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Ap Notes 15-A AP CHEMISTRY NOTES 4-1 THERMOCHEMISTRY: ENTHALPY AND ENTROPY Reaction Rate – how fast a chemical reaction occurs Collision Theory In order for a chemical reaction to occur, the following conditions must be met: 1) The reacting particles must collide – the more collisions, the faster the reaction occurs 2) The reacting particles must collide with enough energy to cause bonds to break and new bonds to form 3) The reacting particles must collide with the correct orientation Factors Affecting the Rate of a Chemical Reaction *Temperature –higher To = faster the motion of particles = more collisions *Concentration – higher concentration = greater number of particles = more collisions *Particle Size (Surface Area) – smaller particle size = greater surface area = more room for collisions *Catalyst – increases the rate of a chemical reaction by lowering the activation energy “hump” (see below); catalysts are not themselves changed in the chemical reaction SPONTANEITY OF CHEMICAL REACTIONS Just because substances are mixed together does not mean they will react. Whether or not a reaction is spontaneous (ie. occurring on its own without the the continual addition of energy) is determined by two factors: 1. Enthalpy (H) – the heat gained or lost during a chemical reaction *endothermic reaction – a chemical reaction in which heat energy is absorbed; products are higher in heat content than the reactants (although the container, etc., surrounding the reaction system is colder since heat has been removed from them to feed the reaction) *exothermic reaction - a chemical reaction in which heat energy is given off; products are lower in heat content than the reactants (although the container, etc., surrounding the reaction system is warmer since heat has been given to them) *Reactions tend to favor the lowest heat or energy state, so exothermic reactions are thermodynamically favored POTENTIAL ENERGY DIAGRAMS: Note that both types of reactions require a certain amount of activation energy (Ea) – the amount of energy needed to begin a chemical reaction. The activation energy for an endothermic reaction is considerably larger than the activation energy for an exothermic reaction. The activated complex is a transition state consisting of unstable “intermediate” forms of the atoms/molecules involved in the chemical reaction as reactants react to form the products. The heat of reaction (Hrxn) is the energy change involved when a chemical reaction occurs. Note that the heat of reaction will be negative (-Hrxn) for an exothermic reaction (since heat is given off) and positive (+Hrxn) for an endothermic reaction (since heat is taken in or absorbed). 2. Entropy (S) – the amount of disorder (energy and positional microstates) in a system *The Second Law of Thermodynamics states that everything in the universe moves toward greater disorder *Reactions are “entropically” favored when the products have a higher state of entropy than the reactants. In chemical reactions: *Entropy of a gas > Entropy of a liquid > Entropy of a solid (more energy and positional microstates) (fewer energy and positional microstates) (more disorder) (less disorder) *Entropy increases when a substance is divided into parts (the more parts / particles, the greater the entropy) *Entropy favors an increase in temperature EXAMPLES: __________ Cu(s) > Cu(l) Entropy ______________________ __________ MgCO3(s) > MgO(s) + CO2(g) Entropy ______________________ 2+ 2- __________ Ba (aq) + SO4 (aq) > BaSO4(s) Entropy ______________________ AP CHEMISTRY NOTES 4-2 THERMOCHEMISTRY: ENTHALPY AND ENTROPY CALCULATIONS The “change in enthalpy” (H) of a reaction can be calculated using the following equation: o o H = Hf (products) - Hf (reactants) o *NOTE: 1. Hf is the standard heat of formation (the enthalpy change involved when a compound is formed from its elements ) o 2. Hf of an element in its free state (including diatomic elements) is “0” EXAMPLE: Calculate the enthalpy change for the following reaction, determine whether it is endothermic or exothermic, and determine whether it is thermodynamically favored or thermodynamically un-favored: __________ CH4(g) + 3/2 O2(g) > 2H2O(l) + CO(g) The “change in entropy” S) can be determined using the following equation: S = So (products) - So (reactants) EXAMPLE: Calculate the entropy change for the following reaction, and determine whether it is entropically favored or not. __________ CH4(g) + 3/2 O2(g) > 2H2O(l) + CO(g) BOTH enthalpy and entropy act together to determine whether the reaction is thermodynamically favored. Unfortunately (depending on your point of view!) these two don’t always work together. A reaction that has a ____ H and a ____ S will always be thermodynamically favored. (spontaneous) A reaction that has a ____ H and a ____ S will always be thermodynamically unfavored. (non-spontaneous) However, in situations in which H and S have the same signs (either both negative or both positive), it is necessary to calculate the change in free energy (G) in order to determine whether the reaction is thermodynamically favored (spontaneous) or not. Gibb’s Free Energy – the maximum amount of energy that can be coupled to another process to do useful work (ie. the total energy available in the reaction system) *Gibb’s Free Energy depends upon the size and direction of enthalpy and entropy changes Exergonic reactions – thermodynamically favored (spontaneous) reactions which release free energy (-G) Endergonic reactions – thermodynamically “un-favored” (non-spontaneous) reactions which absorb free energy (+G) Gibb’s Free energy can be calculated using the following equation: Go = Ho - TSo (Note that temperature is in Kelvins) EXAMPLE: Determine whether the following reaction is thermodynamically favored at 25oC: __________ CH4(g) + 3/2O2(g) > 2H2O(l) + CO(g) AP CHEMISTRY NOTES 4-3 THERMOCHEMISTRY – WORK AND HEAT BASIC TERMS: 1. Thermodynamics – the study of energy and its interconversions 2. Energy – the capacity to do work 3. Kinetic Energy – the energy of motion 4. Potential Energy – energy that can be converted into useful work 5. Heat – the transfer of energy between two objects 6. Work – force times distance 7. State Function – a property independent of pathway (P, V, T, H, S, G, your bank account) 8. System – that which we focus on 9. Surroundings – everything else in the universe 10. Exothermic – energy (as heat) flows out of the system (-H) *number value shown on the products side of a chemical equation 11. Endothermic – energy (as heat) flows into the system (+H) *number value shown on the reactants side of a chemical equation 12. Internal Energy (E) – the sum of all of the potential and kinetic energy of the system 13. Enthalpy (H) – the heat content of the system 14. Entropy (S) – a measure of disorder (or chaos) 15. Free Energy (G) – the energy available in a system to do work *THREE LAWS OF THERMODYNAMICS 1. The First Law of Thermodynamics: The energy of the universe is constant 2. The Second Law of Thermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe 3. The Third Law of Thermodynamics: The entropy of a perfect crystal at 0 Kelvins is zero *ENERGY Internal energy can be changed by a) a flow of heat b) a flow of work c) a flow of both work and heat This change in internal energy (E) can be calculated by using the following equation: E = q + w q = heat (in “cal” or “J”) w = work (in “cal” or “J”) +q = heat absorbed +w = work done on the system - q = heat released - w = work done by the system +E = system gains energy - E = system loses energy EXAMPLES: Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J. Calculate the change in internal energy for a system undergoing an exothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. *GASES AND WORK A common type of work accomplished with chemical processes is that of “work done to a gas” (through compression of the gas) or “work done by a gas” (through expansion of the gas). To determine the amount of work involved when a gas expands or is compressed, the following equation is used: w = - PV where P = pressure (in atmospheres) V = change in volume (in liters) Also recall the following equation: q = mCp∆T where q = heat (in Joules or calories) m = mass (in grams) or moles .o .o Cp = specific heat (J/g C or cal/g C) or molar heat capacity (J/mol.oC) ∆T = change in temperature (oC) Note – “Kelvins” may be used in place of oC with no change in numerical value EXAMPLES: Calculate the work associated with the expansion of a gas from 46.0L to 64.0 L at a constant external pressure of 15atm. Express the answer in both L.atm and Joules. (1 L.atm = 101.3 J) A balloon is being inflated to its full extent by heating the 89.3 kg of helium inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L under a constant pressure of 620 torr. The temperature of the balloon rises from 23.3oC to 79.4oC. Determine “q”, “w”, and E (in kJ) for this process. (1 L.atm = 101.3 J; the molar heat capacity of helium is 20.8 J/mol.oC) AP CHEMISTRY NOTES 4-4 THERMOCHEMISTRY: ENTHALPY & CHEMICAL REACTIONS When the amount of heat involved in a chemical process is calculated, it is often necessary to use the specific heat capacity of a substance in that calculation. Depending upon the information available to us, however, it may be necessary to convert specific heat capacities to “molar heat capacities” (the amount of heat required to change the temperature of 1 mole of a substance by 1oC) and vice vesa.
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