Mr Chiasson Advanced Chemistry 12 / Chemistry 12 1 Unit B: Thermochemical Changes Students will be expected to: Compare the molar enthalpies of several combustion reactions involving organic compounds. (324-7) Write and balance chemical equations for combustion reactions of alkanes (C n H 2n+2 ), including energy amounts. (324-1) Define endothermic reaction, exothermic reaction, specific heat, enthalpy, bond energy, heat of reaction, molar enthalpy, kinetic energy and potential energy. (324-2) Calculate and compare the energy involved in changes of state and that in chemical reactions. (324-3) Design a thermochemical experiment, identifying and controlling major variables. (212-3) Determine experimentally the changes in energy of various chemical reactions. (324-6) Illustrate changes in energy of various chemical reactions, using potential energy diagrams. (324-5) Calculate the changes in energy of various chemical reactions using bond energy, heats of formation and Hess s Law. (324-4) Given the enthalpy notation, ΔH f or ΔH comb, write the appropriate thermochemical equation. Given the sign of ΔH, identify exothermic and endothermic processes. Given a knowledge of the formulas q = mcδt and q = nδh, calculate the heat gained or lost from a system. Given a standard heat of formation table, predict the heat of reaction for a chemical change. The Nature of Energy and Types of Energy Energy is the capacity of something to do work. Kinetic energy is the energy produced by a moving object. Radiant energy, or solar energy, comes from the sun and is Earth s primary energy source. Thermal energy is the energy associated with the random motion of atoms ad molecules. In general, thermal energy can be calculated from temperature measurements. Chemical energy is stored within the structural units of chemical substances; its quantity is determined by the type and arrangement of constituent atoms. Potential energy is energy available by virtue of an object s position. Chemical energy can be considered a form of potential energy because it is associated with the relative positions and arrangements of atoms within a given substance.
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 2 The standard unit for energy is the joule (J). Another standard unit for energy is the calorie. Calories (with a capital C) are used to measure the energy in food; 1 Calorie (capital C) = 1000 calories (lowercase C). Temperature describes the amount of motion that the molecules or atoms in a material have. Fast movement represents high temperature, and slow movement represents low temperature. Interestingly, temperature is a measure of the average kinetic energy of the particles that make up the substance. Common temperature conversions Kelvin = 0 C + 273.15 0 C = K 273.15 Thermodynamics and Thermochemistry Thermodynamics is the study of heat, and how heat can be interconverted into other energy forms. In thermodynamics, the system is a specific part of the universe that is being studied, often the system is a chemical reaction. The surroundings are all parts of the universe that are not the system, typically everything outside the test tube where the chemical reaction is taking place. An open system can exchange both energy and matter with its surroundings. An active volcano is an example of an open system. A closed system can exchange energy but not matter. A sealed greenhouse is a closed system. An isolated system can exchange neither energy nor matter. A sealed thermos bottle is an isolated system. The First Law of Thermodynamics The first law states that energy can t be created or destroyed. In other words, when a system gains or looses energy from the surroundings, the total energy (i.e. the energy of the universe) will be constant. This concept is expressed mathematically as: E = q + w The E (a.k.a. U) represents the internal energy. This includes, but is not limited to, all kinetic and potential energy possessed by all components of a system. E is the change in internal energy of a system as a result of heat flow and work. Heat flow is indicated by q, and the work done by the system is w. The conventions that apply to the above equation are: q is a positive value if heat is absorbed by the system, q is a negative value if heat is lost by the system, w is a positive value if work is done on the system,
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 3 w is a negative value if work is done by the system. The Second Law of Thermodynamics states that heat energy always travels spontaneously from a warmer body (body with higher temperature) to a colder body. You have known since you were a child that if you touch a hot stove you get burned. This is the second law of thermodynamics in action. The stove is warmer than your body and therefore transfers heat to your hand. When you go outside without protection, when it is cold, you may get frostbite. This is another example of the second law of thermodynamics in action. The heat from your body is moving into the surrounding air, when your body can no longer replace the heat as fast as it is being lost, your skin freezes and you get frostbite. Check your understanding 1. What exactly happens to the skin when you get frostbite? 2. Research first aid treatments and determine how frostbite is treated. 3. Show how the 1st and 2nd law of thermodynamics is applied to the first aid treatment for frostbite? 4. Repeat questions 1-3 but this time for burns of the skin.
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 4 Potential Energy Profile (Diagram) of a Physical Change There are energy changes during physical changes - change in state or phase change. During the heating of solid, liquid and gaseous water there is a change in temperature. During the change in state temperature remains constant, the system is still absorbing energy to overcome the intermolecular forces. These heats are hidden (can not see the change with a thermometer); thus the label latent (hidden). The latent heat of fusion is the energy change that occurs during the phase change between solid and liquid. The latent heat of vaporization is the energy change that occurs during the phase change between liquid and gas. Molar heat of fusion is the amount of energy required to melt one mole of substance. Molar heat of vaporization is the amount of energy required to vaporize one mole of substance. Fusion (melting) the substance absorbs energy, ΔH fus is positive Freezing the substance releases energy, ΔH solid is negative ΔH fus = - ΔH solid Vaporization the substance absorbs energy, ΔH vap is positive Condensation the substance releases energy, ΔH cond is negative ΔH vap = - ΔH cond To calculate the energy for the phase change you need to multiply the number of moles (n) by the molar heat for the phase change: q = nδh phase
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 5 Calorimetry Calorimetry is the measurement of the quantity of heat exchanged. A calorimeter is a thermally insulated container where a reaction system is contained and the energy exchange between the system and its surroundings (environment) can be measured. The calorimeter and its contents (usually water) are considered the surroundings. For example, if the energy from an exothermic chemical reaction is absorbed in a container of water, the change in temperature of the water provides a measure of the amount of heat added. Calorimeters are used to determine the energy content of foods by burning the foods in an oxygen atmosphere and measuring the energy yield in terms of the increase in temperature of the calorimeter. The reaction system is a chemical or physical process that occurs within the calorimeter. The amount of energy exchanged between the surroundings and the reaction system can be determined by calorimetric calculations. The specific heat constant (c) of a substance is defined as the energy required to raise the temperature of one gram of a substance by one Celsius degree. Every substance has a characteristic specific heat constant. The relationship between heat and temperature change is usually expressed in the equation below where c is the specific heat constant. The relationship does not apply if a phase change is encountered; because the heat added or removed during a phase change does not change the temperature. The specific heat constant of water is about 4.19 J/gramCº. The First Law of Thermodynamic says that the heat energy of the system is equal to the negative heat energy of the surroundings. q surroundings = - q system The heat, q, entering or exiting can be determined by using the heat equation for the surroundings. q (heat) = (mass) (specific heat) (final temperature - initial temperature) (q = mcδt) The change in heat content during a chemical reaction (change) or a physical change is measured indirectly, by using the energy changes in the surroundings. In a simple coffee cup calorimeter the water (surroundings) changes in temperature from the energy change of the reaction (the system). The thermometer measures the initial and final temperature of H 2 O. The temperature of the water changes as a result of the energy change in the system.
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 6 Question: A sample of 100.grams of water is placed in a coffee cup calorimeter at a temperature of 20.0ºC. Ammonium nitrate is dissolved in the water lowering the temperature of the water to 10.0ºC. Calculate the heat change, q, for the water. Specific heat of water = 4.19 J/gCº. Solution: The variables are: mass of water = 100.grams t i of water = 20.0ºC t f of water = 10.0ºC q = mcδt q water = (100.g) (4.19 J/gCº) (10.0ºC 20.0ºC) = -4190 J The q water is negative indicating that the heat flows from the surroundings toward the solution process. The solution process q = +4190 J and is described as an endothermic heat of solution. H solution = +4190 J or +4.190 kj An equation that includes the heat change is a thermochemical equation. Here are a couple of examples: NH 4 NO 3 (s) + 25kJ NH 4 NO 3 (aq) NaOH (s) NaOH (aq) + 44.2 kj Another way to show heat change is to use the ΔH notation, the heat change for the system is shown outside the equation. A positive ΔH represents an endothermic process (system increases in energy). A negative ΔH represents an exothermic process (system decreases in
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 7 energy). Enthalpy or ΔH is written in terms of the heat change of the system. Enthalpy Enthalpy, H, is a state function used to describe the heat changes that occur in a reaction under constant pressure. When a reaction is allowed to take place in an open container, a quantity of heat proportional to the quantity of matter present, will be released or absorbed. This flow of heat is the enthalpy change, H. The units for H are kj (or kj/mol). Reactions that release heat are termed exothermic, they have negative values of H. Reactions that absorb heat are termed endothermic, they have positive values of H. Generally, processes that feel cold (like an ice pack) are endothermic and processes that feel hot (like a fire) are exothermic. Examples of Potential Energy Diagrams 1. Potential Energy Profile (Diagram) for a Phase Change NH 4 NO 3 (s) NH 4 NO 3 (aq) ΔH= +25kJ NaOH (s) NaOH (aq) ΔH= -44.2kJ
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 8 2. Energy Profile for a Chemical Change (SATP 25 ºC and 101kPa). The reactants and products are at SATP. To show that the ΔH was measured at SATP we use the symbol ΔHº. CaO (s) + H 2 O (l) Ca(OH) 2 (s) + 65.2 kj (exothermic) Another way to show heat change is to use the ΔH notation, the heat change for the system is shown outside the equation. A negative ΔH represents an exothermic process (system decreases in energy). CaO (s) + H 2 O (l) Ca(OH) 2 (s) ΔHº = -65.2 kj/mol This can be represented by the energy profile: 2NaHCO 3 (s) + 129 kj Na 2 CO 3 (s) + H 2 O (g) + CO 2 (g) (endothermic) 2NaHCO 3 (s) Na 2 CO 3 (s) + H 2 O (g) + CO 2 (g) ΔHº = +129 kj
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 9 A positive ΔH represents an endothermic process (system increases in energy). This reaction can be represented by the energy profile: Enthalpy Calculations The endothermic reaction shown below indicates that 92.2 kj are absorbed when 2 moles of NH 3 decompose to form 1 mole of N 2 and 3 moles of H 2. Question: Solution: 2NH 3 (g) N 2 (g) + 3H 2 (g) H = +92.2 kj or 92.2 kj + 2NH 3 (g) N 2 (g) + 3H 2 (g) In the reaction above, 100 g of NH 3 are allowed to react to produce N 2 and H 2, how many kj of heat will be absorbed? The reaction indicates that for every 2 moles of NH 3 consumed, 92.9 kj will be absorbed. This information will be used in the calculation below to convert from moles of NH 3 to kj. 100 g NH 3 X 1 mol NH 3 X _92.2 kj_ = 271 kj 17 g NH 3 2 mol NH 3 When a reaction is reversed, the sign of H is changed: N 2 (g) + 3H 2 (g) 2NH 3 (g) H = -92.2 kj
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 10 When a reaction is multiplied through by a number, H is multiplied by that same number: ½[N 2 (g) + 3H 2 (g) 2NH 3 (g) ½N 2 (g) + 3/2H 2 (g) NH 3 (g) H = -92.2 kj] H = -46.1 kj Hess s Law Hess s law states that the enthalpies of reactions may be added when these reactions are added. Substances appearing on the same side are added, while those on opposite sides are subtracted. Typically, some reactions will need to be reversed and multiplied through by a number, so that when they are combined the desired equation will result. Question: Solution: Find the enthalpy of the equation: C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O (l) given the following information: (1) 2C (graphite) + H 2 (g) C 2 H 2 (g) H = +226.7 kj (2) C (graphite) + O 2 (g) CO 2 (g) H = -393.5 kj (3) H 2 + ½O 2 (g) H 2 O (l) H = -285.8 kj Since C 2 H 2 (g) appears only in equation (1), it is necessary to reverse equation (1) so that C 2 H 2 appears on the left hand side to correspond to the desired equation. Since CO 2 (g) appears only in equation (2), it is necessary to multiply equation (2) by two, so that two CO 2 s will appear on the right hand side. Since H 2 O (l) appears only in equation (3), there is no need to alter this equation since one H 2 O appears on the right hand side of equation (3) and this is exactly what is required in the desired equation. C 2 H 2 (g) 2C (graphite) + H 2 (g) 2C (graphite) + 2O 2 (g) 2CO 2 (g) H = -226.7 kj H = -787.0 kj H 2 + ½O 2 (g) H 2 O (l) H = -285.8 kj C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O (l) H = -1299.5 kj
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 11 Standard Heat of Formation A substance in its most stable form at one atmosphere is said to be in its standard state (the temperature need not be 25 0 C, but this temperature is most commonly used to tabulate data). The enthalpy change which occurs when one mole of a compound is produced from its elements in their standard states is called the standard heat of formation, H 0 f. The equation for the standard heat of formation for ammonia, NH 3, is given below: ½N 2 (g) + 3/2H 2 (g) NH 3 (g) H 0 f = -46.1 kj By definition, all elements in their standard states have a H 0 f equal to zero. Standard Heat of Reaction When all substances in a chemical reaction are in their standard states, the enthalpy change is called the standard heat of reaction, H 0 reaction. The standard heat of reaction may be found by taking the sum, Σ, of the standard heats of formation of the products, and subtract them from the sum of the H 0 f s of the reactants. The coefficients in front of each substance must be multiplied by their respective H 0 f s. Question: Solution: H 0 reaction = [Σ H 0 f (products)] - [Σ H 0 f (reactants)] Using the above formula, find the enthalpy of the equation: C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O (l) given the following information: (1) 2C (graphite) + H 2 (g) C 2 H 2 (g) H = +226.7 kj/mol (2) C (graphite) + O 2 (g) CO 2 (g) H = -393.5 kj/mol (3) H 2 + ½O 2 (g) H 2 O (l) H = -285.8 kj/mol If we let (CO 2 ), (H 2 O), (C 2 H 2 ) and (O 2 ) represent the H 0 f of CO 2, H 2 O, C 2 H 2 and O 2, respectively, we get: H 0 reaction = [ 2(CO 2 ) + 1(H 2 O) ] - [ 1(C 2 H 2 ) + 5/2(O 2 ) ] H 0 reaction = [ 2(-393.5) + 1(-285.8) ] - [ 1(+226.7) + 5/2(zero) ] H 0 reaction = -1299.5 kj
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 12 Bond Dissociation Energy The change in enthalpy required to break a bond is called the bond dissociation energy (or bond energy). When bonds are broken, energy is absorbed. when bonds are formed, energy is released. Here s the equation for the bond dissociation of hydrogen: H 2 (g) 2H (g) H = +435 kj The larger the bond energy, the greater the bond s strength and the shorter the bond s length. The enthalpy of reaction may be determined by subtracting all the bond energies associated with products from those associated with the reactants: H 0 reaction = [Σ bond energies (reactants)] - [Σ bond energies (products)] Note that in this equation, it is the bond energies of the products which are being subtracted from the reactants. Question: Solution: Using the above formula, estimate the enthalpy of the equation: C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O (l) bond energies are given below in kj/mol: C C H C H O C=O C=C C C O=O 347 414 464 803 611 837 498 First we need to work out the structural formulas to clearly identify the number and types of bonds: H C C H + 5/2 O=O 2 O=C=O + H O H Let s list the bonds with bond energies indicated with parenthesis, and coefficients indicating the number of bonds of each type: H 0 reaction = [2(H C) + 1(C C) + 5/2 (O=O)] - [4(O=C) + 2(H O)] Finally, let s put the numerical values in, then crunch the numbers: H 0 reaction = [2(414) + 1(837) + 5/2 (498)] - [4(803) + 2(464)] H 0 reaction = -1230. kj
Mr Chiasson Advanced Chemistry 12 / Chemistry 12 13 Note that two moles of CO 2, contain four moles of C=O bonds. Note that the bond energy of three C C bonds does not equal the bond energy of one C C. Since bond energies are generally averages taken from many molecules, the result obtained for the above question is less accurate than in the two previous methods we used to determine the H.