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JF Basic Chemistry Tutorial Thermodynamics

Shane Plunkett plunkes@tcd.ie. Gas Laws and the behaviour of gases Enthalpy - Exothermic and endothermic reactions - Hess ’ s Law - Calorimetry - Bond enthalpy Entropy - Second and Third Laws of Thermodynamics 3. Spontaneity and Gibbs Free Energy.

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JF Basic Chemistry Tutorial Thermodynamics

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  1. Shane Plunkett plunkes@tcd.ie • Gas Laws and the behaviour of gases • Enthalpy • - Exothermic and endothermic reactions • - Hess’s Law • - Calorimetry • - Bond enthalpy • Entropy • - Second and Third Laws of Thermodynamics • 3. Spontaneity and Gibbs Free Energy JF Basic Chemistry Tutorial Thermodynamics • Recommended reading • M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change, 3rd Ed., Chapter 6 • P. Atkins & L. Jones, Molecules, Matter and Change, 3rd Ed., Chapter 2 • Multiple choice tests: http://www.mhhe.com/silberberg6

  2. Charles’ Law Laws of Thermodynamics - relationships At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to the temperature V  T P and n fixed Boyle’s Law At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the applied (external) pressure V  1/P T and n fixed Avogadro’s Law At a fixed temperature and pressure, equal volumes of any gas contain equal numbers of particles (or moles) T and P fixed V  n Ideal Gas Law Combination of these laws… PV = nRT R is the universal gas constant

  3. Gas Law Problem A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21oC. PV = nRT Don’t forget to change the units!! V = 438 L R = 0.0821 atmLmol-1K-1 n = number of moles Given mass = 0.885 kg n = mass in g/molar mass of O2 n = 885g/32gmol-1 = 27.7 mol of O2 T = 21oC = 294K P = nRT/V P = (27.7mol)(0.0821 atmLmol-1K-1)(294K) (438L) P = 1.53 atm

  4. Behaviour of gases Dalton’s Law of Partial Pressure • Gases mix homogeneously in any proportions • Each gas in a mixture behaves as if it were the only gas present (as long as there is no chemical reaction) Dalton’s Law: In a mixture of unreacting gases, the total pressure is the sum of the partial pressures of the individual gases Ptotal = P1 + P2 + P3 + …. Each component in a mixture contributes a fraction of the total number of moles in the mixture….the mole fraction (X) of that component Pa = Xa x Ptotal Where a is the one component you are interested in Xa = na ntotal

  5. Question A mixture of noble gases consisting of 5.5 g of He, 15.0 g of Ne and 35.0 g of Kr is place in an tank at STP. Calculate the partial pressure of each gas. Xa = na ntotal Pa = Xa x Ptotal The gases are noble gases, therefore they are stable (inert) and will not react with each other. Need to work out mole fraction (X) of each gas. Molar mass of He = 4.003 gmol-1 Mass of He = 5.5 g nHe = 1.37 mol XHe = nHe / ntotal = 1.37 / 2.53 = 0.54 PHe = XHe x Ptotal =0.54 x 1 atm = 0.54 atm XNe = nNe / ntotal = 0.74 / 2.53 = 0.29 PNe = XHe x Ptotal = 0.29 x 1 atm = 0.29 atm XKr = nKr / ntotal = 0.42 / 2.53 = 0.17 PKr = XKr x Ptotal = 0.17 x 1 atm = 0.17 atm Molar mass of Ne = 20.18 gmol-1 Mass of Ne = 15.0 g nNe = 0.74 mol Molar mass of Kr = 83.8 gmol-1 Mass of Kr = 35.0 g nKr = 0.42 mol ntotal = nHe + nNe + nKr = 1.37 mol + 0.74 mol + 0.42 mol = 2.53 mol Ptotal = 1 atm

  6. Behaviour of gases Kinetic-Molecular Theory An observation of the behaviour of gases at the molecular level….looking at their motion and their speed; also helps us to make sense of the gas laws. The theory is based on 3 assumptions. • Particle volume. Any gas consists of a large amount of individual particles. The volume of an individual particle is extremely small compared with the volume of the container – therefore the theory says that the gas particles have mass but no volume. • Particle motion. Gas particles are in contant, random, straight line motion, except when they collide with the container walls or each other. • Particle collisions. Collisions are elastic. This means that the colliding molecules exchange energy (but do not lose energy through friction). Therefore their total kinetic energy (Ek)is contant. In between collisions, the molecules do not influence each other at all.

  7. Behaviour of gases The average kinetic evergy of the molecules is proportional to the absolute temperature. At a given temperature, the molecules of all gases have the same average kinetic energy. Therefore, if two different gases are at the same temperature, their molecules have the same average kinetic energy. If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled. Ek T

  8. Behaviour of gases Although the molecules in a sample of gas have an average kinetic energy and therefore an average speed, the individual molecules move at various speeds, i.e. they exhibit a distribution of speeds – some move fast, others relatively slowly. Collisions can change individual molecular speeds but the distribution of speeds remains the same. Ek T

  9. Behaviour of gases At the same temperature, lighter gases move on average faster than heavier gases.

  10. Behaviour of gases Mean Free Path The mean free path (λ) of any molecule is the average distance travelled by a molecule between two successive collisions.

  11. Every time we carry out a reaction in the lab, we get a change in energy taking place. We study thermodynamics in order to keep track of these energy changes. Open system -can exchange mass or energy with surroundings Our first task is to identify the system in the universe we are interested in observing Surroundings Closed system -can exchange only energy with surroundings System ΔEsys + ΔEsurr = 0 = ΔEuniv × Isolated system - No transfer of either energy or mass Law of conservation of energy states that energy cannot be made or destroyed, but can be coverted from one form to another Conclusion: The total energy of the universe remains constant The First Law of Thermodynamics

  12. Enthalpy: method for measuring energy changes during chemical reactions Standard reaction enthalpy, ΔH°rxn: change in enthalpy when reactants in their standard states are changed to products in their standard states At constant pressure, the change in enthalpy of a system is the amount of energy released or absorbed Standard enthalpy of formation, ΔH°form: change in enthalpy when one mole of product is formed from reactants in their standard states Standard enthalpy of combustion, ΔH°comb: change in enthalpy when one mole of reactant is burned completely in oxygen reactants • Exothermic reactions • release heat to their surroundings • negative enthalpy change, ΔH < 0 Heat given out products • Endothermic reactions • take in heat from their surroundings • positive enthalpy change, ΔH > 0 reactants Heat taken in products

  13. Hess’s Law Example Given that for the combustion of glucose C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔH = -2816 kJ and for the combustion of ethanol, C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)ΔH = -1372 kJ Calculate ΔH (in kJ) for the fermentation of glucose: C6H12O6(s) 2C2H5OH(l) + 2CO2(g)ΔH = ? • The overall reaction enthalpy is equal to the sum of the individual enthalpies for the reactions which make it up C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)ΔH = -2816 kJ 4CO2(g) + 6H2O(l)2C2H5OH(l) + 6O2(g) ΔH = +2744 kJ C6H12O6(s) 2C2H5OH(l) + 2CO2(g)ΔH = -72 kJ Exothermic reaction

  14. Example Step 1: Write equation for reactions given Enthalpy of formation: 3Pb(s) + 2O2(g) Pb3O4(s)ΔH°f = -175.6 kJ mol-1 Given that ΔH° for formation (ΔH°f) of Pb3O4 is -175.6 kJ mol-1 and ΔH° for the reaction 3PbO2 Pb3O4 + O2 is 22.8 kJ mol-1. What is the ΔH°f for PbO2 (in kJ mol-1)? 3PbO2(s) Pb3O4(s) + O2(g) ΔH° = 22.8 kJ mol-1 Step 2: Write down what you are looking for Enthalpy of formation of PbO2: Pb(s) + O2(g) PbO2ΔH°f = ? Step 3: Combine the equations we know to get the answer we require 3Pb(s) + 2O2(g) Pb3O4(s)ΔH°f = -175.6 kJ mol-1 Pb3O4(s) + O2(g) 3PbO2(s)ΔH° = -22.8 kJ mol-1 3Pb(s) + 3O2(g) 3PbO2(s)ΔH°f = -198.4 kJ mol-1 Pb(s) + O2(g) PbO2(s)ΔH°f = -66.13 kJ mol-1

  15. Question • Given the following information: 2C(s) + 3H2(g) C2H6(g)ΔH = -84.68 kJ mol-1 C(s) + O2(g)  CO2(g)ΔH = -393.51 kJ mol-1 H2(g) + ½O2(g)  H2O(l)ΔH = -285.83 kJ mol-1 Calculate the standard enthalpy of combustion of ethane: C2H6(g) + 3½O2  2CO2(g) + 3H2O(l) Answer: -1559.8 kJ mol-1 • Given the following information: • S(s) + O2(g) SO2(g)ΔH = -296.1 kJ mol-1 • C(s) + O2(g)  CO2(g)ΔH = -393.5 kJ mol-1 • CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)ΔH = -1072 kJ mol-1 • Calculate the enthalpy of formation of carbon disulfide, CS2: • C(s) + 2S(s)  CS2(l) Answer: +86.3 kJ mol-1

  16. Mechanical stirrer Thermometer To electrical source By using a bomb calorimeter in the lab, we can determine the reaction enthalpy Material combusted in oxygen The equation to calculate the heat change is q = m s ΔT where q is the heat change m is the mass of the sample s is the specific heat of the sample ΔT is the temperature change during reaction The specific heat of a substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius

  17. Example q = m s ΔT m = 350g s = 4.184 J g-1°C-1ΔT = (15.0 – 10.5)°C = 4.5°C A sample of 350g of water is heated from 10.5°C to 15.0°C. The specific heat of water is 4.184 J g-1 °C-1. Calculate the heat change. q = (350 g)(4.184 J g-1 °C-1)(4.5 °C) = 6589.8 J = 6.59 kJ Question A 560g sample of mercury is heated from 40°C to 78°C. The specific heat of mercury is 0.139 J g-1 °C-1. What is the heat change for the reaction? Answer: 2.96 kJ A 782g sample of water is cooled from 25°C to 1°C. The specific heat of water is 4.184 J g-1 °C-1. What is the heat change for the reaction? Answer: -78.5 kJ

  18. Bond Enthalpy • Measure of stability of molecule • Enthalpy change required to break a given bond in 1 mole of gaseous molecules • Bond formation: exothermic process, negative sign enthalpy • Bond breakage: endothermic process, positive sign enthalpy • Average values since energy of given bond varies from molecule to molecule, e.g. due to electronegative atoms Example Given the following data: H2(g) + I2(s) 2HI(g)ΔH°= +54 kJ mol-1 H2(g)  2H(g)ΔH°= +436 kJ mol-1 I2(g)  2I(g)ΔH°= +214 kJ mol-1 What is the bond dissociation energy for HI?

  19. Step 1: Draw out structures for each molecule and decide what bonds are broken 1. The equation we are interested in is: ΔH°= +54 kJ mol-1 The information we are given is: 2. 2H ΔH°= +436 kJ mol-1 3. ΔH°= +214 kJ mol-1 2I Bonds broken in equation 1: H H ΔH°= +436 kJ mol-1 I I ΔH°= +214 kJ mol-1 Bonds broken, endothermic, ΔH°= +650 kJ mol-1

  20. Step 2: Draw out structure and decide what bonds are formed ΔH°rxn = +54 kJ mol-1 Bonds formed, exothermic, ΔH°= -2 (H I) ΔH°rxn = ΔH°bonds broken + ΔH°bonds formed +54 kJ mol-1 = 650 kJ mol-1 – 2 ΔH° (H I) 2 ΔH° (H I) = (650 – 54) kJ mol-1 2 ΔH° (H I) = 596 kJ mol-1 ΔH° (H I) = 298 kJ mol-1

  21. Question Given the following data: HCl(g) + F2(g) HF(g) + ClF(g) ΔH°= -232 kJ mol-1 ClF(g)  Cl(g) + F(g) ΔH°= 256 kJ mol-1 HCl(g)  H(g) + Cl(g) ΔH°= 427 kJ mol-1 F2(g)  2F(g)ΔH°= 158 kJ mol-1 What is the bond enthalpy for HF? Answer: +561 kJ mol-1

  22. Entropy • Measure of disorder of a system • May be increased by increasing number of ways of arranging components. Explained by Boltzmann equation: S = k lnW where S = entropy of system k = Boltzmann constant W = number of possible arrangements • Has relationship with spontaneous change. Second Law of Thermodynamics: spontaneous processes (those which occur naturally without any external influence) are accompanied by an increase in entropy of the universe • Absolute entropies may be determined from Third Law of Thermodynamics: At zero degrees Kelvin, the entropy of a perfect crystal is zero • Because this starting point exists, can measure standard molar entropies: entropy change for 1 mol of a pure substance at 1 atm pressure (usually 25°C)

  23. Predicting Entropy Changes • An increase in temperature leads to greater kinetic energy of moving particles, more motion and hence greater S° • Going from solid to liquid to gas (i.e. to less ordered systems) leads to an increase in S° • For spontaneous change, ΔS must be greater than zero. For negative ΔS values, the process is spontaneous in the reverse direction Example Predict whether the entropy change for the following reaction will be positive or negative: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) 6 gas molecules  3 gas molecules + 4 liquid molecules A decrease of the more disordered gas system indicates the entropy change for the reaction should be negative

  24. S° (J / mol K) Given the following information, calculate ΔS° for the reaction CO2 213.7 H2O 69.9 = [(3 mol)(213.7 J/molK) + (4 mol)(69.9 J/molK)] – [(1 mol)(269.9 J/molK) + (5 mol)(205.0 J/molK)] C3H8 269.9 O2 205.0 ΔS°rxn = ΣS°products - ΣS°reactants = [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)] – [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)] = 641.1 J/K + 279.6 J/K – (269.9 J/K + 1025 J/K) = 920.7 J/K – 1294.9 J/K = -374.2 J/K

  25. Question S° (J / mol K) Predict the sign of ΔS° and calculate its value from the following: 2NO(g) + O2(g) 2NO2(g) 210.65 NO O2 205.0 NO2 239.9 Answer: should be negative ΔS° = -146.5 J/K

  26. Spontaneity and Gibbs Free Energy • Gibbs Free energy is a measure of the spontaneity of a process • ΔG is the free energy change for a reaction under standard state conditions • At constant temperature and pressure: ΔG = ΔH – TΔS – an increase in ΔS leads to a decrease in ΔG – if ΔG < 0, the forward reaction is spontaneous – if ΔG > 0, the forward reaction is nonspontaneous – if ΔG = 0, the process is in equilibrium – if ΔH and ΔS are positive, then ΔG will be negative only at high T, i.e. forward reaction spontaneous – if ΔH is positive and ΔS is negative, ΔG will always be positive, i.e. forward reaction nonspontaneous – if ΔH is negative and ΔS is positive, ΔG will always be negative, i.e. forward reaction spontaneous – if ΔH and ΔS are negative, ΔG is negative only at low T, i.e. forward reaction spontaneous

  27. Example ΔG = ΔH – TΔS T = 300 K At 27°C, a reaction has ΔH = +10 kJ mol-1 and ΔS = +30 J K-1 mol-1. What is the value of ΔG? ΔG = (+10 kJ mol-1) – (300 K)(+30 J K-1 mol-1) ΔG = +10 kJ mol-1 – 9000 J mol-1 ΔG = +10 kJ mol-1 – 9 kJ mol-1 ΔG = + 1.0 kJ mol-1 Question For the reaction 4KClO3(s) 3KClO4(s) + KCl(s) Calculate ΔG for the process at 298 K if ΔH = -144.3 kJ and ΔS = -36.8 J K-1 Answer: -133.3 kJ

  28. Kinetics Refers to reaction rates – the speed of the reaction, i.e. the change in concentration of reactants (or products) with respect to time Equilibrium Refers to the extent of the reation – when no further change occurs, i.e. the (final) concentration of the product given unlimited time An equilibrium reaction: Reactants Products Rateforward Ratebackward At equilibrium, the rate of the forward reaction = the rate of the backwards reaction ∆G0 = Standard Gibb’s free energy change R = gas contant T = temperature K = equilibrium constant ∆G0 = -RTlnK Le Chatelier’s Principle When a chemical system at equilibrium is disturbed, the system shifts to counteract the change

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