Lecture Presentation Chapter 5 Thermochemistry John D. Bookstaver St. Charles Community College Cottleville, MO
Thermochemistry Thermodynamics is the study of energy and its transformations. Thermochemistry is the study of the relationships between chemical reactions and energy changes involving heat. CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (g) + energy (heat)
Energy Energy is the ability to do work or transfer heat. Work (w) is the energy used to cause an object with mass to move (energy transferred when a force moves an object) w = F d = m g d F is the force, and d is the distance A force is any kind of push or pull exerted on an object.
Energy Heat (q) is a process quantity. Energy used to cause the temperature of an object to rise Energy is transferred from a hotter object to colder object.
Potential Energy Potential energy (E p or PE) is energy an object possesses by virtue of its position or chemical composition. E p = mgh where m is mass; g is the gravitational constant, 9.8 m/s 2 ; and h is the height of the object relative to some reference height.
Kinetic Energy Kinetic energy (E k or KE) is energy an object possesses by virtue of its motion. 1 2 E k = mv 2 where m is mass and v is speed
Units of Energy The SI unit of energy is the joule (J). 1J = 1kg A common, non-si unit is the calorie (cal). amount of energy required to raise the temperature of 1 g of water from 14.5 C to 15.5 C. 1 cal = The nutritional Calorie(capital C), Cal = 1,000 cal = 1 kcal Wheaties that provides 100 nutritional Cal really gives 100 kcal or 100,000 cal m s 2 2 4.184J
System and Surroundings System: The material(s) being studied Surroundings: everything else (not being studied) Energy can be transferred back and forth between a system and its surroundings in the forms of work and heat! Types of systems: Open: matter and energy can be exchanged with the surroundings Closed: exchange energy but not matter with the surroundings Isolated: neither energy nor matter can be exchanged with the surroundings Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward http://universe-review.ca/i13-23-open2.jpg; http://uncrate.com/p/2011/03/thermos-stainless-king.jpg 2012 Pearson Education, Inc.
1 st Law of Thermodynamics Conservation of energy: Energy is neither created nor destroyed. the total energy of the universe is a constant; Energy lost by a system is gained by the surroundings and vice versa. To explore this further we need a more precise definition of energy. Internal Energy, E of a system: sum of all the E k and E p of every component of the system. As the numerical value of a system s internal energy is generally not known, we look at the CHANGE in E, E. E = E final E initial
Internal Energy Due conservation of energy, we can determine E without knowing the values for E final and E initial. A closer look at E. It has 3 parts: A number Together shows magnitude and change A unit A sign: gives direction What does this mean??? E is positive when E final > E initial : system gains energy from its surroundings, ( E > 0). This is an endergonic process: system ABSORBS energy from the surroundings. E is negative when E final < E initial : system lost energy to its surroundings, ( E < 0). This is an exergonic process: system RELEASED energy from the surroundings.
An Example Energy diagram In a chemical rxn: Initial state refers to the reactants Final state refers to the products When H 2 and O 2 (initial state) form H 2 O (final state) at a given temp., the system loses energy to the surrounding, thus E final < E initial. This makes E for the process (rxn) negative.
E, Heat and Work A system can exchange energy with its surroundings as heat (q) and work (w). E of a system changes in magnitude as Heat is added or removed from the system Work done on or by the system An analogy: E is like your money in a bank. You can make deposits and withdrawals in the form of heat or work. Deposits increase E : E > 0 (positive) Withdrawals decrease E : E < 0 (negative) To balance your account (the math): E = q + w Your bank statement reads: When heat is added to a system or work is done on the system, E increases!
Sign Conventions When heat is added to a system or work is done on the system, E increases! E = q + w An Example: Gases A and B are confined in a closed system (see figure) and react to form a solid product C. As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downwards as the rxn takes place. As the volume of the gas decreases under constant atm pressure, the surrounding do 480 J of work on the system. Calc E? A (g) + B (g) C (s)
Terminology Some terminology to indicate the direction of heat transfer: A reaction in which the system absorbs/ gains heat: ENDOTHERMIC The surroundings will feel cool Examples: Mixing water with ammonium nitrate Mixing water with potassium chloride Melting ice cubes A reaction in which the system releases/loses heat: EXOTHERMIC The surroundings will feel hot Examples: Mixing water and calcium chloride Acid-base reactions Making ice cubes Hand warmers http://en.wikipedia.org/wiki/hand_warmer; http://www.fieldandstream.com/blogs/fishing/2010/04/stuff-works-use-toe-warmers-heat-your-hands;
Endothermic and Exothermic Processes When heat is absorbed by the system from the surroundings, the process is endothermic. An endothermic reaction feels cold. N 2(g) + O 2(g) + energy (heat) 2NO (g)
Endothermic and Exothermic Processes When heat is released by the system into the surroundings, the process is exothermic. An exothermic reaction feels hot. CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (g) + energy (heat)
State Functions We cannot determine precisely the value of E but. It does have a fixed value for a given set of conditions. The conditions: temperature and pressure Also, since energy is an extensive property: E of a system is proportional to the total quantity of matter in the system State function: the value of the state function depends only on the present state of the system, not on the path the system took to reach that state.
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Are q and w State Functions? State function: the value of the state function depends only on the present state of the system, not on the path the system took to reach that state. That is: E depends only on E initial and E final. Look at the diagram: E is the same for both (a) and (b) a) Battery shorted out by a wire System loses energy to surroundings as HEAT b) Battery shorted out by a motor (fan) System loses heat to surroundings as Work (to make the fan turn) Heat In both cases q and w are different, hence q and w are not state functions
Enthalpy Measures the flow of heat occurring at constant pressure. The system: gas in a container 2 measurable properties: pressure (P) and volume (V). Like E, are P and V state functions? Yes Why? Depend only on current state of the system; independent of path taken to reach that state. Combining all 3 state functions = enthalpy (H) H = E + PV Do you think H is a state function? Yes
P-V Work Look at the schematic: Zn + HCl produces H 2 (g). Assume piston has no mass. The system before the rxn is at equilibrium: the piston does not move and the pressure acting on the piston is atm P. P is constant. The rxn occurs: H 2 (g) is produced. The piston rises. Increase in volume The gas is doing work on the surrounding for the piston to rise against P. This is called P-V work: the work involved in the expansion or compression of gases. w = -P V
w = -P V Sign Convention P is always positive or zero If volume expands, do you expect V to positive or negative? Why? Positive b/c V final is greater then V initial. The formula is then: w = P V (notice no sign); so why is there one? Now think: in the previous scenario, we said that the gas does work on the surroundings. Does this mean that work was done a) on the system or b) by the system? By the system What s the sign convention for work in this case? Negative! And it must be shown in the formula! So we get: w = -P V The sign means something in the reactions!
Change in Enthalpy, H Recall: H = E + PV Also, E = q + w and w = -P V When a change occurs at constant P, we now talk about the change in enthalpy, H. How do we calc H? H = E + P V H = (q + w) w H = q The H is the heat (q p ) gained or lost at constant P. For most rxns: H and E are small because P V is small. What does this mean for reactions: A process is endothermic when H is positive: H > 0 A process is exothermic when H is negative: H < 0
Sample Problem - Determining the Sign of H Indicate the sign of the enthalpy change, H, in these processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic: (a) An ice cube melts; (b) 1 g of butane (C 4 H 10 ) is combusted in sufficient oxygen to give complete combustion to CO 2 and H 2 O.
Enthalpies of Reaction We know: H = E + P V Can we simplify H? H = H final - H initial How does this relate to a chemical reaction: A + B C + D?? H = H products H reactants The H for a rxn is called enthalpy of reaction or heat of reaction: H rxn When specifying a numerical value for H rxn, also give the reaction that occurred! An example: 2 H 2 (g) + O 2 (g) 2 H 2 O (g) H = -483.6 kj Is the rxn endo- or exothermic?
Thermochemical Equations When the coefficients of balanced equations represent the # of mols of products and reactants producing an enthalpy change. when the specific amounts of chemicals involved is not given. 2 H 2 (g) + O 2 (g) 2 H 2 O (g) H = - 483.6 kj Some guidelines: Enthalpy is an extensive property: magnitude of H is proportional to the amount of reactant consumed in the reaction. What is H when 4 mols of H 2 is consumed? If 2 mols H 2 produces -483.6 kj Then 4 mols H 2 will produce twice as much: - 967.2kJ What is H when 3 mols of O 2 is consumed? If 1 mol O 2 produces -483.6 kj then 3 mols O 2 will produce 3x as much: -1450.8 kj If the rxn to form H 2 O was: H 2 (g) + ½ O 2 (g) H 2 O (g), what would H be? H = - 241.8 kj
Enthalpy Guidelines H for a reaction is equal in magnitude but opposite in sign to H of the opposite reaction. Forward rxn: 2 H 2 (g) + O 2 (g) 2 H 2 O (g) H = - 483.6 kj Reverse rxn: 2 H 2 O (g ) 2 H 2 (g) + O 2 (g) H = + 483.6 kj We must put the + sign! H for a reaction depends on the states of the reactants and products. 2 H 2 (g) + O 2 (g) 2 H 2 O (g) H = - 483.6 kj rxn 1 2 H 2 (g) + O 2 (g) 2 H 2 O (l) H = - 571.6 kj rxn 2 Note that in rxn 2, the water produced is in the liquid state. The enthalpy change is slightly higher. Why? More heat is available for transfer to the surroundings b/c the enthalpy of H 2 O (g) is greater than that of H 2 O (l). 2 H 2 O (g) 2 H 2 O (l) ; H = H products H reactants = -571.6 kj (- 483.6kJ) = -88 kj
Sample Problem - Relating H to Quantities of Reactants and Products How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? ( H rxn = -890 kj)
Solution Analyze Our goal is to use a thermochemical equation to calculate the heat produced when a specific amount of methane gas is combusted. According to Equation 5.18, 890 kj is released by the system when 1 mol CH 4 is burned at constant pressure. Plan Equation 5.18 provides us with a stoichiometric conversion factor: (1mol CH 4 = 890 kj). Thus, we can convert moles of CH 4 to kj of energy. First, however, we must convert grams of CH 4 to moles of CH 4. Thus, the conversion sequence is grams CH 4 (given) moles CH 4 kj (unknown to be found). Solve By adding the atomic weights of C and 4 H, we have 1 mol CH 4 = 16.0 CH 4. We can use the appropriate conversion factors to convert grams of CH 4 to moles of CH 4 to kilojoules: The negative sign indicates that the system released 250 kj into the surroundings. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Practice Exercise Relating H to Quantities of Reactants and Products Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H 2 O 2 (l) 2 H 2 O(l) + O 2 (g) H = 196 kj Calculate the quantity of heat released when 5.00 g of H 2 O 2 (l) decomposes at constant pressure. Answer: 14.4 kj Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward http://rlv.zcache.com/enthalpy_pirate_tshirt-p235742845394872483z89ss_400.jpg 2012 Pearson Education, Inc.
Calorimetry The measurement of heat flow Calorimeter: instrument used to measure heat flow. When an object gains heat, it gets hot; temperature changes. This temp change varies from substance to substance and is determined by a substance s heat capacity. What is heat capacity,c? The amount of heat required to raise the temp of an object by 1 K or 1 C. The greater C is, the greater the heat required to produce a rise in temp. Molar heat capacity, C m : heat capacity for 1 mole of a substance. Units: J/mol K Specific heat capacity, C s : heat capacity for 1 g of a substance. Units: J/g K
Specific Heat Capacity Recall: specific heat capacity (or simply specific heat) is the amount of energy (heat) required to raise the temperature of 1 g of a substance by 1 K (or 1 C). Units for C s is J/gK Heat = specific heat capacity x mass x change in temp. q = m x C s x T Remember that: H = q so H = m x C s x T
Sample Problem - Relating Heat, Temperature Change, and Heat Capacity (a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 C (about room temperature) to 98 C (near its boiling point)? (b) What is the molar heat capacity of water?
Practice Exercise (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 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kj of heat? Answer: (a) 4.9 105 J, (b) 11 K decrease = 11 C decrease Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Constant-Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = H = m x C s x T
(a) -15.2 kj/g (b) -1370 kj/mol Bomb Calorimetry Constant volume calorimetry Combustion rxns Reactions can be carried out in a sealed bomb. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. Must know total heat capacity of the calorimeter, C cal. q rxn = -C cal x T A 0.5865 g sample of lactic acid (HC 3 H 5 O 3 ) is burned in a calorimeter whose heat capacity is 4.8412 kj/ C. The temperature increases from 23.10 C to 24.95 C. calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.
Hess Law 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. 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. We can estimate H using published H values and the properties of enthalpy. We need to find the route for which H is known for every step of a reaction process. For example: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) H = -802 kj 2 H 2 O (g) 2 H 2 O (l) H = -88 kj + CH 4 (g) + 2 O 2 (g) + 2 H 2 O (g) CO 2 (g) + 2 H 2 O (g) + 2 H 2 O (l) H = -890 kj The overall rxn: cancel substances common on both sides of rxn arrow. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) H = -890 kj
Sample Problem - Using Hess s Law to Calculate H The enthalpy of reaction for the combustion of C to CO 2 is 393.5 kj/mol C, and the enthalpy for the combustion of CO to CO 2 is 283.0 kj/mol C: 1. 2. Using these data, calculate the enthalpy for the combustion of C to CO: 3.
Practice Exercise Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is 393.5 kj/mol, and that of diamond is 395.4 kj/mol: C(graphite) + O 2 (g) CO 2 (g) H = 393.5 kj C(diamond) + O 2 (g) CO 2 (g) H = 395.4 kj Calculate H for the conversion of graphite to diamond: C(graphite) C(diamond) H =? Answer: +1.9 kj Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Enthalpies of Formation H f (heat of formation) is the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Magnitude of any enthalpy change is dependent on Pressure Temperature State of reactants and products A set of conditions is therefore required to compare enthalpies of different rxns. This is called standard state and is tabulated for us. Standard enthalpy change of a rxn is the enthalpy change when all reactants and products are in their standard states: H Standard enthalpies of formation, H f, are measured under standard conditions 25 C and 1.00 atm pressure.
Sample Problem - Equations Associated with Enthalpies of Formation For which of these reactions at 25 C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose H is an enthalpy of formation? (a) (b) (c)
Practice Exercise Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl 4 ). Answer: C(graphite) + 2 Cl 2 (g) CCl 4 (l) Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Let s Simplify this H = Σn H f,products Σm H f reactants n and m are the stoichiometric coefficients. C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) H =??? 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
Sample Problem - Calculating an Enthalpy of Reaction from Enthalpies of Formation (a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C 6 H 6 (l), to CO 2 (g) and H 2 O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane with that produced by 1.00 g benzene. Plan (a) We need to write the balanced equation for the combustion of C 6 H 6.We then look up values in Appendix C or in Table 5.3 and apply Equation 5.31 to calculate the enthalpy change for the reaction. (b) We use the molar mass of C 6 H 6 to change the enthalpy change per mole to that per gram. We similarly use the molar mass of C 3 H 8 and the enthalpy change per mole calculated in the text previously to calculate the enthalpy change per gram of that substance.
Practice Exercise Use Table 5.3 to calculate the enthalpy change for the combustion of 1 mol of ethanol: C 2 H 5 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O(l)
Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.