chapter 17 thermochemistry and kinetics. thermochemistry – study of transfer of energy as heat...
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Chapter 17
Thermochemistry
and
Kinetics
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Thermochemistry – study of transfer of energy as heat that accompanies chemical reactions and physical changes
Heat and Temperature- Temperature – measure of the
average kinetic energy of particles (increase T, increase KE; object feels hotter)
- Joules – SI unit of heat (and energy)kJ is also used commonlyEX. 980 kJ 980,000 J
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- Heat – the energy transferred between samples of matter because of a difference in their temperatures
- Energy moves from warmer to cooler objects
- Energy absorbed or released in a reaction (or change) is measure in a calorimeter.
- Calorimeters are sealed and placed in H2O. The T change of H2O is used to determine the energy change inside the calorimeter.
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Heat Capacity and Specific Heat- Objects have different capacities for
absorbing heat (energy).- Specific heat – the amount of energy
required to raise the temperature of one gram of a substance by 1o C.
- Units are J/g oC- Table on pg. 513- Water has a HIGH specific heat
(4.184 J/g oC)
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- That means water can absorb (or release) a lot of energy before its T will begin to change.
- Even on a hot day, lake water can still feel cold.
q = c x m x ΔT
q = energyc = specific heatm = massΔT = change in T (subtract the T’s)
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Ex. A 4.0 g sample of glass heated from 1o C to 41o C, and was found to have absorbed 32 J of energy. What is the specific heat of the glass?
q = c x m x ΔT
ΔT = 41oC – 1oC = 40oC
32 J = c (4.0 g)(40oC)
32 J = c (160 g oC)
c = 0.2 J/goC
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Your Turn!
Determine the specific heat of a material if a 35 g sample absorbed 48J as it was heated from 293K to 313K.
If 980 kJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be?
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Heat of Reaction – amount of energy absorbed or released during a chemical reaction
- When energy is released (a product) in a reaction, it is an exothermic reaction
ex. 2 H2 + O2 2 H2O + 483.6 kJ
If 4 moles of H2O formed then 967.2 kJ would be released (2 x 483.6)
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- When energy is absorbed (added to the reactants) it is an endothermic reaction.
- Energy absorbed or released during a reaction is represented by ΔHrxn
- H stands for enthalpy – heat content of a system
- ΔH is the change in heat during the reaction: ΔH = Hproducts – Hreactants
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ΔH is positive for endothermic reactions (heat added to reactants)
ex. 2 H2 + O2 H2O ΔH = +483.6 kJ
ΔH is negative for exothermic reactions (reactants losing heat)
2 H2O 2 H2 + O2 ΔH = -483.6 kJ
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Heat of Formation – energy absorbed or released when 1 mole of a compound is formed from its elements
- Designated as ΔHfo
Ex. What would ΔHfo be for H2O(g)?
2 H2 + O2 2 H2O ΔH = -483.6 kJ
That’s for 2 moles of H2O; find ΔH for 1 mole:
-483.6 kJ = - 241.8 kJ = ΔHfo
2
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Heat of Combustion – energy released as heat by complete combustion of 1 mole of a substance
(remember combustion is adding O2)ex.
C3H8 + 5O2 3CO2 + 4H2O
ΔHocomb = -2219.2 kJ
Combusting one mole of C3H8 releases 2219.2 kJ
(because ΔHocomb is negative, it releases
energy – exothermic)
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Heat of combustion is defined in terms of one mole of reactant, whereas the heat of formation is defined in terms of one mole of product.
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1. Identify each of the following reactions as exothermic or endothermic.
CH4(s) + 2O2(g) → CO2(g) + 2H2O(l)ΔH° = -890 kJ
2HCl(g) → H2(g) + Cl2(g)ΔH° = 185 kJ
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) ΔH° = -1169 kJ
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Rate of Reactions Kinetics – study of reaction rates Rate Influencing Factors- for reactions to occur, particles must
come into contact with each other (collide) in a favorable way.
- anything that changes the frequency of collisions or the efficiency of collisions will change the rate of the reaction
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1) Surface Area - In homogeneous mixtures
(dissolved) particles mix and collide freely so those reactions occur rapidly
- Heterogeneous mixtures can only react where the 2 phases are in contact with each other; so the surface area of a solid is a factor;
• Increase surface area, increase rate
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2) Temperature - Increasing T increases KE of particles
(move faster); therefore there will be more collisions and rate will increase
- Decreasing T decreases KE and number of collisions, so rate decreases
3) Concentration
- The greater the concentration of a substance, the more particles there will be; the more particles, the more collisions and increase in rate
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- Lowering the concentration has the opposite effect on the rate.
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
MnO2
2 H2O2 ----- > 2 H2O + O2