Chapter 5 Thermochemistry PDF

Chapter 5 Thermochemistry  What is thermodynamics? The study of energy and its transformations. What is thermochemistry? Is one area of thermodynamics that studies quantity of heat absorbed or evolved ( given off) by chemical reactions.  5.1 The Nature of Energy Energy is the abil...

Chapter 5 Thermochemistry  What is thermodynamics? The study of energy and its transformations. What is thermochemistry? Is one area of thermodynamics that studies quantity of heat absorbed or evolved ( given off) by chemical reactions.  5.1 The Nature of Energy Energy is the ability to do work or transfer heat. – Work: is defined as the energy used to cause an object that has mass to move. – Heat: is defined as the energy used to cause the temperature of an object to rise.  Kinetic Energy Kinetic energy is the energy of motion: 1 Ek =  mv2 2 m = mass, and v = speed. A car running on the road has kinetic energy and a molecule in the solution has kinetic energy. Potential Energy Potential energy is energy an object possesses by virtue of its position.  Potential Energy The most important form of potential energy in molecules is electrostatic potential energy, Eel: KQ1Q2 Eel = d It arises from the interactions between charged particles. Κ is a constant, Q1 and Q2 are charges and d, the relative distance separated the two charges.  In this chapter we are going to study the energy changes when substances react and the energy a substance has because of its kinetic energy and its chemical energy (potential energy).  Units of Energy The SI unit of energy is the joule (J): kg m2 1 J = 1  s2 An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = 4.184 J  System and Surroundings A system is the part of the universe we are interested in studying. Surroundings are the rest of the universe. Systems may be:  Open: matter and energy can be exchange with the surroundings.  Closed: only energy is exchanged with the surroundings.  Isolated: neither energy nor matter can be exchanged with the surroundings. Example of closed system 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).  Transferring energy: work and heat Energy used to move an object over some distance is work: w=Fd where w is work, F is the force, and d is the distance over which the force is exerted. Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects.  First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is constant; if the system loses energy, it must be gained by the surroundings, and vice versa.  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.  Internal Energy 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  Changes in Internal Energy If E > 0, Efinal > Einitial – Therefore, the system absorbed energy from the surroundings.  Changes in Internal Energy If E < 0, Efinal < Einitial – Therefore, the system released energy to the surroundings. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is, E = q + w. E, q, w, and Their Signs Which set of values for heat and work will result in a decrease of internal energy? a. q = –150 J; w = +150 J b. q = –150 J; w = +300 J c. q = +150 J; w = –300 J d. q = +300 J; w = –150 J A system absorbs heat during an _______ process. a. exothermic b. isothermal c. adiabatic d. endothermic  Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic (endo = “in”), q > 0. Example:melting of ice. When heat is released by the system into the surroundings, the process is exothermic (exo = out”) q < 0. Example: combustion of gasoline.  State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. However, we do know that the internal energy of a system is independent of the path by which the system achieved that state.  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. State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its E is the same. – But q and w are different in the two cases.  Enthalpy Chemical and physical changes that occur around us (evaporation of water from a lake or a reaction in an open flask in the lab), occur under constant pressure. A system that consists of a gas confined to a container can be characterized by its energy E, pressure P and volume V. Combining these three state functions we get a new state function called enthalpy (H). H = E + PV  Zn (s) + 2H+ (aq) Zn2+ (aq) + H2 (g) Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). The work involved in the expansion or compression of gases is called pressure-volume work ( or P-V work). We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −PV  Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV  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 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.  Enthalpy Since the heat gain or lost by the system can be readily measured/calculated ( unlike internal energy), enthalpy is a more useful function that internal energy. Endothermicity and Exothermicity A process is endothermic when H is positive.( q >0) A process is exothermic when H is negative. (q < 0)  The change in enthalpy for a given reaction Hrxn, is the enthalpy of the products minus the enthalpy of the reactants. Hrxn or rxn heat: defined as enthalpy change that accompanies a reaction Hrxn = H(products) – H(reactants) Consider the equation for the production of water: 2H2(g) + O2(g) 2H2O(g) H = –483.6 kJ An equation like this one is called thermochemical equation. The equation tells us that 483.6 kJ of energy  Properties of enthalpy 1. Enthalpy is an extensive property (just like heat, to melt a 2 kg block of ice requires more heat than to melt half of a kg ice). H2(g) + ½ O2(g) H2O(g) H = –(483.6 kJ)/2 = –241.8 kJ Note this is exactly half of the value for H as compared to previous slide. So we can conclude: H noted at the end of the balanced equation depends on the number of moles of reactants and products. Pay attention to the stoichioemetry of the reaction. To avoid this confusion the concept called – molar enthalpy is used. For example, for the formation of water, H = –241.8 kJ/mol If you have 2 moles, then H = (2 mole) x (-241.8 kJ/mol) = - 483.6 kJ.  he enthalpy change for a reaction is equal in magnitu t opposite in sign to H for the reverse reaction. Hforward = – Hreverse writing this, we mean that the sign of H depends on t ection of the reaction. mple: (g) + 2O2(g) CO2(g) + 2H2O(l) H = –890 kJ erse the direction of the reaction (g) + 2H2O(l) CH4(g) + 2O2(g) H = +890 kJ. 3. H for a reaction depends on the state of the products and the state of the reactants. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = – 890 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = – 802 kJ  Calorimetry H can be determined experimentally by measuring the heat flow accompanying a reaction at constant pressure. Calorimetry is the measurement of heat flow. Calorimeter is an apparatus that measures heat flow. To measure heat, we need to define two concepts: heat capacity and specific heat.  Heat Capacity The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity. Take a block of iron, and a block of ceramic (same size). Heat them up at the same time ---- which one require more heat to raise the temperature to 1 degree higher? --- iron (iron has a higher heat capacity than the ceramic material).  Using the heat capacity has a major problem. For the same substance, like iron blocks, let’s say one is 10 g and the other is 1 kg. The latter absorbs more heat to raise the same degree of temp, because it is bigger in size, requiring more heat. To overcome this problem two quantities are introduced: – molar heat capacity (Cm) is the amount of energy required to raise the temperature of 1 mole of a substance by 1 K (or 1 C). – Specific heat or specific heat capacity (Cs) is the heat capacity of 1 gram of substance. Specific heat is: heat transferred Specific heat = mass  temperature change q Cs = m  T When a sample absorbs heat its temperature increases therefore, the heat a substance gains or loses can be calculated from eq: q = m x T x Cs  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)? T = 98 C – 22 C = 76 C = 76 K q = Cs  m  T = (4.18 J/g–K)(250 g)(76 K) = 7.9  104 J (b) What is the molar heat capacity of water? The molar heat capacity is the heat capacity of one mole of substance. Using the atomic weights of hydrogen and oxygen, we have 1 mol H2O = 18.0 g H2O From the specific heat given in part (a), we have 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 temperature change for the water in the calorimeter.  If two aqueous solutions, each containing a reactant are mixed in the calorimeter, the reactants react to form products. The reactants and products of the reaction are the system and the water in which are dissolved is part of the surroundings. If the calorimeter is perfectly insulated, any heat release or absorbed by the reaction will raise or lower the temperature of the water.  We measure the temperature change of the solution and assume that any changes are due to: Heat transferred from the reaction to the water (exothermic process). Heat is lost by the reaction and gained by the water, so the T of the solution raises. Heat transferred from the water to the reaction ( endothermic process).Heat is gained by the reaction and lost by water, so T of solution decreases.  Therefore, the heat gained or lost by the solution qsol, is equal in magnitude but opposite in sign to the heat absorbed or released by the reaction qrxn: qsoln =(specific heat of soln) x (grams of soln) x T = –qrxn For dilute aqueous solutions, the specific heat of the solution will be close to that of pure water.  Measuring H Using a Coffee-Cup Calorimeter 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 T of the resultant solution increases from 21.0 C to 27.5 C. Calculate the enthalpy change for the reaction per mole of HCl (in kJ/mol), assuming that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g–K. HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) /mL) = 100 g 21.0 C = 6.5 C = 6.5 K m  T J/g–K)(100 g)(6.5 K) = –2.7  103 J = –2.7 kJ Because the process occurs at constant pressure, H = qP = –2.7 kJ To express the enthalpy change on a molar basis, we use the fact that the number of moles of HCl is given by the product of the volume (50mL = 0.050 L) and concentration (1.0 M = 1.0 mol/L) of the HCl solution: (0.050 L)(1.0 mol/L) = 0.050 mol Thus, the enthalpy change per mole of HCl is H = –2.7 kJ/0.050 mol = –54 kJ/mol  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.  Bomb calorimeter is used to carry out constant volume calorimetry measurement. The most common type of reaction studied under these conditions or using this device is combustion. If we know the heat capacity of the calorimeter, Ccalorimeter, then the heat of reaction, qrxn = –Ccalorimeter x T. Ccalorimeter can be determined using a known reaction with known heat output like benzoic acid. Once Ccalorimeter is known, this bomb calorimeter can be used to measure the rxn heat for other unknown rxns. Since the reaction is carried out under constant volume, q relates to E or no PV work (recall that H = E + PV, so the heat is a measure of the change of the internal energy of the reaction)  Measuring qrxn Using a Bomb Calorimeter The combustion of (CH6N2), 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 T of the calorimeter increases from 25C to 39.50 C. If the heat capacity of the calorimeter is measured to be 7.794 kJ/ C, calculate the heat of reaction for the combustion of a mole of CH6N2. T = (39.50 C – 25.00 C) = 14.50 C We can use T and the value for Ccal to calculate the heat of reaction : xn = –Ccal  T = –(7.794 kJ/C)(14.50 C) = –113.0 kJ We can readily convert this value to the heat of reaction for a mole of CH6N2:  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.  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.” 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. Using Hess’s Law to Calculate H The enthalpy of reaction for the combustion of C to CO2 is – 393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol C: 1. 2. Using these data, calculate the enthalpy for the combustion of C to CO: 3.  Enthalpies of Formation Hess’s law can be used to calculate the H of many different reactions. For example: H vaporization, is the enthalpy for converting liquids to gases. One very important H is the 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. Since the H depends on the T, P and physical state of the reactants and products, we must define a set of conditions, called standard conditions if we want to compare enthalpies of different reactions.  Standard state (standard conditions) refer to the pure substance at: 1 atm and 25C (298 K). Standard enthalpy, H, is the enthalpy measured when all reactants and products are in their standard state. Standard enthalpy of formation of a compound, Hf, is the enthalpy change for the formation of 1 mol of compound with all substances in their standard states. If there is more than one state for a substance under standard conditions, the more stable one is used. Example: When dealing with carbon we use graphite because graphite is more stable than diamond. The standard enthalpy of formation of the most stable form of an element is zero. Standard Enthalpies of Formation  Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Imagine this as occurring in three steps: 1. C3H8(g)  3C(graphite) + 4H2(g) 2. 3C(graphite) + 3O2(g) 3CO2(g) 3. 4H2(g) + 2O2(g)  4H2O(l)  Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) The sum of these equations is C3H8(g)  3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) 3CO2(g) 4H2(g) + 2O2(g)  4H2O(l) C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)  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.  Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0kJ)] = [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)] = (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ