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Chapter 11 - Thermochemistry Heat and Chemical Change
Milbank High School
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Section 11.1 The Flow of Energy - Heat
OBJECTIVES: Explain the relationship between energy and heat.
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Section 11.1 The Flow of Energy - Heat
OBJECTIVES: Distinguish between heat capacity and specific heat.
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Energy and Heat Thermochemistry - concerned with heat changes that occur during chemical reactions Energy - capacity for doing work or supplying heat
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Energy and Heat Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.
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Exothermic and Endothermic Processes
In studying heat changes, think of defining these two parts: the system the surroundings
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Exothermic and Endothermic Processes
The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. All the energy is accounted for as work, stored energy, or heat.
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Exothermic and Endothermic Processes
Fig. 11.3a, p heat flowing into a system from it’s surroundings: defined as positive q has a positive value called endothermic system gains heat as the surroundings cool down
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Exothermic and Endothermic Processes
Fig. 11.3b, p heat flowing out of a system into it’s surroundings: defined as negative q has a negative value called exothermic system loses heat as the surroundings heat up
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Exothemic and Endothermic
Every reaction has an energy change associated with it Exothermic reactions release energy, usually in the form of heat. Endothermic reactions absorb energy Energy is stored in bonds between atoms
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Heat Capacity and Specific Heat
A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC. Used except when referring to food a Calorie, written with a capital C, always refers to the energy in food 1 Calorie = 1 kilocalorie = 1000 cal.
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Heat Capacity and Specific Heat
Joule-- the SI unit of heat and energy 4.184 J = 1 cal Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”)
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Heat Capacity and Specific Heat
For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC) Thus, for water: it takes a long time to heat up, and it takes a long time to cool off! Water is used as a coolant! Note Figure 11.7, page 297
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Heat Capacity and Specific Heat
To calculate, use the formula: q = mass (g) x T x C heat abbreviated as “q” T = change in temperature C = Specific Heat Units are either J/(g oC) or cal/(g oC) Sample problem 11-1, page 299
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Section 11.2 Measuring and Expressing Heat Changes
OBJECTIVES: Construct equations that show the heat changes for chemical and physical processes.
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Section 11.2 Measuring and Expressing Heat Changes
OBJECTIVES: Calculate heat changes in chemical and physical processes.
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Calorimetry Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.
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Calorimetry For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system
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Calorimetry Changes in enthalpy = H
q = H These terms will be used interchangeably in this textbook Thus, q = H = m x C x T H is negative for an exothermic reaction H is positive for an endothermic reaction (Note Table 11.3, p.301)
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C + O2 ® CO2 + 395 kJ Energy Reactants Products C + O2 395kJ C O2
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In terms of bonds C O O C O Breaking this bond will require energy. C
Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them.
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Exothermic The products are lower in energy than the reactants
Releases energy
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CaCO3 + 176 kJ ® CaO + CO2 CaCO3 ® CaO + CO2 Energy Reactants Products
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Chemistry Happens in MOLES
An equation that includes energy is called a thermochemical equation CH4 + 2O2 ® CO2 + 2H2O kJ 1 mole of CH4 releases kJ of energy. When you make kJ you also make 2 moles of water
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Thermochemical Equations
A heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions for the reaction is kPa (1 atm.) and 25 oC
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CH4 + 2 O2 ® CO2 + 2 H2O kJ If grams of CH4 are burned completely, how much heat will be produced? 1 mol CH4 802.2 kJ 10. 3 g CH4 16.05 g CH4 1 mol CH4 = 514 kJ
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How many grams of water would be produced with 506 kJ of heat?
CH4 + 2 O2 ® CO2 + 2 H2O kJ How many liters of O2 at STP would be required to produce 23 kJ of heat? How many grams of water would be produced with 506 kJ of heat?
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Summary, so far...
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Enthalpy The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy changes
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Enthalpy Symbol is H Change in enthalpy is DH (delta H)
If heat is released, the heat content of the products is lower DH is negative (exothermic) If heat is absorbed, the heat content of the products is higher DH is positive (endothermic)
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Energy Change is down DH is <0 Reactants Products
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Energy Change is up DH is > 0 Reactants Products
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Heat of Reaction The heat that is released or absorbed in a chemical reaction Equivalent to DH C + O2(g) ® CO2(g) kJ C + O2(g) ® CO2(g) DH = kJ In thermochemical equation, it is important to indicate the physical state H2(g) + 1/2O2 (g)® H2O(g) DH = kJ H2(g) + 1/2O2 (g)® H2O(l) DH = kJ
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Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance Note Table 11.4, page 305
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Section 11.3 Heat in Changes of State
OBJECTIVES: Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.
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Section 11.3 Heat in Changes of State
OBJECTIVES: Calculate heat changes that occur during melting, freezing, boiling, and condensing.
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Heats of Fusion and Solidification
Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies
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Heats of Vaporization and Condensation
Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid. Table 11.5, page 308
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Heats of Vaporization and Condensation
Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses Hvap = - Hcond
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Heats of Vaporization and Condensation
Note Figure 11.5, page 310 The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous You can receive a scalding burn from steam when the heat of condensation is released!
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Heats of Vaporization and Condensation
H20(g) H20(l) Hcond = kJ/mol Sample Problem 11-5, page 311
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Heat of Solution Heat changes can also occur when a solute dissolves in a solvent. Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substance
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Section 11.4 Calculating Heat Changes
OBJECTIVES: Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.
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Section 11.4 Calculating Heat Changes
OBJECTIVES: Calculate heat changes using standard heats of formation.
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Hess’s Law If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s law of heat summation Example shown on page 314 for graphite and diamonds
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Why Does It Work? If you turn an equation around, you change the sign:
If H2(g) + 1/2 O2(g)® H2O(g) DH= kJ then, H2O(g) ® H2(g) + 1/2 O2(g) DH = kJ also, If you multiply the equation by a number, you multiply the heat by that number: 2 H2O(g) ® 2 H2(g) + O2(g) DH = kJ
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Standard Heats of Formation
The DH for a reaction that produces 1 mol of a compound from its elements at standard conditions Standard conditions: 25°C and 1 atm. Symbol is The standard heat of formation of an element = 0 This includes the diatomics
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What good are they? DHo = ( Products) - ( Reactants)
Table 11.6, page 316 has standard heats of formation The heat of a reaction can be calculated by: subtracting the heats of formation of the reactants from the products DHo = ( Products) - ( Reactants)
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Examples CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) CH4 (g) = - 74.86 kJ/mol
O2(g) = 0 kJ/mol CO2(g) = kJ/mol H2O(g) = kJ/mol DH= [ (-241.8)] - [ (0)] DH= kJ
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