Chapter 6. Homework. Thermodynamics vs. Kinetics. Energy. Energy Unit. Chapter 6. Principles of Reactivity: Energy and Chemical Reactions

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1 Homework Chapter 6 Chapter 6 29, 33, 35, 39, 45, 49, 53, 57, 63, 67, 71, 79, 87 Principles o Reactivity: and Chemical Reactions Thermodynamics vs. Kinetics A.E. Kinetics E or H Reaction Coordinate What type o reaction is this one? Thermodynamics H is negative or this reaction so it is an Exothermic process. is deined as the capacity to do work. There are two types o energy Kinetic: the energy o moving objects. Thermal energy Mechanical energy Electric energy Sound Potential: the energy that results rom an object's position. Chemical potential Gravitational Electrostatic Unit can be measured in two dierent units. Calorie: the quantity o energy that is required to raise the temperature o 1.00 g o pure liquid water by one degree Celsius. That is where we get calories when we talk about ood. Except on ood calorie (C) is equal to 1000 calories (c). Joule: the derived SI unit or energy. 1 J = 1 kg m 2 s -2 Conversion actor between Joules and calories 1 cal = J

2 Law o Conservation o can be converted rom one orm to another but can neither be created nor destroyed. (E universe is constant) Temperature There is a common misconception that temperature is a measure o the heat within an object. Temperature is actually a measure o the average kinetic energy o particle in a sample. Temperature is directly proportional to the average kinetic energy o the particles. Heat Heat is a transer o energy between two objects at dierent temperatures. So i we have a hot object that comes in contact with a cold object what happens? Hot object: molecules have a high average kinetic energy (they are moving very ast). Cold object: molecules have a low average kinetic energy (they are moving more slowly). The dierent particles collide with one another. I a hot particle collides with a cold particle their energy will be transerred. The hot particle moves more slowly and the cold particle moves aster. This continues until all o the particles have the same average kinetic energy. This is called conduction. Heat Capacity Is deined as the amount o heat required to raise the temperature o an object by 1 C. This is an awkward measurement because it depends on the size o an object. Thereore we use the speciic heat capacity. Deined as the heat required to produce a given temperature change per gram o material. Example What is the thermal energy required to warm a 10.0 g piece o copper rom 25 C to 325 C. Rearrange the speciic energy equation q = speciic energy mass T q = g (598 K 298 K) = 1160 J Speciic heat capacity (C) = quantity o heat supplied (mass o object) (temperature change) q(j) C = [m(g)] [ T ( K )]

3 What Does the Answer Mean? The sign o the heat transerred (q) tells which direction the heat was transerred. + q: means that the heat is transerred to the substance - q: means that heat is transerred rom the substance. Heat o Fusion & Heat o Vaporization Heat o Fusion The quantity o heat required to melt a solid to orm a liquid Heat o Vaporization The quantity o heat required to convert a liquid to a vapor Examples How much ice can melt with 500 kj o energy? (heat o usion water = J/g) 500 kj = J 1 g ice melted at 0 C J = J g ice melted How much water can be vaporized with 500 kj o energy? (heat o vaporization water = 2256 J/g) 500 kj = J 1 g water vaporized J = = 222 g liquid water vaporized 2256 J First Law o Thermodynamics The Law o Conservation o The total energy o the universe is constant. A mathematical expression o this is E = q + w q is heat w is work Surroundings SYSTEM Exothermic Process Surroundings SYSTEM Endothermic Process Exo- and Endo- thermic process Exothermic process Heat lows rom the system to surroundings. The temperature o the surroundings increases. Endothermic process Heat lows to the system rom the surroundings. The temperature o the surroundings decreases. Enthalpy An enthalpy change is heat energy transerred at constant pressure. We can actually derive this. From the 1 st law o thermodynamics E = q + w Work can be deined as: w = -P V We know that enthalpy is related to energy H = E + PV However we rarely measure energies and enthalpies. We measure changes in these quantities. H = E + P V Plug in what work is equal to H = E + -w Plug in what E is equal to H = q + w + -w H = q Thereore at constant pressure, enthalpy is heat.

4 More about Enthalpy Change Based upon the enthalpy change we can determine i it is an endo- or exo- thermic reaction. I H is positive then the reaction was endothermic. Example: water to vapor I H is negative then the reaction was exothermic. Example: vapor to water Enthalpy Changes or Chemical Reactions Just like changing states, chemical reactions can be exo- or endo- thermic. Remember any reers to products minus reactants. Thereore, the H o a chemical reaction is just the change in heat o the products minus the reactants. Example How much heat can be produced by burning 454 g o propane gas? C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4H 2 O(l) H = kj 454 g o propane = 10.3 mols propane We also know that 220 kj o heat is produced per 1 mole o propane kj 10.3 mols propane = 22,900 kj heat 1.00 mols propane Hess s Law We would like to know the H values or as many reactions as possible. This would take to much time to do and so much space to store this data. Instead we can use Hess s Law, which states that i a reaction is the sum the two or more other reactions, then H or the overall process must be the sum o the H values o the constituent reactions. Application o Hess s Law We want to know the H or the reaction: C(s) + 2 H 2 (g) CH 4 (g) We can use the ollowing reactions to determine the H or the reaction. C(s) + O 2 (g) CO 2 (g) H = kj H 2 (g) + ½ O 2 (g) H 2 O(l) H = kj CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = kj We know we need to 1. Multiple the H 2 (g) equation by 2 2. Need to reverse the CH 4 (g) equation. Application o Hess s Law (cont d) 2 [H 2 (g) + ½ O 2 (g) H 2 O(l)] H = ( kj) 2 = kj CO 2 (g) + 2H 2 O(l) CH 4 (g) + 2O 2 (g) H = kj C(s) + O 2 (g) CO 2 (g) H = kj Add them up we get C(s) + 2H 2 (g) CH 4 (g) H = kj

5 State Functions The only reason why Hess s Law works is because enthalpy is a state unction. A state unction a quantity whose value is determined only by the state o the system. The enthalpy change or a chemical or physical change does not depend on the path you choose rom the initial conditions to the inal conditions. Path independent Standard States Compound Element For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure liquid or solid. The orm [N 2 (g), K(s)] in which it exists at 1 atm and 25C. Standard Enthalpy o Formation Is the amount o energy required to make one mole o a compound rom its elements in their standard states. We denote these enthalpies as H, where indicates the standard condition. We also have the standard molar enthalpy o o ormation, H The subscript indicates that 1 mol o the compound in question has been ormed. So? This means that we can calculate the enthalpy change or a reaction based upon the standard molar enthalpy o ormation. Enthaply change or a reaction = H H rxn = [ H (products)] [ H rxn (reactants)] Example We want the H o H rxn or the ollowing reaction. CaCO 3 (s) CaO(s) + CO 2 (g) [ H (products)] [ H = (reactants)] rxn H rxn = [ ] [ ] = kj Another Example Nitroglycerin is a powerul explosive, giving our dierent gases when detonated. Given that the enthalpy o ormation or nitroglycerin, H, is -364 kj/mol. Calculate the enthalpy change when 10.0 g o nitroglycerin is detonated. 2 C 3 H 5 (NO 3 ) 3 (l) 2 N 2 (g) + ½ O 2 (g) + 6 CO 2 (g) + 5H 2 O(g) 1. Two elements are in their standard states (N 2 and O 2 ) thereore their values are zero. 2. Look up CO 2 ( kj/mol) and H 2 O ( kj/mol)

6 Still Working H rxn = [6(-393.5) + 5(-241.8)] [2(-364)] = kj Now that we know the H rxn, we can calculate how much energy 10.0 g nitroglycerin will produce g C 3 H 5 (NO 3 ) 3 = moles C 3 H 5 (NO 3 ) kj moles C3H5 (NO3) 3 = 62.6 kj 2 mols C H (NO )

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