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The study of energy and the changes it undergoes.
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the ability to do work. conserved. made of heat and work. a state function. independent of the path, or how you get from point A to B.
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is divided into two halves. 1. System - the part you are concerned with. 2. Surroundings - the rest of the universe. Exothermic reactions release energy to the surroundings. Endothermic reactions absorb energy from the surroundings.
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Potential energy Heat
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Potential energy Heat
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Every energy measurement has three parts. 1. A unit ( Joules of calories). 2. A number how many. 3. and a sign to tell direction of energy flow. negative - exothermic +positive- endothermic
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System Surroundings Energy E <0
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System Surroundings Energy E >0
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Heat given off is negative. Heat absorbed is positive. Work done by system on surroundings is negative. Work done on system by surroundings is positive.
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The energy of the universe is constant= the law of conservation of energy. q = heat w = work E = q + w Take the system’s point of view to decide signs.
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Work is a force acting over a distance. w= F x d P = F/ area d = V/area w= (P x area) x (V/area)= P V Work can be calculated by multiplying pressure by the change in volume at constant pressure.
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If the volume of a gas increases, the system has done work on the surroundings. work is negative w = - P V units of liter - atm 1 L atm = 101.3 Joules Expanding work is negative. Contracting, surroundings do work on the system w is positive.
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What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? If 2.36 J of heat are absorbed by the gas above, what is the change in energy?
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Symbol: H H = E + PV (that’s the definition) At constant pressure: H = E + P V The heat at constant pressure q p can be calculated q p + w = q p - P V q p = E + P V = H So, q = H at constant pressure.
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Measuring heat. Use a calorimeter. Two kinds 1. Constant pressure calorimeter (aka coffee cup calorimeter) A coffee cup calorimeter measures H. An insulated cup, full of water. The specific heat of water is 1 cal/gºC or 4.184 J/g ºC. Heat of reaction= H = Cp x mass x T
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heat capacity for a material, C is calculated C= heat absorbed/ T = H/ T molar heat capacity = C/moles specific heat capacity is dependent on the mass of material: C p = C/mass Cp has units of J/gC or J/gK heat = Cp x mass x T OR heat = molar heat x moles x T Make the units work and you’ve done the problem right.
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The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?
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2. Constant volume calorimeter is called a bomb calorimeter. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. The heat capacity of the calorimeter is known and tested. Since V = 0, P V = 0, E = q
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thermometer stirrer full of water ignition wire Steel bomb sample
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Enthalpy is a state function. It is independent of the path. We can add equations to come up with the desired final product, and add the H Two rules 1. If the reaction is reversed the sign o H is changed 2. If the reaction is multiplied, so is H
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The enthalpy change for a reaction at standard conditions: (25ºC, 1 atm, 1 M solutions) Symbol Hº When using Hess’s Law, work by adding the equations up to make it look like the answer. The other parts will cancel out.
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Hº= -394 kJ Hº= -286 kJ Hº= -1300. kJ
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Given Calculate Hº for this reaction Hº= +77.9kJ Hº= +495 kJ Hº= +435.9kJ
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Table of standard heats of formation: Appendix 4 The amount of heat needed to form 1 mole of a compound from its elements in their standard states. Standard states are 1 atm, 1M and 25ºC For an element, the H f is = 0
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Need to be able to write the equations. What is the equation for the formation of NO 2 ? ½N 2 (g) + O 2 (g) NO 2 (g) Have to make one mole to meet the definition.
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We can use heats of formation to figure out the heat of reaction. Ex. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O Heat of reaction = The sum of the heats of formation of the products – sum of the heats of formation of the reactants.
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Spontaneity, entropy and free energy
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A reaction that will occur without outside intervention. We can’t determine how fast. We need both thermodynamics and kinetics (next chapter) to describe a reaction completely. Thermodynamics compares initial and final states. Kinetics describes pathway between.
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1st Law- the energy of the universe is constant. Keeps track of thermodynamics but doesn’t correctly predict spontaneity. WE MUST ALSO CONSIDER ENTROPY Entropy (S) is disorder or randomness 2nd Law - the entropy of the universe always increases.
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Defined in terms of probability. Substances take the arrangement that is most likely. The most likely is the most random. Calculate the number of arrangements for a system.
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2 possible arrangements 50 % chance of finding the left empty Positional entropy
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l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed
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l 4 possible arrangements l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed
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Gases completely fill their chamber because there are many more ways to do that than to leave half empty. There are many more ways for the molecules to be arranged as a liquid than a solid. Gases have a huge number of positions possible = highest entropy. S solid <S liquid <<S gas
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Solutions form because there are many more possible arrangements of dissolved particles than if they stay bound as a solid. Reactions that form gases or liquids are favorable since positional entropy is increased.
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2nd Law S univ = S sys + S surr If S univ is positive the process is spontaneous. If S univ is negative the process is spontaneous in the opposite direction
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For exothermic processes S surr is positive. For endothermic processes S surr is negative. Consider this process H 2 O(l) H 2 O(g) S sys is positive S surr is negative S univ depends on temperature.
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Entropy changes in the surroundings are determined by the heat flow. An exothermic process is favored because by giving up heat, the entropy of the surroundings increases. The size of S surr depends on temperature S surr = - H/T S surr = + when exothermic
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S sys S surr S univ Spontaneous? ---Yes +++No, reverse +-?At high temp. -+?At low temp.
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The entropy of a pure crystal at 0 K is 0. Gives us a starting point. All others must be >0. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed in the appendix. Entropy is a state function. Σ Sº Products – Summation Sº reactants to find Sº More complex molecules have higher Sº.
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Free energy, G, is that energy free to do work. The maximum amount of work possible at a given temperature and pressure = w max Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.
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Another way to determine spontaneity G=H-TS Never used this way. G= H-T S sys at constant temperature Divide by -T G/T = - H/T- S G/T = S surr + S sys G/T = S univ If G is negative at constant T and P, the Process is spontaneous.
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For the reaction H 2 O(s) H 2 O(l) Sº = 22.1 J/K mol Hº =6030 J/mol Calculate G at 10ºC and -10ºC Look at the equation G= H-T S Spontaneity can be predicted from the sign of H and S.
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HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous
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Gº = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Can’t be measured directly, can be calculated from other measurements. Gº= Hº-T Sº Use Hess’s Law with known reactions.
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There are tables of Gº f. It is a state function. Summation Gº f products- summation reactants Gº f reactants. The standard free energy of formation for any element in its standard state is 0. Remember- Spontaneity tells us nothing about rate.
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G tells us spontaneity at current conditions. When will it the reaction stop? It will go to the lowest possible free energy which may be an equilibrium – no net change between products and reactants. At equilibrium G = 0, Q = K Gº = -RTlnK
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delta GK =0=1 0 >0<0
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Gº= -RTlnK = Hº - T Sº ln(K) = Hº/R(1/T)+ Sº/R A straight line of lnK vs 1/T
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