Thermochemistry Jan2010

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Thermochemistry

Chem B Jan 2010 - Ms. Jessvin Sidhu 1


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Introduction In general, all chemical reactions and chemical

processes involves energy change :- Heat energy absorbed from the environment into

system = temp of environment/surrounding to


Heat energy released from the system into theenvironment to = temp of environment

/surrounding to

WHY ???


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EnthalpyEnthalpy => Heat of Reaction=> Heat of Reaction



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http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/enm1s3_4.swf



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Introduction A certain amount of chemical energy is stored

within every atom, molecule or ion.

Thi n r i h m f h n i l n r


(stored energy) + kinetic energy ( energy associated withmovement) of the substance and results from:

The attraction & repulsion btwn protons and electrons

The motion of the electrons

The movement of the atoms.


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Enthalpy

The total energy stored in a substance is calledthe enthalpy or heat content, of the substance.

Symbol for enthalpy is H(unit is Joule)


The First Law of Thermodynamicsstates that:- energy is neither created nor destroyed, but simply

changes from one form into another.

Whenever energy is converted from one form intoanother, the total quantity of energy remains thesame.


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Enthalpy

In all chemical reactions, the enthalpy/ energy of

the reactants and products differ (one is higherthan the other).

Unfortunately, we cannot directly measure the


heat content of a substance. However, we can measure the change in

enthalpy when a substance undergoes a

chemical reaction. ??? = measuring the rise @ fall of temperature

of the surrounding.


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Exothermic Reactions

Chemical reactions that release heat tothe environment = temp

In your opinion, which is higher ? Hreactant


or products (think First Law of Thermodynamics /conservation of energy )


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Exothermic Reaction The total chemical energy of the

reactants is more than theenergy of the products.

Since energy is never lost, thedifference in chemical energy


products is released into theenvironment = as heat energy.

Products are at lower energy level.Hence products are more stable.

Chemical bonds in products arestronger.

The change in enthalpy during a reaction is known as the heat ofreaction and is denoted by H


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Endothermic reaction

The chemical energy of thereactant is less than theenergy of the reactants.

Energy must be absorbed from


reactants for the reaction tooccur.

Temperature of thesurrounding become lower.

Reactants are at lower energylevel. Hence reactants aremore stable.

Chemical bonds in reactants

are stronger.


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Standard Laboratory Conditions The enthalpy change of reaction depends on the amount

of reactants used and the temperature of the reactantscompared to the products.

To remove those variables in enthalpy studies, the


o ow ng con t ons are assume :

This conditions are referred to as standard laboratoryconditions (SLC) Temperature at 25C 1 mole of substance is involved Concentration of solution is 1 mol/L ( 1M)

Pressure is kept at 1 atm

Heat change is measured in kilojoules (kJ)


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Thermochemical equations Thermochemical equations are just like other balanced equations

except they also specify the heat absorbed or produced for the

reaction.

The heat flow is listed to the right of the equation using the symbolH. The most common units are kJ/mol.


CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H= -802 kJ/mol

Coefficients refer to the number of moles. Thus, the equation is readas, when 1 mole of methane gas reacts with 2 moles of oxygen gas,1 mole of carbon dioxide and 2 moles of water vapour forms and

802 kJ of energy is released.

Hrefers to the equation as it is written, even though the unit isexpressed as kJ/mol.


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Certain laws or rules apply when usingthermochemical equations:

H is directly proportional to the quantity of a substance thatreacts or is produced by a reaction.

Amount of energy produced is directly proportional to mass.


Therefore, if you double the coefficients in an equation, then thevalue of H is multiplied by two. For example:

C2H4 (g) + 3O2 (g) 2CO2 + 2H2O (l); H = -1411 kJmol-1

2C2H4 (g) + 6O2 (g)

4CO2 + 4H2O (l);

H

= -2822 kJmol-1


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H for a reaction is equal in magnitude but oppositein sign to H for the reverse reaction.

For example: 2H2 (g) + O2 (g) 2H2O (g) H= - 468 kJ/mol


2H2O (g) 2H2 (g) + O2 (g) H= + 468 kJ/mol

If a reaction is reversed, H is equal to, but opposite in

sign, to that of the forward reaction.


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The physical state of matter must be shown, sincechanges of state require energy changes:



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Sample problem Sodium nitrate dissolves readily in water according to the equation

NaNO3(s) NaNO3(aq); H= +21.0 kJ mol1

a Determine the energy change when 1.0 g of solid NaNO3 isdissolved in water.

b Would ou ex ect the tem erature to rise or fall when sodium


nitrate dissolves in water? Explain your answer.


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n(NaNO3) = =0.0117 mole

1 mole NaNO3 absorbs 21 kJ energy

0.0117 mole absorbs 0.0117 21 kJ = 0.25 kJ

85

1


The temperature will fall. Since the reaction is endothermic theenthalpy of reactants is higher than the enthalpy of products, i.e. thereaction absorbs energy.


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Calculate the energy released when the followingquantities of ethane gas burn according to the equation:

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l); H= 3120 kJ mol1

a 3.00 mol

b 100


c 10.0 L at SLC


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From the equation, use stoichiometry to find energyreleased.

2 mol C2H6 releases 3120 kJ of energy

3.00 mol C2H6 releases 3.00 kJ2

3120

M

m


So energy = 4680 kJ

= 4.68 103 kJ


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Calculate the amount of ethane, using n=

n(C2H6) =

= 3.326 mol

Use stoichiometry to find energy released.

m

1molg30.068

g100


2 mol C2

H6

releases 3120 kJ of energy

3.326 mol C2H6 releases 3120 kJ

So energy released = 5189 kJ

= 5.19 103 kJ

2

326.3


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Calculate the amount of ethane, using n= at SLC.

n(C2H6) =

= 0.4082 mol

Use stoichiometry to find energy released.

mV

V

1molL24.5

L10.0


2 mol C2H6 releases 3120 kJ of energy

0.4082 mol C2H6 releases 3120 kJ

So energy released = 636.7 kJ= 637 kJ

2

4082.0


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Practice Calculate the energy released when the following

quantities of hydrogen gas burn according to theequation:

2H2(g) + O2(g) 2H2O(g); H= 572 kJ mol1

a 1.00 mol


.

c 100 L at SLC


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Sample problem: What volume of hydrogen, measured at STP, must be burnt

according to the equation

2H2(g) + O2(g) 2H2O(g); H= 572 kJ mol1

in order to yield 100 kJ of heat energy?


se s o c ome ry o n e amoun o 2.

572 kJ is released by 2 mol

100 kJ is released by 2 mol

n(H2) = 0.3497 mol

Calculate the volume of H2 at STP, using n=

V(H2) = n Vm

= 0.3497 mol 22.4 L mol1

= 7.83 L

572

100

mV

V


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Butane is used as the fuel in pocket cigarette lighters. It is a liquidwhen stored under pressure in the lighter, but vaporises when thevalve is opened. Combustion of butane is represented by theequation:

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l); H= 5772 kJ mol1

a How much energy is evolved when 10.0 g of butane burnscompletely?

b How much ener is evolved when 0.100 L of butane


measured at SLC, burns completely?

c Calculate the volume of butane, measured at 15C and 108kPa, that must be burnt to yield 1.00 kJ of energy. (Hint: PV = nRT)

Ans: 497 kJ, 11.8 kJ, 7.68 mL


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Answer



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Measuring enthalpy change- Calorimetry

The value of H can be determined

experimentally by measuring the heat flowaccompanying a reaction at SLC.


The measurement of heat flow iscalorimetry; a device used to measure

heat flow is a calorimeter.


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Coffee-cup calorimeter For solution of soluble salts.

The reactants and products of the rxn are thesystem.

The water in which they dissolve as well asthe calorimeter are part of the surroundings.

We assume that calorimeter perfectly


solution. The heat produced by the reaction is entirely

absorbed by the solution; it does not escapethe calorimeter.

Heat lost by rxn = Heat gained by the soln

opposite occurs for endothermic rxn

We also assume that the calorimeter itselfdoesnt absorb heat ~~ low thermal conductivity


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Coffee-cup calorimeter The heat energy gained or

lost by the soln. (water) canbe readily calculated from:

Mass of the water


Heat capacity of the solution(cH2O = 4.18 J g

-1 C-1)

Temperature change

Q = m x c x T

Joule (J) g J

g. C

Tf Ti

C


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Sample problem Suppose 200mL of boiling water is needed to make a

cup of tea. Calculate the energy required for a teh-tarikif the initial water temperature is 18.0C (density of wateris 1g/cm3 )


Ans:


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Sample problem 100mL of 1.0M HCl at 21.0C and 100mL of 1.0 M NaOH at

22.0C are mixed in a coffee cup calorimeter. The highesttemperature of the mixture is recorded to be 45.0C. Calculate theheat change (q) experienced during the reaction and the enthalpyof neutralization for the reaction. (cH2O = 4.18 Jg-1K-1)


Although the usual name for H is heat of reaction, there are

some reactions for which specific names have been given.

Heat of solution is the change in enthalpy when 1 mole of anysubstance dissolves in water.

Heat of neutralisation is the change in enthalpy when an acid reacts

with a base to form 1 mole of water exothermic. Heat of combustion is the change when 1 mole of substance burns

in air, and is always exothermic.


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Sample problem 50mL of 1.0M HCl at 21.0C and 50mL of 1.0 M NaOH at 22.0C

are mixed in a coffee cup calorimeter. The highest temperature ofthe mixture is recorded to be 28.0C. Calculate the heat change(q) experienced during the reaction and the enthalpy ofneutralization for the reaction in kJ/mol. (cH2O = 4.18 J g

-1 C-1)


Ans: -54


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Practice questionWhen 50.0mL of 0.100M AgNO3 and 50.0mL of 0.100M HCl are

mixed in a constant-pressure calorimeter, the temperature of the

mixture increses from 22.20C to 23.11C. Calculate the H for thisreaction in kJ/mol (cH2O = 4.18 J g

-1 C-1).


Ans: -68


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Homework Butane gas burns in oxygen producing carbon dioxide and steam

at 100C as shown in the following equation.

2C4H10 (g) + 13O2 (g)

8CO2 (g) + 10H2O (g)

H = -5728kJ/mol . When the reaction product is returned to SLC (25C) theH has a value of -6168 kJ/mol.

Explain the difference in the H values obtained.


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Thermochemistry Jan2010