Thermochemistry
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Thermochemistry. The study of the transfer of heat energy. Let’s Revisit Intensive and Extensive Properties. Intensive Extensive Video Clip Restate in your own words intensive and extensive properties. Energy Flow. Thermodynamics is the study of the flow of energy. Thermodynamic Energy Flow.

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Thermochemistry

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Thermochemistry

Thermochemistry

The study of the transfer of heat energy


Let s revisit intensive and extensive properties

Let’s Revisit Intensive and Extensive Properties

  • Intensive Extensive Video Clip

  • Restate in your own words intensive and extensive properties


Energy flow

Energy Flow

  • Thermodynamics is the study of the flow of energy.

  • Thermodynamic Energy Flow


Definitions

Definitions

  • Thermochemistry—the study of heat or thermal energy transfer and its application to chemical reactions and physical processes

  • Enthalpy—a measure of the total energy of a system

  • Heat capacity—the amount of heat that is required to change the temperature of that object by 1.00 degree Celsius (or Kelvin)

  • Specific heat capacity (specific heat)—the amount of heat energy needed to change the temperature of 1.00 gram of the substance by 1.00 degree Celsius


Different types of heat with physical processes

Different Types of Heat with Physical Processes

  • There is also a certain amount of energy associated with phase changes.

  • There is a difference in the sensible heat (the heat that is associated with temperature changes) and the latent heat (the heat that is associated with a change in phase of a substance while the temperature remains constant)

  • The heat associated with the phase change of solid to liquid is the latent heat of fusion (Hf). The heat required to change from a liquid to a vapor is called the latent heat of vaporization (Hv).


Heat with temperature change

Heat with Temperature Change

  • The specific heat of a substance can be determined experimentally by determining the quantity of heat transferred by a definite mass of the substance as its temperature rises or falls at constant pressure

  • Specific heat (cp) = quantity of heat transferred (Q) mass of material (m) x temperature change (DT)

  • Better known as:Q = m cpDT


Heating and cooling curve

Heating and Cooling Curve

  • Heating Curve Animation

  • Sketch a heating and cooling curve and label according to the animation


Example 1 temp change

Example 1: Temp Change

q = m c DT

How many joules of energy must be transferred from a cup of coffee to your body if the temperature of the coffee drops from 60.0 oC to 37.0 oC? Assume that the cup holds 250. mL and that the specific heat of coffee is the same as that of water, 4.18 J/goC.


Example 1 solution

Knowns:

DT = 60.0oC – 37.0oC

= 23.0oC

V = 250. mL

cp = 4.18 J/goC

Unknowns:

Q (heat energy)

Formula:

Q = mcpDT

What else do you need to solve it?

Mass!

Example 1 Solution


Example 1 solution1

Example 1 Solution

Q = mcpDT

Density = m/V and the D of H2O is 1.0 g/mL

D x V = m

1.0 g/mL x 250. mL = 250. g

Q = (250. g) (4.18 J/goC) (23.0oC) = 24035 J

= 24.0 kJ


Example 2 phase change

Example 2: Phase Change

  • Q = m Hf for solid to liquid

  • Q = m Hv for liquid to a vapor

    Example 2:

    You have four ice cubes at 0oC. If each cube has a mass of 9.63 g, how much heat is required to melt these ice cubes to liquid water at 0oC? The latent heat of fusion of ice is 333 J/g.


Example 2 solution

Knowns:

4 ice cubes

m of 1 cube = 9.63 g

Hfof ice = 333 J/g

Unknowns:

Q

Formula:

Q = m Hf

Example 2 Solution


Example 2 solution1

Example 2 Solution

q = m Hf

m = 4 x 9.63 g

= 38.5 g

q = (38.5 g) (333 J/g) = 12820.5 J

= 12.8 kJ


Heat transfer

Heat Transfer

  • When heat is transferred into the system, it is an endothermic process

  • When heat is transferred out of the system, the process is exothermic

  • All energy transfers follow the first law of thermodynamics, the law of energy conservation. All of the energy transferred between a system and its surroundings must be conserved

  • The heat transferred into (or out of) a system at a constant pressure is the enthalpy change (DH)

  • Enthalpies are experimentally determined, and can be calculated for individual reactions.


Enthalpy of formation

Enthalpy of Formation

  • One way to determine the DH of a reaction is to use a value called the enthalpy of formation (Hof)

  • The enthalpy of formation is defined as the energy needed to form one mole of the compound from its elements

  • Example: The energy for H2 & O2 to form H2O

  • The values are always in units of kJ/mol

  • The Hof of an element is 0 kJ/mol because element cannot be formed from anything simpler than itself

  • Enthalpies of formations are usually found in a reference table and are not something students memorize

  • The enthalpy of an entire reaction can be calculated from the Hof of the reactants and products

  • The formula is: DH = DHof(products) – DHof(reactants)


Example 3

Example 3

DH = DHof(products) – DHof(reactants)

  • Given:DHof CO2(g)= -393.5 kJ/mol

    DHof H2O(l) = -285.8 kJ/mol

    DHof C5H12 (g) = -145.7 kJ/mol

    Calculate the heat of the combustion reaction:

    C5H12(g) + 8O2(g)  5CO2 (g) + 6H2O(l)


Example 3 solution

Example 3 Solution

DH = DHof(products) – DHof(reactants)

Given:DHof CO2(g)= -393.5 kJ/mol

DHof H2O(l) = -285.8 kJ/mol

DHof C5H12 (g) = -145.7 kJ/mol

C5H12(g) + 8O2(g)  5CO2 (g) + 6H2O(l)

DH = [5(-393.5 kJ) + 6(-285.8 kJ)]

– [1(-145.7 kJ) + 0]

= -3536.6 kJ


Graphing enthalpy of reactions

Graphing Enthalpy of Reactions

  • An endothermic reaction means that the enthalpies of formation of the reactants is lower than the enthalpy of formations of the products

  • An exothermic reaction that the enthalpies of formation of the products is lower than the enthalpy of formations of the reactants


Reaction energy diagrams

Reaction Energy Diagrams

  • Note that the energy of each reaction does not go directly from reactants to products

  • There is an energy “hump” that must be overcome in order for a reaction to occur

  • This energy barrier is also called the energy of activation (or activation energy, Ea)

  • It is the amount of energy needed for a reaction to go from reactants to the transition complex (or activated complex) that must occur before the products can form. The energy associated with this activation step must be invested before the reaction will go to completion, no matter whether the net reaction is endothermic or exothermic.


Endothermic pathway

Endothermic Pathway

  • (a) = Energy of Activation (Ea)

  • (c) = Enthalpy of reaction (DHrxn)

  • Note that Enthalpy of reaction DHrxnis a positive number


Exothermic pathway

Exothermic Pathway

  • (a) = Energy of Activation (Ea)

  • (c) = Enthalpy of reaction (DHrxn)

  • Note that Enthalpy of reaction DHrxnis a negative number


Enthalpy signs

Enthalpy Signs

A positive enthalpy of reaction indicates that the reaction is endothermic, and the energy term can be written as if it is a reactant into the chemical equation. Any reaction that contains an energy term is known as a thermochemical equation

A negative enthalpy value indicates that the reaction is exothermic. The energy term is written as a product in the thermochemical reaction. Note that when the exothermic enthalpy is written into the products, it does not carry its negative sign with it

The positive or negative sign in front of an enthalpy term simply indicates whether the reaction is endothermic or exothermic. By convention, the math is not performed with negative numbers.


Example 4

Example 4

Carbon dioxide can be decomposed into carbon monoxide and oxygen.

2CO2 2 CO + O2

The DH for this reaction is 43.9 kJ

  • Is the reaction endothermic or exothermic?

  • If there is only 22.2 kJ of heat transferred, how many moles of CO2 can be decomposed?


Example 41

Example 4

2CO2 2 CO + O2

The DH for this reaction is 43.9 kJ

  • Is the reaction endothermic or exothermic?

  • Endothermic because it is a positive value

  • Therefore it can be written into the reactants side of the equation

    43.9 kJ + 2CO2 2 CO + O2


Example 4 solution

Example 4 Solution

43.9 kJ + 2CO2 2 CO + O2

Known: 22.2 kJ

Unknown: mol CO2

22.2 kJ x =


Example 4 solution1

Example 4 Solution

43.9 kJ + 2CO2 2 CO + O2

Known: 22.2 kJ

Unknown: mol CO2

22.2 kJ x 2 mol CO2 =

43.9 kJ


Example 4 solution2

Example 4 Solution

43.9 kJ + 2CO2 2 CO + O2

Known: 22.2 kJ

Unknown: mol CO2

22.2 kJ x 2 mol CO2 =1.01 mol CO2

43.9 kJ


Measuring heat transfer

Measuring Heat Transfer

  • Calorimetry is the process by which we measure heat transfer and its associated energy changes in the lab.

  • Calorimetry


Example 5

Example 5

  • Suppose you heat a 55.0-g piece of silver in the flame of a Bunsen burner to 425.0oC and then you plunge it into a beaker of water. The beaker holds 600. g of water, and its temperature before you drop in the hot silver is 25.0oC. The final temperature of the silver and water mixture is 44.0oC. What is the specific heat of the silver? The specific heat of water is 4.18 J/goC.


Example 51

Example 5

  • Suppose you heat a 55.0-g piece of silver in the flame of a Bunsen burner to 425.0oC and then you plunge it into a beaker of water. The beaker holds 600. g of water, and its temperature before you drop in the hot silver is 25.0oC. The final temperature of the silver and water mixture is 44.0oC. What is the specific heat of the silver? The specific heat of water is 4.18 J/goC.

  • Before solving, here is a hint: Keep the data about silver separate from the data about water


Example 5 solution

Example 5 Solution

  • Knowns:

    • Ag: m = 55.0 g; Ti = 425.0oC; Tf = 44.0oC

    • H2O: m = 600. g; cp = 4.18 J/goC; Ti = 25.0oC; Tf = 44.0oC

  • Unknown: cp for Ag

  • Formula: Q = mcpDT


Example 5 solution1

Example 5 Solution

  • Notice that you do NOT have enough information to solve for the cp of Ag because you do not know Q for Ag.

  • So what is Q? The heat the silver lost. Where did silver lose its heat? To the water! The water got the heat energy to warm up from the hot silver.

  • Because of heat transfer, if we can calculate the q of the water, we will know q for silver!


Example 5 solution2

Example 5 Solution

  • Solve for Q of water:

  • Q = mcpDT (*Be sure to use the data for WATER!)

    DT (H2O) = 44.0oC – 25.0oC = 19.0oC

  • Q (H2O) = 600 g x 4.18 J/goC x 19.0oC

  • Q (H2O) = 47652 J


Example 5 solution3

Example 5 Solution

  • From the first law of Thermodynamics:

    • Heat loss = Heat gained

  • Therefore Q (Ag) = Q (H2O)

  • Q (H2O) = 47652 J = Q (Ag)

  • So, for Ag, Q = mcpDT and now we have everything we need to solve for cp of Ag


Example 5 solution4

Example 5 Solution

  • Q = mcpDT & our knowns for Ag are

    m = 55.0 g; Ti = 425.0oC; Tf = 44.0oC; Q = 47652 J

    DT = 425.0oC – 44.0oC = 381.0oC

  • 47652 J = 55.0 g x cp x 381.0oC

  • cp = 2.27 J/goC


Example 5 solution5

Example 5 Solution

  • So the specific heat of Ag is about half that of H2O. This means that it takes twice as much heat to change the temperature of 1 g of H2O by 1oC as it does Ag metal. The high specific heat of water is one of its special properties!


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