class05 chemistryg12 notes and homework

8/4/2019 Class05 ChemistryG12 Notes and Homework

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GRADE12CHEMISTRY

OLYMPIADSCHOOL

WEDNESDAY2:304:30

FRIDAY2:304:30


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Chapter 5-Energy and Change

(Studying Energy Changes) e aw o conserva on o energy s a es a e o a

energy of the universe is constant.

,created. This idea can be expressed by the followingequation: Euniverse = 0

Energy can, however, be transferred from one substanceto another.

.interpret energy changes, scientists must clearly define

what part of the universe they are dealing with.


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The s stem is defined as the art of the universe that is

being studied and observed. In a chemical reaction, the system is usually made up of

. By contrast, the surroundings are everything else in the

universe. The two equations below show the relationship, ,

surroundings. Universe = S stem + Surroundin s

Euniverse = Esystem + Esurroundings = 0

From the relationship, we know that any change in thesys em s accompan e y an equa an oppos e c angein the surroundings.E = E

The surroundings are considered to be only the part ofthe universe that is likely to be affected by the energy


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Heat and Temperature

Heat, Q, refers to the transfer of kinetic energy. Heat is .

Heat is transferred spontaneously from a warmer object to acooler ob ect. When ou close the door of our home on a cold

day to prevent the cold from getting in, you are actuallypreventing the heat from escaping. You are preventing the

objects, including the cold air, outside.


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, ,kinetic energy of the particles that make up asubstance or system. You can think oftem erature as a wa of uantif in how hot orcold a substance is, relative to another substance.Temperature is measured in either Celsiusde rees (C) or kelvins (K).

The Celsius scale is a relative scale. It wasdesigned so that waters boiling point is at

100

C and waters melting point is at 0

C.

The Kelvin scale, on the other hand, is anabsolute scale. It was designed so that 0 K is thetemperature at which a substance possesses no

kinetic energy.

Temperature in Kelvin degrees = Temperature inCelsius degrees + 273.15


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Enthalpy and Enthalpy Change Chemists define the total internal energy of a substance at a

constant pressure as its enthalpy, H.

Chemists do not work with the absoluteenthalpy of the

reactants and products in a physical or chemical process. ns ea , ey s u y e en a py c ange, , a accompan es

a process. That is, they study the relativeenthalpy of thereactants and roducts in a s stem.

This is like saying that the distance between your home andyour school is 2 km. You do not usually talk about the absolutepos on o your ome an sc oo n erms o e r a u e,longitude, and elevation. You talk about their relativeposition, in

relation to each other. The enthal chan e of a rocess isequivalent to its heat change at constant pressure.


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Enthalpy Changes in Chemical

In chemical reactions, enthalpy changes result from chemical bondsbein broken and formed. Chemical bonds are sources of storedenergy.

Breaking a bond is a process that requires energy. Creating a bond isa process that releases energy.

For example, consider the combustion reaction that takes place whennitrogen reacts with oxygen.

N2(g) + O2(g) 2NO(g)

In this reaction, one mole of nitrogen-nitrogen triple bonds and onemole of oxygen-oxygen double bonds are broken. Two moles ofnitrogen-oxygen bonds are formed. This reaction absorbs energy. In

-,bonds than is used to break one nitrogen-nitrogen bond and oneoxygen-oxygen bond. When a reaction results in a net absorptionof

energy, it is called an endothermic reaction. On the other hand, when a reaction results in a net releaseof energy,

it is called an exothermic reaction. In an exothermic reaction, moreenergy is released to form bonds than is used to break bonds.

ere ore, energy s re ease .


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The relationship between bond breaking, bond

, reactions.


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Representing Enthalpy Changes The enthalpy change of a chemical reaction is known as the

enthalpy of reaction, Hrxn.

The enthalpy of reaction is dependent on conditions such as

temperature and pressure. Therefore, chemists often talk about the standard enthalpy of

reaction,Hrxn: the enthalpy change of a chemical reaction

a occurs a an a . Often,Hrxn is written simply asH. The symbol is called

.state or under standard conditions.

reactionin other chemistry books.


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Representing Exothermic Reactions There are three different ways to represent the enthalpy change

of an exothermic reaction. e s mp es way s o use a ermoc em ca equa on: a

balanced chemical equation that indicates the amount of heatthat is absorbed or released by the reaction it represents. Forexample, consider the exothermic reaction of one mole ofhydrogen gas with half a mole of oxygen gas to produce liquidwater. For each mole of hydrogen gas that reacts, 285.8 kJ of

heat is produced. Notice that the heat term is included with theproducts because heat is produced.

. You can also indicate the enthalpy of reaction as a separate

expression beside the chemical equation. For exothermic, .

H2(g) + O2(g) H2O(l) Hrxn = 285.8 kJ A third wa to re resent the enthal of reaction is to use an

enthalpy diagram.


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Representing Endothermic Reactions The endothermic decomposition of solid magnesium carbonate

produces solid magnesium oxide and carbon dioxide gas. For, .

of energy is absorbed.

As for an exothermic reaction, there are three different wa s torepresent the enthalpy change of an endothermic reaction.

You can include the enthalpy of reaction as a heat term in thec em ca equa on. ecause ea s a sor e n anendothermic reaction, the heat term is included on the reactantside of the e uation.

117.3 kJ + MgCO3(s) MgO(s) + CO2(g)

You can also indicate the enthalpy of reaction as a separateexpression beside the chemical reaction. For endothermicreactions, the enthalpy of reaction is always positive.

rxn = .

Finally, you can use a diagram to show the enthalpy of reaction.


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H=1g/mol,O=16g/mol,P=31g/mol


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Heat Changes and Physical Changes

Enthalpy changes are associated with physical changes as well aswith chemical reactions.

You have observed examples of these enthalpy changes in your daily

life. Suppose that you want to prepare some pasta. You put anelement causes the water to become steadily hotter, until it reaches100C (the boiling point of water at 100 kPa). At this temperature,heat is still bein added to the water. The avera e kinetic ener of

the liquid water molecules does not increase, however. Instead, theenergy is used to break the intermolecular bonds between the watermolecules as they change from liquid to vapour. The temperature of

e qu wa er rema ns a un a e wa er as eenvaporized. If you add heat to the vapour, the temperature of the

vapour will increase steadily. en you eat ce t at s co er t an 0 , a s m ar process occurs.The temperature of the ice increases until it is 0C (the melting pointof water). If you continue to add heat, the ice remains at 0C buteg ns o me , as e on s e ween e wa er mo ecu es n e so

state begin to break.


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You can represent the enthalpy change that accompanies aphase changefrom liquid to solid, for examplejust like youre resented the enthal chan e of a chemical reaction. You

can include a heat term in the equation, or you can use aseparate expression of enthalpy change.

, , .of energy.

H2O(s) + 6.02 kJ H2O(l) H2O(s) H2O(l) H= 6.02 kJ Normally, however, chemists represent enthalpy changes

.

These symbols are described below. Enthalpy of vaporization, Hvap : the enthalpy change for the

Enthalpy of condensation, Hcond: the enthalpy change for the

hase chan e of a substance from as to li uid Enthalpy of melting, Hmelt: the enthalpy change for the phase

change of a substance from solid to liquid

, fre

change of a substance from liquid to solid


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Vaporization and condensation are opposite. ,

these processes have the same value but

o osite si ns. For example, 6.02 kJ is needed to vaporize

one mole of water.

Therefore, 6.02 kJ of energy is releasedwhen one mole of water freezes.

Hvap = Hcond

Similarl meltin and freezin are o ositeprocesses. H = H


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Determining Enthalpy

f R i n Ex rim n ifi H

Capacity)

one gram of a substance 1C (or 1 K) is the specific heat capacity,c, of the substance. Specific heat capacity is usually expressed in

-1 o -1 You can use the specific heat capacity of a substance to calculate

the amount of energy that is needed to heat a given mass a.

You can also use the specific heat capacity to determine theamount of heat that is released when the temperature of a given

. .J/g C. This relatively large value indicates that a considerable

amount of energy is needed to raise or lower the temperature of. All samples of the same substance have the same specific heat

capacity. In contrast, heat capacity, C, relates the heat of asamp e, o ec , or sys em o s c ange n empera ure. ea

capacity is usually expressed in units of kJ/C.


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full of water at room temperature. All the

,

but the two samples have different heat.

heat to raise the temperature of the water in

raise the temperature of the water in theteacu b 10C. Therefore the water in the

bathtub has a higher heat capacity.


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C


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Specific Heat Capacity and Heat

You can use the following equation to calculate the heat change of asubstance, based on the mass of the substance. You can also use

and the change in its temperature.

Q=

mx c x

T , ,C), T= Tf (final temperature) Ti (initial temperature)(C or K)

Water is often used in controlled surroundings to measure the heat ofa reaction. For exam le ou can use the e uation above to

determine the amount of energy that is needed to heat 1.00 x 10^2 gof water from 20.0C to 45.0C. Q= mx c xT The mass of the water is 1.00 x 10^2 g. The specific heat capacity of

water is 4.184 J/g C. The temperature of the water increases by25.0C.

= . x g . g . = 1.05 x 10^4 J To raise the tem erature of 1.00x 10^2 of water b 25C, 1.05 x

10^4 J of heat is needed.

M i H T f i


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Measuring Heat Transfer in a

A calorimeter is used to measure enthalpy changes for.

insulating a system from its surroundings. By measuring the

temperature change of the system, you can determine theamoun o ea a s re ease or a sor e y e reac on. orexample, the heat that is released by an exothermic reactionraises the tem erature of the s stem.

Qreaction = Qinsulated system


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There are various types of calorimeters. Forinstance, a bomb calorimeterallows chemists to

determine energy changes under conditions ofconstant volume. We have so far learned that an enthalpy change

reactants at a constant pressure. Therefore, thecalorimeter we use to determine an enthalpy

change should allow the reaction to be carried outopen to the atmosphere.

To determine enthalpy changes in high schoollaboratories, a coffee-cup calorimeter provides

. -

composed of two nested polystyrene cups (coffeecups). They can be placed in a 250 mL beaker foradded stability. Since a coffee-cup calorimeter is

,constant-pressure calorimeter.

As with any calorimeter, each part of the coffee-cup

calorimeter has an associated heat capacity. ,however, and because a coffee-cup calorimeter isnot as accurate as other calorimeters, the heatcapacity of a coffee-cup calorimeter is usually

.

value of 0 J/C.

U i C l i D i h


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Using a Calorimeter to Determine the

n a py o a eac on A coffee-cup calorimeter is well-suited to determining the.

The water in the calorimeter absorbs (or provides) the energy

that is released (or absorbed) by a chemical reaction. Whencarry ng out an exper ment n a ute so ut on, t e so ut on tseabsorbs or releases the energy.

released by the solution using the equation mentioned earlier,Q= mx c xT, , ,

absorbs the heat.

When a dilute aqueous solution is used in a calorimeter, youcan assume t at t e so ut on as t e same ens ty an spec cheat capacity as pure water. You can also assume that the heatcapacity of the calorimeter is negligible. Therefore, you canassume that all the heat that is released or absorbed by the

reaction is absorbed or released by the solution.


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Hesss Law of Heat Summation

Chemists can determine the enthalpy change of any reaction,

summation.

This law states that the enthalpy change of a physical or

chemical process depends only on the beginning conditions(reactants) and the end conditions (products). The enthalpy

number of intermediate steps in the process.

It is the sum of the enthalpy changes of all the individual stepsthat make up the process.

l


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Example

For example, carbon and oxygen can form carbon dioxidevia two pathways. 1. Carbon can react with ox en to form carbon monoxide.

The carbon monoxide then reacts with oxygen to producecarbon dioxide. The two equations below represent this

pathway. C(s) + O2(g) CO(g) H= 110.5 kJ CO(g) + O2(g) CO2(g) H= 283.0 kJ

.

dioxide directly. C(s) + O2(g) CO2(g) H= 393.5 kJ n ot cases, t e net resu t s t at one mo e o car on

reacts with one mole of oxygen to produce one mole of

carbon dioxide. In the first athwa all the carbonmonoxide that is produced reacts with oxygen to formcarbon dioxide.) Notice that the sum of the enthalpy

change for the second pathway.


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Combining Chemical Equations


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Combining Chemical Equations

According to Hesss law, the pathway that is taken in a

reaction. How can you use Hesss law to calculate the enthalpy

change of a reaction? One way is to add equations forreactions with known enthalpy changes, so that their net resultis the reaction you are interested in.

e up equa ons an as s own. e pro uc s an

the reactants. Then cancel any substances that appear on.

For example, you can combine thermochemicali (1) d (2) b l fi d h h l


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p , yequations (1) and (2) below to find the enthalpy

peroxide, equation (3). 1 H2O2 H2 + O2 H = +188 kJ (2) H2(g) + O2(g) H2O(l) H = 286 kJ

(3) H2O2() H2O(l) + 1/2 O2(g) H

= ? Carefully examine equation (3), the targetequation.Notice that H2O2 is on the left (reactant) side,

side. Now examine equations (1) and (2). Noticewhich sides of the equations H2O2 and H2O areon. ey are on e correc s es, ase onequation (3). Also notice that hydrogen does not

a ear in e uation 3 . Therefore it must cancelout when equations (1) and (2) are added. Sincethere is one mole of H2(g) on the product side of

side of equation (2), these two terms cancel.


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.

Therefore, you know that the enthalpy change for thedecom osition of h dro en eroxide is the sum of the enthalchanges of equations (1) and (2).

H2O2 l H2O l + O2 H

= 188 kJ 286 kJ = 98 kJ


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In the previous example, you did not need to manipulate

the two equations with known enthalpy changes. Theyadded to the target equation as they were written. In, ,

equations before adding them. There are two key ways in

which ou can mani ulate an e uation: 1. Reverse an equationso that the products become

reactants and the reactants become roducts. When ou

reverse an equation, you need to change the sign ofH(multiply by -1).

2. Multiply each coefficient in an equationby the sameinteger or fraction. When you multiply an equation, you

.


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Calculating Enthalpy Changes


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Calculating Enthalpy Changes You can calculate the enthalpy change of a chemical reaction

b addin the heats of formation of the roducts andsubtracting the heats of formation of the reactants.

The following equation can be used to determine the enthalpychange of a chemical reaction.

H = (n Hf products) (n Hf reactants)

In this equation, nrepresents the molar coefficient of eachcompound in the balanced chemical equation and means thesum o .

As usual, you need to begin with a balanced chemical equation.

1, you need to multiply its Hf by the same molar coefficient.

f .

Consider, for example, the complete


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, p , p

, .

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

Using the equation for the enthalpy change and

calculate the enthalpy change of this reaction.

u s u e e s an ar en a p es o orma on

to get the following calculation.

H= [(393.5 kJ/mol) + 2(241.8 kJ/mol)] [(

. .CH4

H d thi th d f ddi h t f


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How does this method of adding heats of

formation relate to Hesss law? Consider theequations for the formation of each compoundthat is involved in the reaction of methane with

oxygen. (1) H2(g) + O2(g) H2O(g) Hf = 241.8

(2) C(s) + O2(g) CO2(g) Hf = 393.5 kJ s + g g f = .

There is no equation for the formation ofoxygen, because oxygen is an element in itsstandard state.

By adding the formation equations, you can obtain the targetequation Notice that you need to reverse equation (3) and


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equation. Notice that you need to reverse equation (3) andmulti l e uation 1 b 2.

2 x (1) 2H2(g) + O2(g) 2H2O(g) Hf = 2(-241.8) kJ (2) C(s) + O2(g) CO2(g) Hf = 393.5 kJ

-1 x (3) CH4(g) C(s) + 2H2(g) Hf = 1(74.6) kJCH4(g) + 2O2(g) + C(s) + 2H2(g)

or

CH4 + 2O2 2H2O + CO2

Add the manipulated

Hf values:

H=

2(241.8) kJ- 393.5kJ + 74.6 kJ

= . This value of His the same as the value you obtained using

H data. When ou used the addition method, ou erformedthe same operations on the enthalpies of formation before

adding them. ,

enthalpy of a reaction is consistent with Hesss law.


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class05 chemistryg12 notes and homework

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