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Chapter 17 Energy and Rate of Reactions.  Thermochemistry – study of the transfer of energy as heat that accompanies chemical reactions and changes 

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Presentation on theme: "Chapter 17 Energy and Rate of Reactions.  Thermochemistry – study of the transfer of energy as heat that accompanies chemical reactions and changes "— Presentation transcript:

1 Chapter 17 Energy and Rate of Reactions

2  Thermochemistry – study of the transfer of energy as heat that accompanies chemical reactions and changes  Heat and Temperature - Temperature – measure of the average kinetic energy of particles (as T increases, KE increases; and the hotter an object feels) - Joules – SI unit of heat (and energy) kJ is also used commonly - Heat – the energy transferred between samples of matter because of a difference in their temperatures

3 - Energy moves from warmer to cooler objects - Energy absorbed or released in a reaction is measured in a calorimeter - Calorimeters are sealed and placed in water; the change in the temperature of the water is used to determine the energy change inside the calorimeter

4  Specific Heat - Objects have different capacities for absorbing energy (heat) - specific heat – amount of energy required to raise the temperature of one gram of a substance buy 1 o C - units are J/g o C - water has a high specific heat (4.184 J/g o C) - that means water can absorb or release a lot of energy before it’s temperature begins to change - that’s why even on a hot day, a lake feels cold

5 q = c x m x ΔT q = energy (heat); c = specific heat m = mass; ΔT = change in Temp. (subtract the Temps) Ex. A 4.0 g sample of glass is heated from 1 o C to 41 o C, and absorbed 32 J of energy. What is the specific heat of the glass? ΔT = 41-1 = 40 o C 32 J = c (4.0 g) (40 o C) 32 J = c (160 g o C) c = 0.2 J/g o C

6 Example 1  How much heat energy is needed to raise the temperature of a 55 g sample of aluminum from 22.4 C to 94.6 C? The specific heat of aluminum is.89 J/c g

7  Heat of Reaction – amount of energy absorbed or released during a reaction - when energy is released (it’s a product) it is an exothermic reaction 2 H 2 + O 2  2 H 2 O + 483.6 kJ - if 4 moles of H 2 O form, then (2 x 483.6) = 967.2 kJ released - when energy is absorbed (it’s a reactant) it is an endothermic reaction - Energy absorbed or released during a reaction is represented by ΔH rxn

8 - H stands for enthalpy – heat content of a system - ΔH rxn is the change in heat during the reaction ΔH rxn = H products – H reactants ΔH rxn is positive for endothermic reactions (heat is added to the reactants) ΔH rxn is negative for exothermic reactions (reactants are losing heat) Ex. 2 H 2 + O 2  2 H 2 O ΔH rxn = -483.6 kJ Reverse the equation and reverse the sign: 2 H 2 O  2 H 2 + O 2 ΔH rxn = +483.6 kJ

9  Heat of Formation - molar heat of formation – energy absorbed or released when 1 mole of a compound is formed from its elements - designated as ΔH o f What would ΔH o f be for the formation of H 2 O 2 H 2 + O 2  2 H 2 O ΔH rxn = -483.6 kJ That’s for 2 moles of H 2 O. -483.6 / 2 = -241.8 kJ for 1 mole of H 2 O

10 Calculating Enthalpy  Ex. 1 Calculate the Δ H for the following reactions and determine if they are exothermic or endothermic. 

11 - If a compound has a very negative ΔH o f then it is a very stable compound (it’s more stable as a compound than the elements were by themselves) - The more positive the ΔH o f the more unstable the compound is and the more likely it will be to decompose into its elements

12  Heat of Combustion – energy released by complete combustion of one mole of a substance (the reactant) Remember: combustion is addition of O 2 Ex. C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O ΔH o c = -2219.2 kJ Combusting 1 mole of C 3 H 8 releases 2219 kJ (because ΔH o c is negative, its exothermic)

13  Rate of Reactions  Kinetics – study of reaction rates  Rate Influencing Factors - For reactions to occur, particles must come into contact with each other (collide) - Anything that change the frequency or efficiency of the collisions will change the rate of the reaction 1) Surface Area - In homogeneous mixtures (particles are dissolved), they are able to mix and collide freely so those reactions occur rapidly

14 - In heterogeneous mixtures, reaction can only take place where the 2 phases are in contact with each other - so the surface area of a solid is a factor; the greater the surface area, the faster the rate of reaction Ex. Zn + HCl; powdered zinc will react faster than a cube of zinc 2) Temperature - Increasing Temp increases the kinetic energy of particles (making them move faster)

15 - If the move faster, there will be more collisions and therefore the rate will increase - Decreasing the Temp has the opposite effect 3) Concentration - the greater the concentration of a substance, the more particles there will be - the more particles there are, the greater then number of collisions and therefore the rate will increase

16 - Lowering the concentration has the opposite effect 4) Presence of Catalysts - Catalyst – changes the rate of a reaction without being used up (speeds up the reaction without interfering) - they are written over the arrow in the equation Ex. MnO2 2 H 2 O 2 -------  2 H 2 O + O 2


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