Unformatted text preview: Thermodynamics What is Thermo dynamics ? Movement of “Heat” Related to heat Related to [email protected] Historically: Developed by engineers interested in steam engines Determine the amount of work that could be obtained from the heat put into the engine Today: Physicists, Chemists and Biologists Thermodynamics Thermochemistry (1st law) Study of ENERGY [email protected] How much work can be obtained from a chemical [email protected] Inside this ba?ery there is a chemical [email protected] that produces work!!!! Calorimetry Study of HEAT measurements Measure the temperature changes in a system as it absorbs or releases heat. Energy [email protected] Energy: The capacity to do work or to produce Heat [email protected] against an opposing force Process for the transfer of energy How many of you have seen Finding Nemo? Work is when all the ﬁsh in the net swim in the same [email protected] They create a force [email protected] over a distance and save themselves from being caught! Heat is when the ﬁsh in the net swim in random [email protected] They do not oppose the force pulling the net up! Energy [email protected] Energy: The capacity to do work or to produce Heat [email protected] against an opposing force Process for the transfer of energy What does [email protected] of energy mean? Energy cannot be created nor destroyed Energy can be converted (transferred) from one form to another [email protected] Energy: energy due to [email protected] (KE = 1/2mv2) [email protected] Energy: energy due to [email protected] (PE = mgh) Consider tossing a Ball into the air When tossed the ball rises ([email protected] energy) Energy was transferred to the ball by the man The man did WORK on the ball The ball then falls back to the earth ([email protected] energy), but stops moving when it hits the ﬂoor. Does the ball have any energy now? Where did the energy go? Energy was transferred from the ball as heat Why don’t we talk about heat transfer from the man and to the ground? NO Consider tossing a Ball into the air Why don’t we talk about heat transfer from the man and to the ground? In thermo problems we must always deﬁne a SYSTEM and its SURROUNDINGS. We are deﬁning the ball to be our SYSTEM Considering only the ball – is seems that energy was not conserved We gave the ball energy – by WORK The ball lost that energy – as HEAT Both WORK and HEAT are ways to transfer energy Statement of the ﬁrst law ΔE = Q + W For the Universe: ΔE = 0 How do we deﬁne a system? Q = HEAT W = WORK For a chosen system: ΔE = 0 Technically arbitrary…but must be consistent Imagine that I liZ a book into the air. I do work on the book to li[ it. If the book is the system: ΔE > 0 If I am the system: ΔE < 0 If the book and I are the system : ΔE = 0 Statement of the ﬁrst law ΔE = Q + W Q = HEAT W = WORK We will deﬁne the chemicals in a chemical rxn as the system Reactants Products System: System: only the chemicals only the chemicals Surroundings: everything else – including the solvent Statement of the ﬁrst law ΔE = Q + W Q = HEAT W = WORK HEAT and WORK are ways to transfer energy between the system and the surroundings The only change in energy for a system is through transfer to/from the surroundings as either HEAT or WORK Statement of the ﬁrst law ΔE = Q + W Q = HEAT W = WORK Lets apply this ides to a chemical system: The burning of a candle When we light the candle (system) we give it energy by doing work on it. Once lit the candle (system) looses energy to the surrounding as heat. ΔEUNIV = Q + W = <0 + >0 = 0 The total change in energy of the universe = 0 Sign [email protected] Force [email protected] over a distance PV work is the most common Electrical work Work done ON the system WORK Process for energy transfer Some rxns get cold: Endo Some rxns get hot: Exo Heat transferred INTO system HEAT W > 0 Energy of the system INCREASES Q > 0 Energy of the system INCREASES Work done BY the system Heat transferred OUT of system W < 0 Energy of the system DECREASES Q < 0 Energy of the system DECREASES Sign [email protected] SURROUNDINGS SYSTEM Work done ON the system Work done BY the system W > 0 Energy of the system INCREASES W < 0 Energy of the system DECREASES Heat transferred INTO system Heat transferred OUT of system Q > 0 Energy of the system INCREASES Q < 0 Energy of the system DECREASES ENDOTHERMIC E = Q + W EXOTHERMIC Endothermic [email protected] ENDOTHERMIC ΔE system > 0 Surroundings loose heat and feel cold Exothermic [email protected] EXOTHERMIC ΔE system < 0 Surroundings take in heat and feel hot Some Examples Ball Toss First do WORK on the ball when you toss it W > 0 Then the ball looses HEAT when it hits the ground Q < 0 ΔE = Q + W = <0 + >0 = 0 Heat a gas enclosed in a 1L container GAS What happens to its temp? What about its KE? When we heat the gas we transfer energy to the gas as HEAT. Q > 0 The change in energy of the system then: ΔE > 0 1st law is an [email protected] system to keep track of ENERGY transfers Ways to transfer Energy WORK Baseball bat hits a baseball Compress a gas Expand a gas against an external pressure Pick up a book HEAT Heat a gas (ﬂame/hair dryer) ‐ Heat from contact The sun [email protected] your skin ‐ Heat with NO contact ‐ [email protected] A Gas is a good system to examine heat and work energy transfers because we have a molecular level understanding of gas behavior Do WORK on the gas: compressing the gas HEAT the gas: external heat source (hairdryer) Compression of a Gas Consider a gas compression in a syringe Sealed end Work is being done to the gas (system) Compression of a Gas Consider a gas compression in a syringe Sealed end Work is being done to the gas (system) If work is energy, and energy is conserved, where did the energy go when the gas was compressed? The energy from the work must have gone into the gas! Just like it went into the book when we liZed it Remember: KE = 3/2RT This means the temperature of the gas must have INCREASED. Temperature is a measure of the internal KE of a system Expansion of a Gas Consider a chemical [email protected] occurring inside a syringe [email protected] plunger 2Na + 2H2O 2NaOH + H2 Expansion of a Gas Consider a chemical [email protected] occurring inside a syringe [email protected] plunger 2Na + 2H2O 2NaOH + H2 The chemical [email protected] generates H2 gas which causes a build up in the pressure. The pressure forces the plunger outwards against the atmospheric pressure This is WORK: A force [email protected] over a distance Expansion of a Gas Consider a chemical [email protected] occurring inside a syringe [email protected] plunger PΔV = (force/area)(ΔV) = (force)(distance) = WORK Newton’s 3rd law: A force cannot be harder than its opposing force. Here the opposing force = Patm WORK = PatmΔV For expanding gas: ΔV > 0, but work is being done BY THE GAS Work must be [email protected] WORK = ‐ PatmΔV HEAT What is heat? Is there a [email protected] between heat and temperature? Remember that KE = 3/2RT Temperature is some measure of the amount of internal energy Energy is a mix of [email protected] and [email protected] energy ‐ MOTION Molecular [email protected] Thermometer immersed in a hot gas Hot gas molecules ﬂy around and bump into one another. Energy is transferred from one molecule to another Excited Gas molecules excite the Hg atoms causing an increase in Energy – MOTION – Increase in the volume HEAT and WORK What is heat? Is there a [email protected] between heat and temperature? Heat is not a thing! HEAT is a process for the transfer of energy Heat can be transferred through [email protected] WORK is also not a thing! WORK is also a process for the transfer of energy ΔE = Q + W There are ONLY 2 ways to transfer energy. The 1st law is an [email protected] [email protected] for this process. State [email protected] Consider 2 people of the same mass climbing a Hill 1 person climbs straight up. The other winds around the hill on a trail. Both begin at the bo?om of the hill and end at the top of the hill If we consider their KE unchanged ΔPE = ΔE = mgΔh Because Δh is the same (hﬁnal – hinital) they have the same ΔE The change in Energy is path independent. It depends only the INITIAL and FINAL states of the system State [email protected] Consider raising the temp of a gas from 300 K to 400 K Path 1: 300 K 250 K 400 K Path 2: 300 K 550 K 400 K Both begin at 300 K and end at 400 K: ΔE = 3/2RT = 3/2R(100) Because ΔT is the same (Tﬁnal – Tinital) they have the same ΔE The change in Energy is again path independent. It depends only the INITIAL and FINAL states of the system State [email protected] For a Gas T, E, V, P are all [email protected] of the state of the gas T, E, V, P are all state [email protected] A state [email protected] depends only the INITIAL and FINAL states of the system It is a path independent [email protected] Q and W are mechanisms to transfer energy in or out of a state ‐ They are not [email protected] of the state ‐ They are ways to change the state Q and W are NOT state [email protected] A new state [email protected] – ENTHALPY (H) To chemists: Enthalpy is the heat of a [email protected] [email protected] H equals Q (the heat of [email protected]) A [email protected] deﬁ[email protected] for H: H = E + PV Why do we care about heat of [email protected]? Engineers: exploring new fuel [email protected] to harvest energy Chemists: strength of chemical bonds Biologists: fuel for the human body A new state [email protected] – ENTHALPY (H) Enthalpy is a state [email protected] It is a [email protected] of the Temp, pressure, volume and number of moles of a system. H = E + PV H(T, P, V, n) = E + PV Because PV = nRT We can only vary 3 of these at a @me A new state [email protected] – ENTHALPY (H) What is the physical meaning of enthalpy? ΔH = ΔE + P(ΔV) For an expanding gas W = ‐PΔV ΔH = q + w + P(ΔV) ΔH = q + ‐ PΔV + P(ΔV) For constant pressure and PV work only: ALL bench top chemistry ΔH = q ALL gases [email protected] the Enthalpy of [email protected] Because H is a state [email protected] ΔH depends on only [email protected] and ﬁnal states ΔH = Hproducts – Hreactants We can predict ΔH values for a chemical [email protected] if we know the H values for the reactants and productsof the [email protected] A + B C ΔH = HC – (HA + HB) This is known as Hess’s Law [email protected] the Enthalpy of [email protected] Consider the [email protected]: CO(g) + 1/2O2(g) CO2(g) Because ΔH = q (constant P, PV work only) we could simply measure the heat of [email protected] for each process in a calorimeter The H’s of these individual compounds are tabulated! ΔHRxn = HCO2(g) – 1/2HO2(g) – HCO(g) Where did these H values come from? A small digression – A Zero point We must deﬁne an enthalpy zero point Consider [email protected] energy Ep = V = mgh ΔV = Vf ‐ Vi = mg(hf – hi) = mg(1m) = mg(2m) 1m This Zero point must be deﬁned to be consistent 2m A small digression – A Zero point For Height: Sea Level For Temperature: Depends on the temp scale Fahrenheit: coldest temp you can reach by adding salt to water [email protected]: Temp at which pure water freezes. Kelvin: Absolute zero What about for ΔH? Zero point for ΔH: Any pure element in its most stable form at 298K has a deﬁned H = 0. i.e. metallic zinc, diatomic oxygen gas, liquid mercury Hess’s Law By deﬁ[email protected]: the H for any pure element in its most stable state is deﬁned to be = 0. Consider the [email protected] to form water from its elements: H2(g) + 1/2O2(g) H2O(l) ΔHRxn = Hproducts – Hreactants = HH2O(l) – 1/2HO2(g) – HH2(g) = HH2O(l) – 1/2(0) – 0 = HH2O(l) H can be found for any compound by considering its [email protected] [email protected] [email protected] [email protected] Special kind of [email protected]: A compound is made from its elements in their most stable states. The heat of [email protected] (Hf): Heat absorbed or released when a compound is formed in this way For any compound: ΔH can be found from its [email protected] [email protected] and it is denoted as ΔHf These values are tabulated. Examples of [email protected] [email protected] ΔHf of H2O: ΔHf of CO2: ΔHf of HCl: H2(g) + 1/2O2(g) H2O(l) C(graph) + O2(g) CO2(g) 1/2H2(g) + 1/2Cl2(g) HCl(g) ΔHf of ZnCl2? ΔHf of NH3? ΔHf of CO? Examples of [email protected] [email protected] Recall this [email protected]: CO(g) + 1/2O2(g) CO2(g) ΔHRxn = ΔHf(CO2(g)) – 1/2ΔHf (O2(g)) – ΔHf(CO(g)) From the tables we can ﬁnd: ΔHRxn = ‐393.51 KJ/mol – 1/2(0) –(‐110.53 KJ/mol) = ‐282.98 KJ/mol Lets Review what we know: Because Enthalpy is a state [email protected] ΔH depends only on the [email protected] and ﬁnal states. The path by which you go from start to ﬁnish is irrelevant. ΔH = Hproducts – Hreactants H values for products and reactants can be found from tabulated ΔHf values ‐ These are based on the zero point assignment on pure elements in their most stable state having enthalpy of zero. To ﬁnd ΔH for a [email protected], simply look up the ΔHf values and use the [email protected] ‐ Be sure to include the [email protected] stoichiometry in your [email protected]! ...
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 Summer '08
 PETERPASTOS
 pH, Energy, ΔH

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