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University of Alberta
Sai Yiu

UNIVERSITY OF ALBERTA INTRODUCTORY UNIVERSITY CHEMISTRY 2 (CHEM 102) Thermochemistry Introduction Why study thermodynamics? The concepts in thermodynamics can be applied to different branchesof science such as physics, chemistry, biology, engineering e tc. In chemis try, it provides rules and predictions that gove rn chemical and physicaltransitions. Nature of Energy • A system is that part of the region of interest. For example, a gas inside a container, a biological cell, a reaction mixture inside a flask. • The surroundings are regions outside the system. An open system –Exchange of matter and energy with the surroundings are allowed A closed system –Exchange of energy with the surroundings only An isolated system –Neither matter nor energy exchangewith the surroundings. The flow of energy between system and surroundinga sre in two forms: 1. Heat, q, which is the flow of energy between two objects due to a difference in temperature. 2. Work, w, force acting over a distance. It becomes compression and expan sion for gaseous system (P –V work). -2 Work by force: w = force x dist-2ce = mass (k2) -2acceleration (m s ) x distance (m) = kg x m s x m = kg m s Work by a gas: w = P ∆V = (Force / area) m 3 = (kg m s -2/ m ) m = kg m s2 -2 Energyunits: 2 2 Joules (J) 1 J = 1 kg m / s Calorie (cal) It is the quantity of energy needed to raise the temperature of 1 g of water by 1oC. 1cal = 4.184 J Calculate the work done when 7.3 L of a gas is expanded to 8.8 L under a constant pressure of 1.5 atmospheres. (1 L atm = 101.3 J) 1.5 atmosphere 1.5 atmosphere 7.3 Litre 8.8 Litre 1 The internal energy of a system Law of conservation of energy – energy can be converted from one form to another but cannot be created nor destroyed. (Energy in an isolated system is constant and since the universe is an isolated system, energy in the universe is constant) Example: The potential energy stored in gasoline is converted to heat energy on combustion which in turn becomes the kinetic energy of the car. The energy becomes heat energy of the road surface because of friction. The internal energy ( E) of a system is the total energy (both potential and kinetic energies) of the system Kinetic energy – movement of the molecules (translationalenergy) Potential energy –vibrational, rotational, electrostatic energies and bond energies of the molecules. The absolute value E of a system cannot be determined . However, the difference of internal energy can be measured: ∆E = E finalE initial First Law of Thermodynamics ∆E = q + w ∆E is the change of internal energy or ∆ E = E final Einitial q = heat w = work. The equation states that the change in internal energy of a system is equal to the energy that passes through its boundary as heat or work. The first law is an application of the Law of Conservation of energy: Energy can be converted from one form to another but can neither be created nor destroyed. Sign convention: The sign reflects from the system’s point of view. A plus sign means going into the system. For example, +q mea ns heat goes into the system. The minus sign means out of the system. Thus–w means work done by the system. +∆E means internal energy of the system increases, - ∆E means internal energy of the system decreases 2 State function Internal energy is a state function. The value of a state function depends on its present state ofthe systemnot on the path taken. Initial energy (initial ∆E depends on the two different states E Energy ofthe system Final energy (Efinal The state of a system is a system where all the physical properties h ave specified values. - Pressure, volume, temperature and its composition (allotropic form) For example: 10 g of water at o Heating 25 C under one Cooling atmospheric pressure 10 g of ice at 0 C 10 g of steam at under one 100 C under one atmospheric pressure atmospheric pressure Hydrogen and oxygen reacted to form10 g of water and allows to cool down Those properties do not depend on how the water was formed, how the pressure or the temperature were changed. Althoughinternal energyis a state function, heat and work are not state functions. ∆E = q + w There may be many different combina tions for q and w to give the same ∆E Therefore, q and w are steps dependent. For example: 3 One litre of gasoline + air Heat -q Burning Motor car -q Energy Work -w Carbon dioxide and water Examples of first law: Calculate the change in internal energy of a system in which 15.6 kJ of work is done to it but 1.4 kJ of heat is lost to the surroundings. Apply the First Law of Thermodynamics: ∆E = q + w ∆E = + (-1.4 kJ) + 15.6 kJ = 15.6 kJ - 1.4 kJ = 14.2 kJ A balloon is being inflated to its full extent by heating theair inside it. In the finaltsages 6 6 of this process, the volume of the balloon changes from 4.00 x 10 L to 4.50 x 10 L by the addition of 1.3 x 10 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ∆E for the process. (1 L atm = 101.3 J) ΔV = 4.50 x 10 L – 4.00 x 106 L = 0.50 x 10 L 6 Work is done to the surroundings (expansion) Work done = -PΔV = - 1.0 atm x 0.50 x 10 L 6 = - 0.5 x 10 atm L = - 0.5 x 10 atm L x 101.3 J /L atm = - 5.1 x 10 J Therefore, ΔE = w + q = - 5.1 x 10 J + 1.3 x 10 J 8 = 7.9 x 10 J Enthalpy We now focus on heat changes in chemical reactions. For a chemical reaction where gas is liberated such as: + 2+ CaCO (3) + 2H (aq) Ca (aq) + CO (g2 +H O(l2 Carbon dioxide gas is liberated and there is work done to the surroundings . w = - P∆V 4 Measurement of heat at constant volume (bomb calorimeter) CaCO +3HCl Closed system Open system If the reaction is carried out in a closed system (bomb calorimeter), ∆V = 0 and there is no work done. ∆E = q +v0 qv= Heat change at constant volume as measured by the bomb calorimeter The measured heat at constant volume is equal to the change in the internal energy Measurement of heat at constant pressure If the reaction is carried out in an open system i(n a beaker), work is done to the surroundings, then ∆E = q p P∆V qp = Heat change measured at constant pressure. But ∆E is a state function and so it must be the same for both in an open and a closed system. Therefore, q =vq - p∆V And q v ≠ q pnd │q │ v │q │ p Thus the heat change measured at constant volume and constant pressure is different. 5 I iti lst teE i iti l w y r e e l qv n q e p nt I Fin alst te E fial = E = q + w E q v p q = q + w v p qv =q p Thus q v q pnly when no gas is involved in a chemical reaction. For example: HCl (aq) + NaOH (aq) → NaCl (aq) + H O (l2 For chemical reaction when gases are involved, qp≠ qv For example: Mg(s) + 2HCl(aq) → MgCl (aq)2+ H (g) 2 The meaning of Enthalpy Most chemical reactions are carried out in an open system.(under constant pressure), we need to use a thermodynamic property for measuring energy change At constant volume, ∆E = q v At constant pressure, ∆E = q + w p ∆E = q v = qp+ w q = q + w v p If w = -P∆V (gases are involved) q = q - P∆V v p ∆E = q p P∆V q p ∆E + P∆V Since U, P and V are state functions, q is a state function p q p= ∆H H = Enthalpy qp= ∆E + P∆V 6 At constant pressure, ∆H = ∆E + P∆V Therefore enthalpy is: H = E + PV Enthalpy is defined as the sum of the internal energy and the pressure-volume product of a system Thus, for a reaction at constant pressure and only P -V work is allowed, the enthalpy change in a system is equal to the heat change. We can therefore write ∆H = ∆E + P∆V, and if n moles of gas formed in the reaction, ∆H = ∆E + RT∆n at constant temperature Which of the following exothermic reactions would have ∆ H = ∆E? NH 4O (s3 N 2(g) + 2H O(2) Δn = +1 C(s) + O 2g) CO 2g) Δn = 0 C H O (l) + 6O (g) 4CO (g) + 5H O(l) 4 10 2 2 2 Δn = 4 – 6 = -2 Importance of enthalpy • Since chemical reactions are carried out under constant pressure and the only work done is against atmospheric pressure, the enthalpy change is an important thermodynamic property. • Since enthalpy is a state function and is independent on the path, the enthalpy changes for different reactions can therefore be predicted. • The enthalpy change of some chemical reactions can be measured experimentally by a calorimeter. Example: Calculate both ∆H and ∆E for the reaction: 2SO 2(g) + O 2g) 2SO (3) P = 1.00 atm., T = 25 C, heat released from the reaction = 198 kJ -1 -1 R = 8.314 J mol K Is the heat released ∆H or ∆E? ΔH = - 198 kJ SO ΔH = ΔE + PΔV 2 SO 3 ΔE = ΔH - PΔV + O 2 In the reaction, PΔV = RTΔn Δn = 2 mol –3 mol = -1 mol Therefore, PΔV = (-1 mol) x 8.314 J mol K x (273 + 25) K 7 = - 2480 J = - 2.48 kJ Therefore, ΔE = -198 kJ – (- 2.48 kJ) = - 196 kJ More heat is rel
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