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McMaster University

THERMOCHEMISTRY → 1 Law of Thermodynamics → Enthalpies of Reaction - Hess' Law, etc. → Bond Energies Why Thermochemistry? We've discussed some physical processes - eg. dissolution of ionic solids (salts) in water - & some chemical reactions - precipitation, acid-base & redox ... in terms of changes of state and changes in molecular formula - making/breaking of bonds, transfer of electrons, etc. Energy flow also accompanies these processes. THERMODYNAMICS provides the framework to understand energy flow & its role in determining whether a process actually occurs or not. eg. 2 Na(s) + 2 H O(l) → 2 NaOH(aq) + H (g) 2 2 occurs spontaneously, liberating a good deal of heat energy However, the reverse reaction never happens. 25 More egs. MgSO (s4 & NH NO (s)4 3 dissolve spontaneously in water. When MgSO (s) d4ssolves in water, heat is liberated. → used in hot packs When NH NO (4) d3ssolves in water, heat is consumed. → used in cold packs Thermochemistry deals with the accounting of heat transfer. We defer the question of what makes a process spontaneous till later when we take up entropy & Gibbs free energy. Thermodynamics hinges on being very precise in our description of a "system" and its processes eg. Types of Systems: isolated closed open system system system system is sealed off system has system has permeable, outside world impermeable walls, but non-thermally- - no heat or matter is not thermally insulating walls flow across boundary insulated - heat and matter flow - rigid walls ⇒ - heat flow, but - walls can be rigid or no mechanical work - no matter flow non-rigid done on or by system - walls can be rigid or non-rigid 26 Properties of Systems 1. Discrete labels: - chemical species identity - phase identity eg. CH (4) Zn(s) Br 2l) 2. Continuous variables: p, T, n, V pressure, temperature, # of moles, volume U, H, S, G energy, enthalpy, entropy, Gibbs free energy State Functions - they depend on the state of the system eg. pi= 1 atm p f 3 atm T i= 300 K T f= 450 K ni = 1 mole some process n f= 1 mole 3 3 V i= 10 m V f = 5 m U i, Hi Si G i U f, Hf, Sf, Gf initial system state final state Processes cause changes in system state & corresponding changes in system state functions. Change in a state function depends only on the initial & final state - not on how the change was achieved ⇒ ∆p = p −fp = 2iatm ∆T = T − f = 15i K 3 ∆n = n −fn = 0imol ∆V = V − f = −5im 27 Work - mechanical or electrical We will focus on mechanical work distance through which spring is compressed compress spring Apply a force to a spring to compress it → work is done on the spring work = force × distance The spring stores this energy. Work is done on the system, which increases the energy of the system. Compressing a gas in a cylinder is similar: Compressed gas has higher energy V apply force on i gas in cylinder via piston V f 28 Temperature & Heat Hot vs. Cold Temperature measures the average kinetic energy per atom T1 T2 T1< T 2 Heat flows from system 2 (on the right) to system 1 (on the left). ⇒ T 1ncreases & T d2creases until the two systems achieve the same temperature – thermal equilibrium q Heat flowing to system T increases the temperature q > 0 Endothermic process q Heat flowing from system decreases the temperature T q < 0 Exothermic process Note – other things (i.e. other than heat flow) can change the temperature ⇒ the sign of q does not always correspond to the sign of ∆T Calorimetry - measuring heat transfer 29 Consider transferring 100 J to … a) 18 g of H O(l) ⇒ ∆T = 1.33 K 2 b) 18 g of Cu(s) ⇒ ∆T = 14.5 K The resulting temperature change depends on the type of material – in addition to the amount of material. In general, Heat Capacity q = C ∆T = m s ∆T mass Specific Heat −1 −1 Units of s : J g K s = Amount of heat required to raise the temperature of 1 g of the substance by 1 K Note, s is always > 0 s[H2O(l)] = q = 100 J = 4.18 J g K1 −1 m ∆T (18 g) × (1.33 K) s[Cu(s)] = q = 100 J = 0.378 J g K −1 m ∆T (18 g) × (14.5 K) 30 Calorimetry utilizes a bulk quantity of a substance with a well known specific heat in thermal equilibrium with a system of interest. A physical or chemical process occurs in the system. Any heat produced or consumed enters or is drawn from the bulk. The heat produced or consumed can be deduced from the change in temperature – there is no heat flow to or from entire calorimeter, since it is thermally isolated. However, there must be no physical or chemical processes in the bulk, as the resulting temperature change would not correspond to the heat. eg. 1 mol solid ice at 1 mol liquid water 0° C & at 0° C & 1 atm pressure 1 atm pressure q = 6008 J Here, there is a substantial heat transfer resulting in no temperature change → Heat of fusion 31 1 mol liquid water 1 1 mol liquid water at 25.00° C & at 26.33° C & 1 atm pressure 2 1 atm pressure Two different processes can produce the same final state from a given initial state. 1. q1= 100 J & w1= 0 J 2. q2= 0 J & w2= 100 J 1. a flame transfers 100 J of heat to system 2. a submerged propeller delivers 100 J of work to system q ≠ q & w ≠ w 1 2 1 2 However, q 1 w 1 = q 2 w 2 It turns out this is true for all processes connecting a particular initial state to a particular final state. ⇒ q + w is the change in a state function … the energy ∆U = q + w The first law of thermodynamics 32 How does one measure ∆U ? ∆U = q + w Arrange for either q or w to be zero & measure the other. We’ve seen how to measure q. - just be sure that w = 0 & you will measure ∆U - this is accomplished by enclosing the system in rigid walls, with no internal moving parts ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ We can also measure w for a process for which the system is thermally isolated. external pressure eg. pext - we assume a constant pressure moving wall distance surface area = A = d thermal isolulation V ensures q = 0 Force = p extA Mechanical work done on system = Force ×distance Pressure is = (p ext A) d force per unit area = p ext (A d) = pext(−∆V) work done on system = − p ext∆V reversible = − p ∆V process - i.e. do it slowly 33 Knowing the work done on the system & ensuring that q = 0, we get ∆U = w = − p ext∆V if pextis constant ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ In the case of w = 0, ∆U = q V - the subscript V denotes a constant volume process - this is required to make w = 0 - A heat measurement at constant volume determines ∆U ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Let’s go back to the constant pressure process - more easily managed – eg. just vent the system to the atmosphere (mechanical equilibrium is needed) - a constant volume requires a bomb calorimeter which must be very strong to withstand high internal pressures which are possible – depending on the process being studied 34 For a constant pressure process, we still have ∆U = q + w - note that we are not now considering the case of q = 0 If we measure q, we measure at constant pressure qp = ∆U − w But w = − p ∆V if p = p ext is constant. So, q p = ∆U + p ∆V holds because = ∆( U + p V) p is constant Because constant pressure processes are so important, the state function, H = U + p V is given a name, enthalpy, and used extensively. 35 ♣ U final Uinitial ∆U<0 ∆U<0 Ufinal U initial ♣ Vf> V i ⇒ ∆V > 0 ⇒ w = −p∆V < 0 i.e. work is done by the system also, q = 0 thermal insulation ensuring that q = 0 So, ∆U = q + w < 0 U initial ∆U<0 U final 2 HCl(aq) + Zn(s) → H (g) + 2 ZCl () 36 Exothermic Endothermic q < 0 q > 0 constant V : constant V : ∆U = q V < 0 ∆U = q V > 0 constant p : constant p : ∆H = q < p ∆H = q > 0p For a chemical re
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