thermochemistry - chapter 6

8/13/2019 Thermochemistry - Chapter 6

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ThermochemistryChapter 6


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Energy

Energyis defined as the capacity to do work

or produce heat.

The Law of Conservation of Energy states that

energy can be neither created or destroyed it

can only change form.

Therefore, the energy of the universe is constant.


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Potential or Kinetic?

Energy can be classified as either potential or

kinetic.

Potential energy is stored energy resulting from

an objects position or composition.

Potential energy due to position results from gravity.

Potential energy due to composition results from

attractive forces between atoms (bonds). Kinetic Energy is the energy of motion.

K.E. = mv2 where m=mass and v=velocity


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Heat vs Temperature

Temperatureis a property that reflects the

random motion of the particles in a substance.

It is a measure of the avg. kinetic energy of

the particles.

Heatinvolves the transfer of energy between

two objects due to a temperature difference.


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Work

Workis defined as force acting over a

distance.

There are 2 ways to transfer energy:

Through Work

Through Heat

The way that energy transfer is divided

between work and heat is dependent upon

the specific conditions, called the pathway.


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The total energy transferred is constantregardless of the pathway.

The amounts of work and heat, however, will differ.

Energy change is independent of the pathway;however, work and heat are both dependent onthe pathway.

A state function refers to a property of thesystem that depends only on its present state. Itdoes NOT depend in any way on the systems past

or future. A change in a state function is independent of the

pathway taken between the two states.

Energy is a state function, work and heat are NOT.


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Chemical Energy

Consider the combustion of methane which is usedto heat many homes.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy (heat)

To discuss this reaction we divide the universe into 2

parts: the system and the surroundings.

The systemis the part of the universe on which we wish to

focus attention.

The surroundingsinclude everything else in the universe.

In the above example we would define the system as

the reactants and products of the reaction. The

surroundings consists of everything else.


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Exothermic vs Endothermic

When a reaction results in the evolution of

heat, it is said to be exothermic; that is,

energy flows out of the system.

In the combustion of methane energy flows out of

the system as heat.

When a reaction absorbs energy from the

surroundings it is said to be endothermic; thatis, energy flows into the system.


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Where does the energy, released as heat, come from in anexothermic reaction?

Since energy must be conserved we know that the energy

gained by the surroundings must be equal to the energy lost bythe system.

In the combustion of methane, the energy content of thesystem decreases, which means that 1 mole of CO2and 2 molesof H2O (the products) possess less potential energy than do 1

mole of CH4and 2 moles of O2(the reactants). In any exothermic reaction, some of the potential energy stored

in the chemical bonds is being converted to thermal energy viaheat.

For exothermic rxns: the P.E. of the products is less than the

P.E. of the reactants. In any endothermic reaction, the situation is reversed. Energy

that flows into the system is used to increase the potentialenergy of the system.

For endothermic rxns: the P.E. of the products is greater than

the P.E. of the reactants.


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Combustion of Methane


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PE PE represents the change in potential energy stored in the

bonds of the products as compared with the bonds of thereactants.

Energy is required or absorbed when bonds are broken.

Energy is released when new bonds are formed.

In exothermic reactions more energy is released by formingthe new bonds in the products than is used to break the bonds

in the reactants.

Thus, the surplus energy is released to the surroundings as

heat. In endothermic reactions more energy is used to break the

bonds in the reactants than is released by forming the new

bonds in the products.

Thus, energy must be absorbed from the surroundings.


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Endothermic Reaction


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Thermodynamics

The study of energy and its interconversions is calledthermodynamics.

The Law of Conservation of Energy is often called theFirst Law of Thermodynamics and is stated as follows:The energy of the universe is constant.

The internal energy E of a system can be defined as thesum of the kinetic and potential energies of all theparticles in the system.

The internal energy of a system can be changed by a

flow of work, heat, or both.E = q + w

Where E represents the change in internal energy, qrepresents heat, and w represents work.


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Thermodynamic quantities always consist of a

number and a sign, indicating the direction of

flow. The sign reflects the systems point of view.

If energy flows into the system via heat

(endothermic reaction), q is assigned a positivesign.

When energy flows out of the system (exothermic

reaction), q is assigned a negative sign.E is always positive for endothermic rxns.

E is always negative for exothermic rxns.


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Sample Problem #1

Calculate E for a system undergoing anendothermic process in which 15.6 kJ of heatflows and where 1.4 kJ of work is done on the

system.We use the equation: E = q + w

q = +15.6 kJ, since the process is endothermic

W = +1.4 kJ, since work is done on the system.

E = 15.6 kJ + 1.4 kJ = 17.0 kJ

The system has gained 17.0 kJ of energy.


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Work Applied to Gases

A common type of work associated with chemicalprocesses is work done by a gas (through expansion) orwork done to a gas (through compression).

Work = -Pressure x Change in Volume

or

w = -PV

For a gas expanding against an external pressure, wis a

negative quantity since work flows out of the system.For a gas being compressed by an external pressure, wis a positive quantity since work flows into the system.


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Sample Problem #3

A balloon is being inflated to its full extent byheating the air inside it. In the final stages of

this process, the volume of the balloon

changes from 4.00 x 10

6

L to 4.50 x 10

6

L bythe addition of 1.3 x 108J of energy as heat.

Assuming that the balloon expands against a

constant pressure of 1.0 atm, calculateEfor

the process. (To Convert between L atm and

J, use 1 L atm = 101.3 J)


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Sample Problem # 3 - Solution

We use the equation:

E = q + w

Since the problem states that 1.3 x 108J of energy isadded as heat,

q = 1.3 x 108JThe work can then be calculated from:

w = -P V

w = -(1.0 atm) x (4.50 x 106L - 4.00 x 106L)

w = -5.0 x 105L atm

To calculate E, we must sum qand w. Since theunits are not the same we must change the unit ofwork to J.


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Sample Problem #3 (Solution cont.)

w = -5.0 x 105L atm

w = -5.0 x 105L atm x 101.3 J

1 L atmW = -5.1 x 107J

E = q + w

E = (+1.3 x 108J) + (-5.1 x 107J)E = 8 x 107J


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Enthalpy

Enthalpy, H, is defined as:

H = E + PV

Where E is the internal energy of the system, Pis the

pressure of the system, and Vis the volume of thesystem.

Enthalpy is a state function.

For a process carried out at constant pressure and

where the only work allowed is that from a volumechange, we can say that the change in enthalpy of thesystem is equal to the energy flow as heat:

H = q


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For this reason, the terms heat of reactionand change in enthalpy are usedinterchangeably for reactions studied atconstant pressure.

For a chemical reaction, the enthalpychange is given by the equation:

H = H products - H reactants


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In a case in which the reactants of areaction have a greater enthalpy than the

products, H will be negative.The overall decrease in enthalpy isachieved by the generation of heat whichhas flowed out of the system and thereaction is exothermic.

In a case in which the products of areaction have a greater enthalpy than the

reactants,H will be positive.Thus, heat has been absorbed by thesystem and the reaction is endothermic.


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Sample Problem #4

When 1 mole of methane (CH4) is

burned at constant pressure, 890 kJ

of energy is released as heat.CalculateHfor a process in which a

5.8 g sample of methane is burned at

constant pressure.


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Sample Problem #4 Solution

5.8 g CH4 x 1 mol CH4 x -890 kJ =

16.0 g CH4 1 mol CH4

H = -320 kJ


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Calorimetry

The instrument used to determine the heat

associated with a chemical reaction is called a

calorimeter.

Substances respond differently to being

heated.

The heat capacity, Cp, of a substance is defined

as Cp = heat absorbed = H

increase in temperature T


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Specific Heat Capacity

When heat capacity is

given per gram of

substance it is called

the specific heatcapacity and its units

are J/C g


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Constant Pressure Calorimetry

The measurement of

heat using a simple

calorimeter.

Pressure remainsconstant.

Used to determine

changes in enthalpy for

reactions occurring in

solution.


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Calorimetry Calculations

Energy released by the reaction =

Energy absorbed the solution =

mass of solution (m) x specific heat capacity

(c) x change in temp. (T)

q = m x c x T

The specific heat capacity of water is 4.18 J/C gand is used for calculations involving aqueous

solutions.


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Calorimetry Calculation #1

When 1.00 L of 1.00 M Ba(NO3)2solution at

25 C is mixed with 1.00 L of 1.00 M Na2SO4

solution at 25 C in a calorimeter, the white

solid BaSO4forms and the temperature of themixture increases to 28.1 C. Assume that the

specific heat of solution is 4.18 J/C g and the

density of solution is 1.0 g/mL. Calculate theenthalpy change per mole of BaSO4formed.


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Ba(NO3)2 + Na2SO4 BaSO4 + 2NaNO3Sodium and nitrate are spectator ions so the net ionicequation is

Ba2+(aq) + SO4-(aq) BaSO4(s)

We use the equation: q = m x c x T

We first need to find the mass of solution

The total volume of solution is 2.0 L, so

2.0 L x 1000 mL x 1.0 g = 2.0 x 103g

1 L 1 mL

q = (2.0 x 103g) x (4.18 J/C g) x (28.1C-25.0C)

q = -2.6 x 104J

Since, 1.0 L of 1.0M barium nitrate contains 1.0 mol ofBa2+, and 1.0L of 1.0M sodium sulfate contains 1.0 molof SO4

2-, 1.0 mol of BaSO4is formed. Therefore

H = -2.6 x 104J/mol or -26 kJ/mol


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A 110. g sample of copper (specific heat

capacity = 0.20 J/C g) is heated to 82.4 C and

then placed in a container of water at 22.3 C.

The final temp. of the water and metal is24.9 C. What is the mass of the water in the

container assuming all heat was transferred

from the metal to the water?


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Heat In = Heat Out

Heat lost by copper = Heat gained by water

q(Cu) = qH2O

m x c x T = m x c x T

(110. g) x (0.20 J/C g) x (57.5C) = (m) x (4.18 J/C g) x(2.6C)

116 g = mass of water


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Constant Volume Calorimetry

Energy changes in

reactions at constant

volume are performed

in a bomb calorimeter. Energy change is

determined by

measuring the increase

in the temp. of thewater and other

calorimeter parts.


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A 0.1964 g sample of quinone (C6H4O2) is

burned in a bomb calorimeter that has a heat

capacity of 1.56 kJ/C. The temperature of the

calorimeter increases by 3.2 C. Calculate theenergy of combustion of quinone per gram

and per mole.


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In a bomb calorimeter

q = (heat capacity) x (change in temp) so

q = (1.56 kJ/C) x (3.2 C)

q = 5.0 kJ = energy of combustion5.0 kJ / 0.1964 g = 25 kJ/g

25 kJ x 108 g = 2700 kJ/mol

1 g 1 mol


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Hesss Law

When going from a particular set of reactants toa particular set of products, the change inenthalpy is the same whether the reaction takesplace in one step or multiple steps.

When using Hesss Law its important to note:

If a reaction is reversed the sign of H is reversed.

The magnitude of H is directly proportional to the

quantities of reactants and products. If thecoefficients in a balanced rxn are multiplied by aninteger, the value of H is multiplied by the sameinteger.


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Hints for Hesss Law

Calculations involving Hesss Law typically requirethat several reactions be manipulated andcombined to finally give the reaction of interest.

It is often easiest to work backwards from therequired reaction using the reactants andproducts to determine how to manipulate theother reactions at your disposal.

Reverse any reactions as needed and multiplythose reactions by an integer if necessary to getthe required amounts.


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Standard Enthalpy of Formation

The standard enthalpy of formation (Hf) of

a compound is defined as the change in

enthalpy that accompanies the formation of

one mole of a compound from its elementswith all substances in their standard states.


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Definitions of Standard State

For a Compound

The standard state of a gaseous substance is apressure of 1 atm.

For a pure substance in a condensed state, thestandard state is the pure liquid or solid.

For a substance present in solution, the standard stateis a concentration of exactly 1M.

For an Element The standard state for an element is the form in which

the element exists under conditions of 1 atm and 25C


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thermochemistry - chapter 6

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