Chemistry: A Molecular Approach - Chapter 7 - Thermochemistry PDF

Chemistry: A Molecular Approach Chapter 7 Thermochemistry This chapter introduces you to thermochemistry, a branch of chemistry that describes the energy changes that occur during chemical reactions. Nature of Energy Energy is defined as the capacity to do work or to produce heat. W...

Chemistry: A Molecular Approach Chapter 7 Thermochemistry This chapter introduces you to thermochemistry, a branch of chemistry that describes the energy changes that occur during chemical reactions. Nature of Energy Energy is defined as the capacity to do work or to produce heat. Work (w) is a force acting over a distance. – Energy = work = force × distance Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them. – only changes can be detected! – flows from warmer → cooler object Classification of Energy Potential energy is energy that is stored in an object, or energy associated with the position of the object. A cat sitting on the highest branch of a tree has what is known as potential energy. If he falls off the branch and falls to the ground, his potential energy is now being converted into kinetic energy. Electrical energy, nuclear energy, and chemical energy are different forms of potential energy  Kinetic energy-the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant Exothermic vs Endothermic Endothermic – Heat flowing into a system from its surroundings –Surroundings get cooler but the system gets warmer –q has a positive value Exothermic - Heat flowing out of a system into it’s surroundings –System gets colder but the surroundings get warmer –q has a negative value  In an endothermic reaction: The reaction vessel cools. Heat is absorbed. Energy is added to the system. q is positive. melting boiling sublimation In an exothermic reaction: The reaction vessel warms. Heat is evolved. Energy is subtracted from the system. q is negative. freezing condensation deposition Units of Energy ▪ Measured with 2 common units calorie – quantity of heat needed to raise the temperature of 1 g of pure water 1 oC. Joule - SI unit of heat and energy James Prescott ▪ How are the units similar? Joule (1818-1889) 1 cal = 4.184 J ▪ 1 Calorie (Cal) or kilocalorie (kcal) = 1000 cal = 4184 J Conservation of Energy ▪ The law of conservation of energy states that energy cannot be created or destroyed. ▪ When energy is transferred between objects, or converted from one form to another, the total amount of energy present at the beginning must be present at the end. ▪ To heat your home, chemical energy needs to be converted to heat. ▪ Sunlight is converted to chemical energy in green plants. The First Law of Thermodynamics: Law of Conservation of Energy ▪ Thermodynamics is the study of energy and its interconversions. ▪ The first law of thermodynamics is the law of conservation of energy. ▪ This means that the total amount of energy in the universe is constant. ▪ Therefore, you can never design a system that will continue to produce energy without some source of energy. ▪ Humans can convert the chemical energy in food, like this ice cream cone, into kinetic energy by riding a bicycle. ▪ Plants can convert electromagnetic radiation (light energy) from the sun into chemical energy. Internal Energy The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system. The change in the internal energy of a system depends only on the amount of energy in the system at the beginning and end. – A state function is a mathematical function whose result depends only on the initial and final conditions, not on the process used. E = E final − Einitial Erxn = Eproducts − Ereactants State functions ▪ State functions do not depend on the path taken ▪ Suppose you have $1000 in your savings account. You withdraw $500 from your savings account. It does not matter whether you withdraw the $500 in one shot or whether you do so at a rate of $50. ▪ At the end when you receive your monthly statement, you will notice a net withdrawal of $500 and will see your resulting balance as $500. ▪ Thus, the bank balance is a state function because it does not depend on the path or way taken to withdraw or deposit money. State functions ▪ Temperature, pressure, volume, and potential energy are all state functions. ▪ The temperature of an oven is independent of however many steps it may have taken for it to reach that temperature. ▪ Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. ▪ Heat and work are not state functions because they are path dependent. ▪ For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students ▪ The amount of work expended to get it there, however, can differ greatly depending on the path chosen. ▪ If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp. Energy Exchange ▪ The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system ▪ Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy ΔE of the system is the sum of the heat transferred (q) and the work done (w): ΔE = q + w  Conventions of q and w q > 0: Heat is transferred from the surrounding to the system q < 0: Heat is transferred from the system to the surrounding w > 0: Work is done by the surroundings on the system It pushed a piston, or it expanded a balloon or something. w < 0: Work is done by the system on the system (you push a piston back in...or you squeeze a balloon back down.) Calculate the overall change in internal energy for a system that absorbs 125 J of heat and does 141 J of work on the surroundings. q is + (heat absorbed) w is − (work done) E = q + w = (+125 J) + (−141J) = −16 J Quantity of Heat Energy Absorbed: Heat Capacity ▪ When a system absorbs heat, its temperature increases. ▪ The increase in temperature is directly proportional to the amount of heat absorbed. ▪ The proportionality constant is called the heat capacity, C. – Units of C are J/°C or J/K. elsius ▪ The greater the capacity of a body, the more heat it requires to raise its temperature. ∆T = Tfinal - Tinitial Specific Heat Capacity Specific Heat Capacity, C sub s (J/g dot degree Celsius asterisk Cs (J / g · C) * The specific heat capacity Substance Elements Blank is the amount of heat Lead 0.128 energy required to raise the Gold 0.128 temperature of one gram of Silver 0.235 a substance by 1 degree Copper 0.385 Celsius (°C). Iron 0.449 Aluminum 0.903 𝐉/(𝐠 ⋅ °𝐂𝐞𝐥𝐬𝐢𝐮𝐬) Compounds blank Ethanol 2.42 Water 4.18 Materials blank Glass (Pyrex) 0.75 Granite 0.79 Sand 0.84 *At 298 K.  Some things heat up or cool down faster than others. Land heats up and cools down faster than water C water = 4.18 J / g 0C C sand = 0.84 J / g 0C Water has a very high specific heat. That means it needs to absorb a lot of energy before its temperature changes. Sand and asphalt, on the other hand, have lower specific heats. This means that their temperatures change more quickly.  Quantifying Heat Energy ▪ The heat capacity of an object is proportional to the following: ▪ Its mass ▪ The specific heat of the material ▪ So, we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object. ▪In order to understand two important concepts related to heat capacity, imagine putting a steel saucepan on a kitchen flame. ▪The saucepan’s temperature rises rapidly as it absorbs heat from the flame. However, if you add some water to the saucepan, the temperature rises more slowly. Why? ▪The first reason is that when you add the water, the same amount of heat must now warm more matter, so the temperature rises more slowly. In other words, heat capacity is an extensive property—it depends on the amount of matter being heated  If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? Cs(Al)= 0.901 J/g x oC q = m × Cs × ΔT ∆T = Tfinal - Tinitial q = (25.0 g)(0.901 J/g x oC)(37 - 310) oC q = - 6149 J= -6.149 kJ Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al. Suppose you find a penny (minted before 1982, when pennies were almost entirely copper) in the snow. How much heat is absorbed by the penny as it warms from the temperature of the snow, which is –8.0 ºC, to the temperature of your body, 37.0 ºC? Assume the penny is pure copper and has a mass of 3.10 g. Specific heat of copper is 0.385 J/g ºC. q= 0.385 J/g ºC x 3.10 g x (37.0 + 8.0) ºC = 53.7J Determine the final temperature when a 25.0g piece of iron at 85.0°C is placed into 75.0grams of water at 20.0°C. The specific heat of iron is 0.45 J/g°C. The specific heat of water is 4.18 J/g°C. a) The colder water will warm up (heat energy "flows" into it). The warmer metal will cool down (heat energy "flows" out of it). b) The whole mixture will wind up at the SAME temperature. c) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water) − qmetal = qwater − mCΔTmetal = mCΔTwater Determine the final temperature when a 25.0g piece of iron at 85.0°C is placed into 75.0 grams of water at 20.0°C. The specific heat of iron is 0.45 J/g°C. The specific heat of water is 4.18 J/g°C. − mCΔTmetal = mCΔTwater ∆T = Tfinal - Tinitial So, by substitution, we then have: − (25.0) (0.45) (x − 85.0)= (75.0) (x − 20.0) (4.18) Solve for x The answer is 22.25 °C=22 °C Enthalpy ▪ The enthalpy, H, deals with the amount of heat absorbed or released during a chemical reaction under constant pressure. ▪ The enthalpy change, Δ𝐻rxn , of a reaction is the heat involved in a reaction at constant pressure. Δ𝐻rxn = 𝑞reaction at constant pressure ▪ When ΔH is positive, heat is being absorbed by the system. ▪ This is called an endothermic reaction. ▪ When ΔH is negative, heat is being released by the system. ▪ This is called an exothermic reaction. Identify each process as endothermic or exothermic and indicate the sign of ΔH. 1. sweat evaporating from skin 2. water freezing in a freezer 3. wood burning in a fire 1. Sweat evaporating from skin cools the skin and is therefore endothermic, with a positive ΔH: The skin must supply heat to the perspiration in order for it to continue to evaporate. 2.Water freezing in a freezer releases heat and is therefore exothermic, with a negative ΔH: The refrigeration system in the freezer must remove this heat for the water to continue to freeze. 3.Wood burning in a fire releases heat and is therefore exothermic, with a negative ΔH. Stoichiometry Involving ΔH: Thermochemical Equations The enthalpy change for a chemical reaction, abbreviated ΔHrxn, is also called the enthalpy of reaction or heat of reaction and depends on the amount of material undergoing the reaction. Determine the amount evolved when 9.07 ×105 g of ammonia is produced according to the following equation. Assume that the reaction occurs at constant pressure. N2 𝑔 + 3H2 𝑔 2NH3 𝑔 ; Δ𝐻rxn = –91.8 kJ  N2 𝑔 + 3H2 𝑔 2NH3 𝑔 ; Δ𝐻rxn = –91.8 kJ The conversion factor is obtained from the second step of the thermodynamic equation. The amount of heat evolved is -2.45 ×106 kJ 1 mol NH3 –91.8 kJ 9.07  10 g NH3  5  = – 2.45  106 kJ 17.0 g NH3 2 mol NH3  Hess’s Law (Father of thermochemistry) Russian chemist (1802–1850) ▪ The change of enthalpy in a chemical reaction is independent of the route by which the chemical change occurs. ▪This is true because enthalpy is a state function, which is a value that does not depend on the path taken. ▪ If a reaction can be expressed as a series of steps, then the ΔHrxn for the overall reaction is the sum of the heats of reaction for each step. Hess’s Law ▪If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. ▪Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations. ▪For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written: ▪C(s)+O2(g)⟶CO2(g) ΔH°=−394kJ For example: C + O2 → CO2 occurs as 2 steps C+ ½O2 → CO H = – 110.5 kJ CO + ½O2 → CO2 H = – 283.0 kJ C + CO + O2 → CO + CO2 H = – 393.5 kJ C + O2 → CO2 H = – 393.5 kJ Standard Conditions ▪ The standard state is the state of a material at a defined set of conditions. ▪ Pure gas at exactly 1 atm pressure ▪ Pure solid or liquid in its most stable form at exactly ▪ 1 atm pressure and temperature of interest ▪ Usually, 25 °Celsius ▪ Substance in a solution with concentration 1 M Standard Conditions  The standard enthalpy of formation,  H f , is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements. The elements must be in their standard states. The  H f ,for a pure element in its standard state = 0 kJ/mol. Standard Enthalpies of Formation  H f , (kJ / mol)  H f , (kJ / mol) delta H naught sub reaction, (k J/mol) delta H naught sub reaction, (k J/mol) Formula Formula Bromine Blank CO(g) −110.5 Br(g) 111.9 CO2(g) −393.5 Br2(l) 0 CH4(g) −74.6 HBr(g) −36.3 CH3OH(l) −238.6 Calcium Blank C2H2(g) 227.4 Ca(s) 0 C2H4(g) 52.4 CaO(s) −634.9 C2H6(g) −84.68 CaCO3(s) −1207.6 C2H5O H(l) −277.6 Carbon Blank C3H8(g) −103.85 C(s, graphite) 0 C3H6O(l, acetone) −248.4 C(s, diamond) 1.88 C3H8O(l, isopropanol) −318. Calculating Standard Enthalpy Change for a Reaction ▪ The standard enthalpy change of reaction ΔH °𝒓𝒙𝒏 ▪ is equal to the sum of the standard enthalpies  H f ▪ of formation of the products minus the sum of the standard enthalpies of formation of the reactants. ΔH °𝒓𝒙𝒏 = Σnp ΔH ºf (products) – Σnr ΔH ºf (reactants) Δ - change in enthalpy o - degree signifies that it's a standard enthalpy change. f - The f indicates that the substance is formed from its elements  means sum n is the coefficient of the reaction Use the standard enthalpies of formation to determine ΔH º for the reaction: To calculate ΔHrxn º from standard enthalpies of formation, subtract the heats of formation of the reactants multiplied by their stoichiometric coefficients from the heats of formation of the products multiplied by their stoichiometric coefficients. ΔH °𝒓𝒙𝒏 = Σnp ΔH ºf (products) – Σnr ΔH ºf (reactants) The answer is negative, which means that the reaction is exothermic. When a reaction is multiplied by a factor, ΔHrxn is multiplied by that factor. C ( s ) + O2 ( g ) → CO2 ( g ) H = − 393.5 kJ 2 C ( s ) + 2 O2 ( g ) → 2 CO2 ( g ) H = 2 ( −393.5 kJ ) = − 787.0 kJ. If a reaction is reversed, then the sign of ΔH is changed. CO2 ( g ) → C ( s ) + O2 ( g ) H = + 393.5 kJ ▪ Consider the following data: 1 3 NH3 (g ) ⎯⎯ → N2 (g ) + H2 ( g ) H = 46 kJ 2 2 2 H2 (g ) + O2 (g ) ⎯⎯ → 2 H2O( g ) H = − 484 kJ ▪ Calculate ΔHrxn for the reaction 2 N2 (g ) + 6 H2O(g ) ⎯⎯ → 3 O2 (g ) + 4 NH3 (g )  1 3 NH3 (g ) ⎯⎯ → N2 (g ) + H2 ( g ) H = 46 kJ 2 2 2 H2 (g ) + O2 (g ) ⎯⎯ → 2 H2O( g ) H = − 484 kJ Reverse the two reactions: 1 3 N2 (g ) + H2 (g ) ⎯⎯ → NH3 (g ) H = − 46 kJ 2 2 2 H2O(g ) ⎯⎯ → 2 H2 (g ) + O2 ( g ) H = +484 kJ Desired reaction: 2 N2 (g ) + 6 H2O(g ) ⎯⎯ → 3 O2 (g ) + 4 NH3 (g )  1 3 N2 (g ) + H2 (g ) ⎯⎯ → NH3 (g ) H = − 46 kJ 2 2 2 H2O(g ) ⎯⎯ → 2 H2 (g ) + O2 ( g ) H = +484 kJ Multiply reactions to give the correct numbers of reactants and products: 4(1 N2 (g ) + 3 H2 (g ) ⎯⎯→ NH3 (g )) 4( H = − 46 kJ) 2 2 3(2 H2O(g ) ⎯⎯→ 2 H2 (g ) + O2 (g )) 3(H = +484 kJ ) Desired reaction: 2 N2 (g ) + 6 H2O(g ) ⎯⎯ → 3 O2 (g ) + 4 NH3 (g ) Final reactions: 2 N2 ( g ) + 6 H2 ( g ) ⎯⎯ → 4 NH3 ( g ) H = − 184 kJ 6 H2O( g ) ⎯⎯ → 6 H2 ( g ) + 3 O 2 ( g ) H = +1452 kJ Desired reaction: 2 N2 (g ) + 6 H2O(g ) ⎯⎯ → 3 O2 (g ) + 4 NH3 (g ) Δ Hrxn =-184+1452= +1268 kJ  Test preparation The combustion of isopropyl alcohol, common rubbing alcohol, is represented by the equation 2 (CH3)2CHOH(l) + 9 O2(g) -> 6 CO2(g) + 8 H2O(l) ΔHrxn°= –4011 kJ. What is ΔHf°of isopropyl alcohol (CH3)2CHOH? ΔHf° [CO2(g)] = −393.5 kJ/mol ΔHf° [H2O(l)] = −285.8 kJ/mol ΔHrxn°= 6(-393.5) + 8(-285.8) – [9(0) +2(ΔHf°isopr)]= - 4011kJ ΔHf°isopr = -318 kJ/mol  Given the following equation, C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔHrxn°= − 2803 kJ calculate the energy released when 45.00 g of glucose is burned in oxygen. 1 mol C6 H12O6 - 2803 kJ 45.00 g C6 H12O6   = −700.0 kJ 180.2 g C6 H12O6 1 mol C6 H12O6 Given the thermochemical equation 2SO2(g) + O2(g) → 2SO3(g), ΔHrxn°= -198 kJ, how much heat is evolved when 600. g of SO2 is burned? Ans: - 928 kJ Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔHrxn°= = −3857 kJ.  f H° [CO2(g)] = −393.5 kJ/mol and  f H° [H2O(l)] = −285.8 kJ/mol, calculate the enthalpy of formation of glycine. Ans.(-537.3 kJ)