? 19. A system does 975 J of work on the surroundings. How much heat does the system absorb at the same time if its energy charge is –350 J? __________________________________________________________________________ CONCEPTUAL QUESTIONS 1. The volume of a balloon is a state function. Explain by making changes to the volume of the balloon why volume is a state function. 2. Hess's law of heat summation works because enthalpy is a state function. Explain. 3. Why is the lattice energy always a positive quantity? Why is the hydration energy always negative? 4. Li+ ion is a smaller cation than K+ ion. Which of the two has the higher hydration energy? Explain. PRACTICE TEST 1. Define energy, work, and heat. Consider a 9V "alkaline" battery. Which of these three does it contain? Name a device that uses a battery to do work. Name a device that uses a battery to generate heat. 2. What is a state function? Name three state functions used in thermochemistry. 3. Given the thermochemical equation: 1 o SO2 (g) + 2 O 2 (g) → SO 3 (g) ∆ Hrxn = –99 kJ How much heat is liberated when a. 1/2 mole of SO2 reacts b. 2 moles of SO2 reacts Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 23 4. Given the thermochemical equation H2 + I2 → 2HI o ∆ Hrxn = 52 kJ What is ∆ H for the reaction 1 1 HI → 2 H 2 + 2 I 2 o ∆ Hrxn = ? 5. Given the following reactions and their associated enthalpy changes: 2H(g) → H2 (g) Br2 (g) → 2Br(g) H2 (g) + Br2 (g) → 2HBr(g) o ∆ Hrxn = –436 kJ o ∆ Hrxn = +224 kJ o ∆ Hrxn = –72 kJ o Calculate ∆ Hrxn in kJ for the reaction: HBr(g) → H(g) + Br(g) 6. Nitrogen and oxygen react according to the following thermochemical equation: N2 (g) + O 2 (g) → 2NO(g) o ∆ Hrxn = 180 kJ How many kJ of heat are absorbed when 50.0 g of N2 reacts with excess O2 to produce 107 g of NO? 7. How much energy is required to raise the temperature of 180 g of graphite from 25°C to 500°C? The specific heat of graphite is 0.720 J/g·°C. 8. 0.500 g of ethanol, C2 H5 OH(l), was burned in a bomb calorimeter containing 2.000 × 10 3 g of water. The heat capacity of the bomb calorimeter was 950 J/°C, and the temperature rise was found to be 1.60°C. a. Write a balanced equation for the combustion of ethanol. b. Calculate the amount of heat transferred to the calorimeter. c. Calculate ∆ H for the reaction as written in part (a). 9. The enthalpy of combustion of sulfur is S(rhombic) + O 2 (g) → SO 2 (g) ∆ H° = –296 kJ What is the enthalpy of formation of SO2 ? 10. The combustion of methane occurs according to the equation o CH4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ∆ Hrxn = –890 kJ Use enthalpies of formation for CO2 and H2 O to determine the enthalpy of formation of methane. 11. During expansion of its volume from 1.0 L to 10.0 L against a constant external pressure of 2.0 atm, a gas absorbs 200 J of energy as heat. Calculate the change in internal energy of the gas. 12. How much work must be done on or by the system in a process in which the internal energy remains constant and 322 J of heat is transferred from the system. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 24 / Thermochemistry ANSWERS Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 116 kJ 162 g 37.9 kJ 3340 kJ 498°C 5.77 kJ 1150 J 2.91 kJ/°C 941 J/°C 48.2 kJ/g 464 kJ/mol o ∆ Hf = –250 kJ. o ∆ Hf = –75.0 kJ –862.7 kJ –37 kJ/mol ∆ Hhydr = 1038 kJ/mol w = +202 L·atm ∆ E = –350 J q = 625 J Conceptual Questions 1. 2. 3. 4. Back Let's say that the balloon is partially inflated and has a certain volume V 1 . After further inflation it reaches a volume V 2 . Because volume is a state function, the change in volume from V1 to V 2 is just the difference V2 – V1 . The difference V2 – V1 would be the same even if we started at V1 and inflated the balloon beyond V2 , to a greater volume V 3 , and then let air out until the volume came back down to V2 . As long as you go from the same initial volume to the same final volume, no matter by what path, V2 – V1 will be the same. According to Hess's law of heat summation if a reaction is the sum of several reaction steps, then the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes of the intermediate steps. The law works because enthalpy is a state function. Starting with the reactants and forming the products will give the same enthalpy change whether we go by a set of intermediate steps or by a direct path. Only the difference in initial and final states determines the change in a state function. The lattice energy is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. The lattice energy is dependent on the strength of attraction between the cations and anions. The lattice energy is a positive quantity because oppositely charged particles attract one another, therefore an input of energy is required to separate them. The hydration energy is a negative quantity because there are strong attractive forces between ions and polar water molecules that lead to ion hydration. According to Coulomb's law the smaller the separation between two oppositely charged particles, the lower their potential energy. The smaller the cation, the closer it can get to a water molecule, and the greater the amount of energy given off during hydration. The Li+ ion has the smaller ionic radius, and therefore the higher hydration energy. (The larger negative value.) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Thermochemistry / 1 25 Practice Test 1. The battery contains energy which is the capacity to do work. Work is energy lost or gained by mechanical means, and heat is energy transferred because of a temperature difference. A toy truck or car can be moved (displaced) when energy in the battery is used to do work. A flashlight uses energy in the battery to produce a hot glowing filament in the bulb which transfers heat to the surroundings. 2. A state function is a property of a system that depends only on its present condition (state) and not on how it got that way. Energy, enthalpy, and temperature are state functions. 3. a. 5.0 × 10 1 kJ b. 198 kJ o 4. ∆ Hrxn = –26 kJ o 5. ∆ Hrxn = 366 kJ 6. 321 kJ (q = 321 kJ) 7. 61.6 kJ 8. a. C 2 H5 OH + 3O 2 → 2CO 2 + 2H2 O b. q = 2000(4.18)(1.6) + 950(1.6) = 14,900 J o c. ∆ Hrxn = –1370 kJ o 9. ∆ Hf (SO 2 ) = – 296 kJ o 10. ∆ Hf (CH4 ) = – 75 kJ 11. ∆ E = –1618 J 12. w = +322 J ___________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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