Chapter 5 Thermochemistry Prof. Dr. Nizam M. El-Ashgar Chemistry Department, IUG

Thermochemistry Thermochemistry: is the study of the relationship between ENERGY and chemical reactions. What is energy? Energy: is the ability to do work or transfer heat. Work: Energy used to cause an object that has mass to move . Heat: Energy used to cause the temperature of an object to rise.

Potential Energy Potential energy: is energy of an object possesses by virtue of its position or chemical composition. chemical energy : stored within structural units of chemical substances.

Potential Energy under the microscope Gravitational forces play little to no role in interactions between atoms and molecules. Predict what plays a much more important role. Electrostatic potential energy, Eel : One of the most important form of P.E. This energy is proportional to electrical charges on two interacting objects, Q1 and Q2, and inversely proportional to the distance, d. Predict an equation.

Electrostatic Potential Energy (Eel) Chemical Energy is Mainly Potential Energy Electrostatic Potential Energy (Eel) The most important form of potential energy in molecules is electrostatic potential energy, Eel: κ is a constant of proportionality = 8.99 x 109 J.m/C2 How can we make electrostatic potential energy absolute zero? Answer: if d is  (infinite)

At finite separation distances for two charged particles: Eel is positive for like charges and negative for opposite charges. As the particles move farther apart, their electrostatic potential energy approaches zero.

Attraction between Ions Electrostatic attraction is seen between oppositely charged ions. Energy is released when chemical bonds are formed; energy is consumed when chemical bonds are broken.

Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion (depends on mass and speed). 1 2 KE =  mv 2

Thermal energy is kinetic energy in the form of random motion of particles in any sample of matter. As temperature rises, thermal energy increases. Heat is energy that causes a change in the thermal energy of a sample. Addition of heat to a sample increases its temperature.

kg m2 1 J = 1  s2 Units of Energy The SI unit of energy is the joule (J). A mass of 2 kg moving at a speed of 1m/s possesses a kinetic energy of 1 J: An older, non-SI unit is still in widespread use: the calorie (cal). Originally defined as: amount of energy require to raise the temperature of 1 g of water from 14.5 C to 15.5 C. 1 cal = 4.184 J exactly In nutrition is the nutritional Calorie (note the capital C): 1 Cal = 1000 cal = 1 kcal 1 J = 1  kg m2 s2

Definitions: System and Surroundings The system: The portion of universe under study. The surroundings: Everything else is called (remain of universe) 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). Note that this is a CLOSED system.

Types of systems Open Closed Isolated Open system: matter and energy can be exchanged with surroundings (ie: boiling water) Closed System: can exchange energy but not matter with surroundings (ie: soft drink bottle) Isolated System: no exchange of matter or energy. (ie: coffee thermos).

Definitions: Force , Work & Heat Force F: is any push or pull exerted on an object. Work w : 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 q: Energy transferred (as heat) from warmer objects to cooler objects.

Example Problem 5.1 A bowler lifts a 5.4 kg (12 lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it. a) What happens to the potential energy of the ball as it is raised? Because the bowling ball is raised to a greater height above the ground, its potential energy increases. b) What is the quantity of work, in J, used to raise the ball? (Note: The force due to gravity is F = m x g, where m is the mass of the object and g is the gravitational constant; g = 9.8 m/s2.)

c) After the ball is dropped, it gains kinetic energy c) After the ball is dropped, it gains kinetic energy. If all the work done in part b has been converted to K.E at the point of impact with the ground, what is the speed of the ball at the point of impact? (5.6 m/s) PE converted to KE:

THE FIRST LAW OF THERMODYNAMICS Energy can neither be created nor destroyed Any energy lost by a system must be gained by the surroundings (vice versa) In other words, the total energy of the universe is a constant (Energy is conserved) Energy can be converted from one type to another. Potential energy is converted to kinetic energy.

Internal Energy: IE The internal energy of a system is the sum of all kinetic (molecular motion) and Potential energies (attractive/repulsive interactions) of all components of the system; we call it E. Note: Numerical values of a systems IE are not known In thermodynamics, we are mainly concerned with the changes in E of the system (ΔE).

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 Efinal & Einitial (can’t be determined exactly) for a practical system we need only E to apply the law To deal with thermodynamic values (Ex. E ) we need three things: 1- number 2- unit 3- sign

Changes in Internal Energy 2 H2(g) + O2(g)  2 H2O(l) If E > 0, Efinal > Einitial Therefore, the system absorbed energy from the surroundings. This energy change is called endothermic E is +ve

Changes in Internal Energy If E < 0, Efinal < Einitial Therefore, the system released energy to the surroundings. This energy change is called exothermic E is -ve

Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). We algebraically represent a change in internal energy by: ∆E = q + w

E, q, w, and Their Signs

Sample Exercise 5.2 ∆E = q + w Two gases, A(g) and B(g), are confined in a cylinder-and-piston arrangement like that in the Figure below. Substances A and B react to form a solid product: A(g) + B(g)  C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? ∆E = q + w q = -1150 J (heat lost from sys, to surr.) W = + 480 J (work done by sys. On surr.) So, ∆E = -1150 J + 480 J = -670 J The system transfers heat to surrounding

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic. Ice melting, Evaporation of acetone Combustion of Gasoline

State Functions Ti Tf T = Tf – Ti = 120 – 50 = 70 oC A state function is a property of a system that is determined only by the present state of the system, not the path that system took. 50 oC 150 oC - 90 oC 30 oC -50 oC 57 oC 39 oC 120 oC Ti Tf T = Tf – Ti = 120 – 50 = 70 oC Examples of state functions: V, E, P, H, n =

State Functions The internal energy of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached room temperature from either direction. And so, E depends only on Einitial and Efinal. , (state function) E = Efinal ─ Einitial

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. Path a: E1 = q1+w1 Path b: E2 = q2+w2 E1 = E2 q1  q2 and w1  w2 q max w max

Work In an open container: The only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

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 (Expansion) Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)

Enthalpy If a process takes place at constant pressure (as the majority of processes we study do), ie open systems. The only work done is this pressure-volume work. We can account for heat flow during the process by measuring the enthalpy (H) of the system Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

Pressure – Volume Work Work involved in expansion or compression of gas When pressure is constant: W = P∆V ∆V is +ve (expansion) and -ve (compression) Q: If a system does not change its volume during the course of a process, does it do pressure- volume work? No work done ie w = 0 because ∆V = 0

qp = E + PV---------------- (2) Enthalpy When the system changes at constant pressure, the change in enthalpy, H, is derived as follows: H = (E + PV) This can be written H = E + PV------------------(1) Since E = q + w and w = -PV, we can substitute these into the enthalpy expression: E = qP -PV qp = E + PV---------------- (2) From (1) and (2) : H = qp So, at constant pressure, the change in enthalpy is the heat gained or lost. H = E + PV

Endothermicity and Exothermicity A process is endothermic when H is positive. A process is exothermic when H is negative.

Enthalpy of Reaction ,H H is a state function In reactions: The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts - Hreactants This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Sample Excercise 5.3 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: An ice cube melts: Endothermic 1 g of butane is combusted in sufficient oxygen to give complete combustion to CO2 and H2O: Exothermic

Thermochemical Reactions A thermochemical equation is a balanced chemical equation that shows the associated enthalpy change but does not specify amount of chemical involved. Consider the thermochemical equation The thermochemical equivalencies are 1 mol C3H8(g) react to give off 2.22×103 kJ 5 mol O2(g) react to give off 2.22×103 kJ 3 mol CO2(g) are formed and give off 2.22×103 kJ 4 mol H2O() are formed and give off 2.22×103 kJ C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O() DH = -2.22×103 kJ Mass of CO2 Moles of CO2 Enthalpy change Molar mass of CO2 Thermochemical equivalent

The Truth about Enthalpy 1- Enthalpy is an extensive property. Directly proportional to the amount of material Ex: CH4(g) + 2O2(g) → CO2(g) + 2H2O() DH = -890 kJ 1 mol CH4(g)  -890 kJ 2 mol CH4(g)  -1780 kJ 2- H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. CO2(g) + 2H2O() → CH4(g) + 2O2(g) DH = +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() DH = -890 kJ CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) DH = -802 kJ As: 2H2O() → 2H2O(g) DH = +88 KJ 4- Multiplying or dividing the equation by a factor (n) the same thing should be for the value of DH 2H2O() → 2H2O(g) DH = +88 KJ H2O() → H2O(g DH = +44 KJ

Sample Exercise 5.4 How much heat is released when 4.50 g of methane gas is burned in a constant pressure system? CH4(g) + 2O2(g) → CO2(g) + 2H2O() DH = -890 kJ 4.50 g ?? Heat 1mol CH4(g)  = -890 kJ

Example: Given the following equation: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH =  2803 kJ/mol calculate the mass in grams of glucose burned in oxygen for releasing 700.0 KJ of energy.

2 H2O2(l)  2 H2O(l) + O2(g) ΔH = –196 kJ Practice Exercise Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2 H2O2(l)  2 H2O(l) + O2(g) ΔH = –196 kJ Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure. Answer: –14.4 kJ

Chemistry 111/112 Chapter 9 Calorimetry We cannot know the exact enthalpy of the reactants and products. We can measure H through calorimetry. Calorimetry is the experimental measurement of heat released or absorbed by a chemical reaction. A calorimeter is the device used to measure heat. The quantity of heat transferred by the reaction causes a change of temperature inside the calorimeter. A simple calorimeter can be made from styrofoam coffee cups. 1 Copyright (c) 2003 Thomson Learning Reprinted with permission

Heat Capacity and Specific Heat Heat capacity C: The amount of energy required to raise the temperature of a substance by 1 K (1C). Specific heat CS: The amount of energy required to raise the temperature of 1 g of a substance by 1 K (1C). Molar heat capacity Cm: The amount of energy required to raise the temperature of 1 mole a substance by 1 K (1C).

Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change cs = q m  T where : q is heat, m is mass, C is heat capacity and T = change in temp (T = Tfinal  Tinitial)‏ Units : Heat capacity, C : J/C or J/K Specific Heat Cs: J/g C or J/g K Molar Heat Capacity Cm: J/mol C or J/mol K

Sample Exercise 5.5 (a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room temperature) to near its boiling point, 98 °C? The specific heat of water is 4.18 J/g-K. (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? Answers: (a) 4.9  105 J, (b) 11 K decrease = 11 °C decrease.

Constant Pressure Calorimetry

Constant Pressure Calorimetry The calorimeter and its contents are the surroundings (qsurr). The chemical reaction is the system (qsys). The law of conservation of energy requires that qsys =  qsurr qsurr = m  cs  T All dilute, aqueous solutions have the same density (1.00 g/mL) and specifi c heat capacity (4.18 J/(g∙°C) as water . Note that: qsys means qrxn and qsurr = qsoln So q rxn = -q soln

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HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Sample Exercise 5.6 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 temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, 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) 50 mL, 1.0M 50 mL, 1.0M Total m:

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) Sample Exercise 5.6 When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 °C to 23.11 °C. The temperature increase is caused by the following reaction:. AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) Calculate ΔH for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g °C. Answer: –68,000 J/mol = –68 kJ/mol

Bomb Calorimetry Constant-Volume Calorimetry Studies Combustion rxn’s May only use after a standardization measurement Difference between ∆E and ∆H is very small q rxn = q cal q cal = qwater + q walls Once you have heat capacity of calorimeter, you may use: qrxn = -Ccal X ∆T

Typical procedure used in a bomb calorimeter Known amount of sample placed in steel container and then filled with oxygen gas Steel chamber submerged in known amount of water Sample ignited electrically Temperature increase of water is determined

Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. H = E + PV As V = 0 So H = E For most reactions, the difference is very small.

Sample Exercise 5.7 The combustion of methylhydrazine with oxygen 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 temperature of the calorimeter increases from 25.00 °C to 39.50 °C. In a separate experiment 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.

Practice Exercise A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 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. Answers: (a) –15.2 kJ/g, (b) –1370 kJ/mol.

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. Used for calculating enthalpy for a reaction that cannot be determined directly

Sample Exercise 5.8 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 CO: Solution:

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: Calculate for the conversion of graphite to diamond: Answer: ΔH3 = +1.9 kJ

Sample Exercise 5.9 2 C(s) + H2(g) → C2H2(g) Calculate ΔH for the reaction: 2 C(s) + H2(g) → C2H2(g) given the following chemical equations and their respective enthalpy changes Solution:

Practice Exercise Calculate ΔH for the reaction NO(g) + O(g) → NO2(g) given the following information: Answer: –304.1 kJ

Enthalpies of Formation 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. Hf depends on: T, P and State (G, L and S, crystalline) of reactants and products.

Standard Enthalpies of Formation Standard enthalpies of formation, Hf°, The enthalpy change that results when 1 mole of a compound is formed from its elements in their standard states (25 °C and 1.00 atm pressure). Hf for an element in its standard state is defined as zero. Hf: O2 (g) = 0 O3 (g )  0 O2 (l)  0 Hf : I2 (s)= 0 I2(g)  0 I2(l)  0 Hf : C(Graph)= 0 C(diam)  0 C(60)  0 Hf : Cl2(g) = 0 Cl2(s)  0 Cl2(l)  0 Values of Hf for substances are found in reference tables Used to calculate the Hrxn

Sample Exercise 5.10 Solution: The eq. Represents Hf of Na2O(s). For which of the following reactions at 25°C would the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose ΔH is an enthalpy of formation? Solution: The eq. Represents Hf of Na2O(s). Does not represent Hf of KCl(s) Correction: K(s) + 1/2Cl2 (g)  KCl(s) c) Does not represent Hf of C6H12O6(s) Correction: 6 C(graphite) + 6 H2(g) + 3 O2(g) →C6H12O6(s)

Practice Exercise Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4). Answer: C(graphite) + 2 Cl2(g) →CCl4(l)

C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Calculation of H We can use Hess’s law in this way: Horxn = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients. Horxn = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)] Horxn = -2220 kJ C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

Sample Exercise 5.11 Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane to that produced by 1.00 g benzene.

Practice Exercise Using the standard enthalpies of formation listed in Table 5.3, calculate the enthalpy change for the combustion of 1 mol of ethanol: Answer: –1367 kJ.

Sample Exercise 5.12 The standard enthalpy change for the reaction: is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) and CO2(g) given in Table 5.3, calculate the standard enthalpy of formation of CaCO3(s).

Practice Exercise Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s): Answer: –156.1 kJ/mol

Bond Enthalpy The enthalpy associated with breaking one mole of a particular bond in a gaseous substance.

Bond Enthalpy The bond enthalpy is always positive because energy is required to break chemical bonds. Energy is always released when a bond forms between gaseous fragments. The greater the bond enthalpy, the stronger the bond.

Bond Enthalpies and Enthalpy of Reaction ADD bond energy for ALL bonds made (+) SUBTRACT bond energy for ALL bonds broken (–) The result is an estimate of ΔH. ΔHrxn = Σ (bond enthalpies – Σ (bond enthalpies of bonds broken) of bonds formed)

So, we can predict whether a chemical reaction will be endothermic or exothermic using bond energies.

Energy in Foods The energy released when one gram of food is combusted is its fuel value. Most of the fuel in the food we eat comes from carbohydrates and fats.

Sample Exercise 5.14 This corresponds to 160 kcal: a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values of these kinds of substances, estimate the energy value (caloric content) of this serving. This corresponds to 160 kcal: (b) A person of average weight uses about 100 Cal/mi when running or jogging. How many servings of this cereal provide the energy value requirements for running 3 mi?

Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels.

Alternative Renewable Energy Sources UNIT 3 Chapter 5: Energy Changes Section 5.4 Alternative Renewable Energy Sources Renewable energy sources in Ontario: account for about 25% of energy production are projected to increase to as high as 40% by 2025 Include: Hydroelectric power: (major source). Wind energy: (currently ~ 1% and projected to 15% in 2025), Solar energy: (currently low but may be as high as 5% in 2025). Much lower contributors: Biomass, wave power, and geothermal energy. Image source: MHR, Chemistry 12 © 2011. ISBN 07-106010-3; page Biomass example

UNIT 3 Chapter 5: Energy Changes Section 5.4 What Is a “Clean” Fuel? Different fuels have differing impacts on the environment. One way this impact is measured is through emissions. For example, CO2(g) emissions per kJ of energy produced. Fuel kg CO2/ kJ energy Anthracite coal 108.83 Oil 78.48 Natural gas 56.03 Nuclear 0.00 Renewables Image source: MHR, Chemistry 12 © 2011. ISBN 07-106010-3; page 333 Table 5.6

HW Problems 24, 25, 30, 36, 37, 39, 41, 42, 43, 44, 45, 48, 51, 53, 55, 57, 59, 63, 65, 69, 72, 73, 78, 80

في هذه القصة عبرة إن رجلاً عجوزاً كان جالسا مع ابن له يبلغ من العمر 25 سنة في القطار. وبدا الكثير من البهجة والفضول على وجه الشاب الذي كان يجلس بجانب النافذة. اخرج يديه من النافذة وشعر بمرور الهواء  وصرخ "أبي انظر جميع الأشجار تسير وراءنا"!! فتبسم الرجل العجوز متماشياً مع فرحة إبنه. وكان يجلس بجانبهم زوجان ويستمعان إلى ما يدور من حديث بين الأب وابنه، وشعروا بقليل من الإحراج فكيف يتصرف شاب في عمر 25 سنة كالطفل!! فجأة صرخ الشاب مرة أخرى:  "أبي، انظر إلى البركة وما فيها من حيوانات،  أنظر..الغيوم تسير مع القطار".. واستمر تعجب الزوجين من حديث الشاب مرة أخرى. ثم بدأ هطول الامطار، وقطرات الماء تتساقط  على يد الشاب، الذي إمتلأ وجهه بالسعادة وصرخ مرة أخرى : "أبي إنها تمطر، والماء لمس يدي، انظر يا أبي". وفي هذه اللحظة لم يستطع الزوجان السكوت وسألا الرجل العجوز: "لماذا لا تقوم بزيارة الطبيب والحصول على علاج لإبنك؟ “ هنا قال الرجل العجوز:" إننا قادمون من المستشفى حيث أن إبني قد أصبح  بصيراً لاول مرة في حياته” تذكر دائماً:  "لا تستخلص النتائج حتى تعرف كل الحقائق"