Chapter 6 Energy and Chemical Reactions

Energy and Energy Changes • Capacity to do work – chemical, mechanical, thermal, electrical, radiant, sound, nuclear • Energy may affect matter – e.g. raise its temperature, eventually causing a state change – All physical changes and chemical changes involve energy changes

Heat Transfer and The Measurement of Heat • SI unit J (Joule) • calorie 1 calorie = 4.184 J

• English unit = BTU • Specific Heat amount of heat required to raise the T of 1g of a substance by 1oC SH = units = J/goC

1

Example - Converting Calories to Joules Convert 60.1 cal to joules

1 cal = 4.184 joules 4.184 J 60.1cal × = 251J 1 cal

Energy and the Temperature of Matter

• The amount the temperature of an object increases depends on the amount of heat added (q).

– If you double the added heat energy the temperature will increase twice as much.

• The amount the temperature of an object increases depends on its mass – If you double the mass it will take twice as much heat energy to raise the temperature the same amount.

Heat Transfer and the Measurement of Heat • Heat capacity amount of heat required to raise the T of a substance by 1oC • HC = J / oC

2

Heat Transfer and the Measurement of Heat • Heat transfer equation necessary to calculate amounts of heat amount of heat = amount of substance x specific heat x ∆T

q = m × SH × ∆T •Example – Calculate the amt. of heat to raise T of 200 g of water from 10.0 oC to 55.0 oC

Heat Transfer and the Measurement of Heat • Heat transfer equation necessary to calculate amounts of heat amount of heat = amount substance x specific heat x ∆T

q = m × SH × ∆T ? J = 200g H2O×

4.184J × (55.0o C −10.0o C) 1 g H2O

= 3.76×104 J or 37.6 kJ

Specific Heat Capacity • Specific Heat (s) is the amount of energy required to raise the temperature of one gram of a substance by one Celsius degree By definition , the specific heat of water is 4.184

J g °C

Amount of Heat = Specific Heat x Mass x Temperature Change Q = SH x m x ∆T

3

Example – Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C Specific Heat of Water = 4.184

JJ gg -°°CC

Mass = 7.40 g Temperature Change = 46.0°C – 29.0°C = 17.0°C

Q = SH x m x ∆T J Heat = 4.184 × 7.40g × 17.0°C = 526 J g °C

Example – A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold? Q = SH

× m × ∆T Q SH = m × ∆T ∆ T = 41 ° C - 23 ° C = 18 ° C 5.8 J J s = = 0.20 1.6 g x 18 ° C g °C

Specific heat of gold is 0.13 g J°C Therefore the metal cannot be pure gold.

The First Law of Thermodynamics •

Thermodynamics is the study of the changes in energy and transfers of energy that accompany chemical and physical processes. 1. Will two (or more) substances react when they are mixed under specified conditions? 2. If they do react, what energy changes and transfers are associated with their reaction? 3. If a reaction occurs, to what extent does it occur?

4

The First Law of Thermodynamics • Exothermic reactions release energy in the form of heat. • For example, the combustion of propane is exothermic. C 3H 8 (g) + 5 O 2(g) → 3 CO 2(g) + 4H 2 O ( l ) + 2.22 ×103 kJ •

The combustion of n-butane is also exothermic. 2 C 4 H10(g) + 13 O 2(g) → 8 CO 2(g) + 10 H 2 O ( l ) + 5.78 × 103 kJ

The First Law of Thermodynamics • Exothermic reactions generate specific amounts of heat. • This is because the potential energies of the products are lower than the potential energies of the reactants.

The First Law of Thermodynamics •

There are two basic ideas of importance for thermodynamic systems. 1. Chemical systems tend toward a state of minimum potential energy.  Some examples of this include:  H2O flows downhill.  Objects fall when dropped.

 The energy change for these two examples is:  Epotential = mgh  ∆Epotential = mg(∆h)

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The First Law of Thermodynamics 2. Chemical systems tend toward a state of maximum disorder. • Common examples of this are: – A mirror shatters when dropped and does not reform. – It is easy to scramble an egg and difficult to unscramble it. – Food dye when dropped into water disperses.

The First Law of Thermodynamics • This law can be stated as, “The combined amount of energy in the universe is constant.” • The first law is also known as the Law of Conservation of Energy. – Energy is neither created nor destroyed in chemical reactions and physical changes.

Some Thermodynamic Terms • The substances involved in the chemical and physical changes under investigation are called the system. – In chemistry lab, the system is the chemicals inside the beaker.

• The environment around the system is called the surroundings. – The surroundings are outside the beaker.

• The system plus the surroundings is called the universe.

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Some Thermodynamic Terms • The set of conditions that specify all of the properties of the system is called the thermodynamic state of a system. • For example the thermodynamic state could include: – – – –

The number of moles and identity of each substance. The physical states of each substance. The temperature of the system. The pressure of the system.

Some Thermodynamic Terms • The properties of a system that depend only on the state of the system are called state functions. – State functions are always written using capital letters.

• The value of a state function is independent of pathway. • An analog to a state function is the energy required to climb a mountain taking two different paths. – – – – –

E1 = energy at the bottom of the mountain E1 = mgh1 E2 = energy at the top of the mountain E2 = mgh2 ∆E = E2-E1 = mgh2 – mgh1 = mg(∆h)

Some Thermodynamic Terms • Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be! • Some examples of state functions are: – T, P, V, ∆E, ∆H, and S •

Examples of non-state functions are: – n, q, w

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Some Thermodynamic Terms •

In thermodynamics we are often interested in changes in functions. – We will define the change of any function X as: – ∆X = Xfinal – Xinitial

• •

If X increases ∆X > 0 If X decreases ∆X < 0.

Enthalpy Change • Chemistry is commonly done in open beakers on a desk top at atmospheric pressure. – Because atmospheric pressure only changes by small increments, this is almost at constant pressure.

• The enthalpy change, ∆H, is the change in heat content at constant pressure. – ∆H = qp

Enthalpy Change • ∆Hrxn is the heat of reaction. – This quantity will tell us if the reaction produces or consumes heat. – If ∆Hrxn < 0 the reaction is exothermic. – If ∆Hrxn > 0 the reaction is endothermic.

• ∆Hrxn = Hproducts - Hreactants – ∆Hrxn = Hsubstances produced - Hsubstances consumed – Notice that this is ∆Hrxn = Hfinal – Hinitial

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Calorimetry • A coffee-cup calorimeter is used to measure the amount of heat produced (or absorbed) in a reaction at constant P – This is one method to measure qP for reactions in solution.

Calorimetry • If an exothermic reaction is performed in a calorimeter, the heat evolved by the reaction is determined from the temperature rise of the solution. – This requires a two part calculation.  Amount of heat   Amount of heat   Amount of heat    =   +    released by reaction   absorbed by calorimete r   absorbed by solution 

• Amount of heat gained by calorimeter is called the heat capacity of the calorimeter or calorimeter constant. – The value of the calorimeter constant is determined by adding a specific amount of heat to calorimeter and measuring the temperature rise.

Calorimetry • Example 15-1: When 3.425 kJ of heat is added to a calorimeter containing 50.00 g of water the temperature rises from 24.00oC to 36.54oC. Calculate the heat capacity of the calorimeter in J/oC. The specific heat of water is 4.184 J/g oC. • This is a four part calculation.

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Calorimetry 1. Find the temperature change.

∆T = (36.54 - 24.00) 0 C = 12.54 0 C 2. Find the heat absorbed by the water in going from 24.000C to 36.540C.

q P = mC∆T

(

= (50.00 g) 4184 .

J

. C) )(1254 0

g0 C

. J ≈ 2623 J = 262337

Calorimetry 3. Find the heat absorbed by the calorimeter.  Take the total amount of heat added to calorimeter and subtract the heat absorbed by the water.

3.425 kJ = 3425 J

(3425 J

- 2623 J ) = 802 J

4. Find the heat capacity of the calorimeter.  (heat absorbed by the calorimeter)/(temperature change)

802 J = 63.955 12.54 0 C

J

0

C

≈ 64.00

J

0

C

Calorimetry • Example 15-2: A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)

When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC.

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Calorimetry The heat capacity of the calorimeter has previously been determined to be 27.8 J/0C. Assume that the specific heat of the mixture is the same as that of water, 4.18 J/g0C and that the density of the mixture is 1.02 g/mL.

Calorimetry • This is a three part calculation. 1.Calculate the amount of heat given off in the reaction.

temperature change ∆T = (25.947 - 23.000) 0 C = 2.9470 C heat absorbed by calorimeter

(

)

. q = 2.9470 C ( 278

J0

) = 819. J

C

Calorimetry temperature change ∆T = (25.947 - 23.000)0 C = 2.9470 C heat absorbed by calorimeter

(

)

q = 2.9470C (278 .

J0

) = 819. J

C

mass of solution in calorimeter 102 . g = 510 . g (25.00 mL + 25.00 mL) mL

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Calorimetry heat absorbed by solution q = mC∆T

(

q = (51.0 g ) 418 .

J

)(2.947 C) = 628 J 0

g0C

total amount of heat produced by reaction q = 81.9 J + 628 J = 709.9 J

Calorimetry 2. Determine ∆H for the reaction under the conditions of the experiment. •

We must determine the number of moles of reactants consumed which requires a limiting reactant calculation.

CH 3COOH ( aq ) + NaOH ( aq ) → NaCH 3COO ( aq ) + H 2 O ( l )

(25.00 mL NaOH )

0.500 mmol NaOH  ×  1 mL NaOH

 1 mmol NaCH 3COO  = 12 .5 mmol NaCH COO 3  1 mmol NaOH 

Calorimetry

C H 3C O O H

(a q )

+ N aO H

(a q )

→ N a C H 3C O O

(a q )

+ H 2O

(l )

0 .5 0 0 m m o l N a O H   × ( 2 5 . 0 0 m L N a O H ) 

1 m L N aO H  1 m m o l N a C H 3 C O O  = 1 2 .5 m m o l N a C H C O O   3   1 m m ol N aO H

(2 5 . 0 0

 0 .6 0 0 m m o l C H 3 C O O H  m L C H 3 C O O H )  ×  1 m L C H 3C O O H 

 1 m m o l N a C H 3C O O    = 1 5 .0 m m o l N a C H 3 C O O  1 m m o l C H 3C O O H 

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Calorimetry 3. Finally, calculate the ∆Hrxn based on the limiting reactant calculation.

∆H rxn

12.5 mmol = 0.0125 mol 709.9 J = = 56792 J / mol ≈ 56.8 kJ / mol 0.0125 mol

Thermochemical Equations • Thermochemical equations are a balanced chemical reaction plus the ∆H value for the reaction.

C5 H12(l ) + 8 O 2(g) → 5 CO 2(g) + 6 H 2 O ( l ) ∆H orxn = - 3523 kJ • ∆H < 0 designates an exothermic reaction. • ∆H > 0 designates an endothermic reaction

Standard States and Standard Enthalpy Changes • Thermochemical standard state conditions – The thermochemical standard T = 298.15 K. – The thermochemical standard P = 1.0000 atm. • Be careful not to confuse these values with STP.

• Thermochemical standard states of matter – For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. – For gases the standard state is the gas at 1.00 atm of pressure. • For gaseous mixtures the partial pressure must be 1.00 atm.

– For aqueous solutions the standard state is 1.00 M concentration.

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Standard Molar Enthalpies of Formation, ∆Hfo

• The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements. – The symbol for standard molar enthalpy of formation is ∆ H f o.

• The standard molar enthalpy of formation for MgCl2 is:

Mg (s ) + Cl 2( g ) → MgCl 2(s ) + 6418 . kJ ∆H of MgCl 2 ( s ) = −6418 . kJ / mol

Standard Molar Enthalpies of Formation, ∆Hfo • Standard molar enthalpies of formation have been determined for many substances and are tabulated in Appendix L • Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. • The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for which ∆Horxn = -1281 kJ. P in standard state is P4 Phosphoric acid in standard state is H3PO4(s)

Standard Molar Enthalpies of Formation, ∆Hfo 3

2

H2( g) + 2 O2( g) + 14 P4(s) → H3PO4(s) + 1281 kJ ∆Hof H3PO4( s ) = −1281 kJ / mol

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Standard Molar Enthalpies of Formation, ∆Hfo • Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere. H

2 (g )

+

F2 (g )

→

s td . s ta te s td . s ta te f o r th is rx n . ∆ H

o 298

= 2 ∆H

f r o m A p p e n d ix K , ∆ H ∆H

o 298

=

2 H F( g ) s td . s ta te

o f

o f

= −271 kJ / m ol

( 2 m o l )( − 2 7 1 k J / m o l ) = − 5 4 2 k J

Standard Molar Enthalpies of Formation, ∆Hfo • Example 15-6: Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.

Standard Molar Enthalpies of Formation, ∆Hfo 4 Al ( s) + 3 O2( g) → 2 Al 2 O3( s) from Appendix K, ∆H of Al2O3 = −1676 kJ / mol 1 mol Al 2 mol Al 2 O3 × × 27.0 g Al 4 mol Al −1676 kJ = −466 kJ 1 mol Al 2 O3

? kJ = 15.0 g Al ×

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Hess’s Law • Hess’s Law of Heat Summation states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. – Hess’s Law is true because ∆H is a state function.

• If we know the following ∆Ho’s

[1] 4 FeO(s) + O 2(g) → 2 Fe 2O3(s) [2] 2 Fe(s) + O 2(g) → 2 FeO(g) [3] 4 Fe(s) + 3 O 2(g) → 2 Fe 2O3(s)

∆H o = −560 kJ ∆H o = −544 kJ ∆H o = −1648 kJ

Hess’s Law • •

For example, we can calculate the ∆Ho for reaction [1] by properly adding (or subtracting) the ∆Ho’s for reactions [2] and [3]. Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product. – Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product. • Each reaction can be doubled, tripled, or multiplied by half, etc. • The ∆Ho values are also doubled, tripled, etc. • If a reaction is reversed the sign of the ∆Ho is changed.

2 x [ − 2 ] 2 (2 F e O (s ) → 2 F e (s ) + O 2 ( g ) )

∆H 0 2( + 544 ) kJ

[ 3 ] 4 F e (s ) + 3 O 2 ( g ) → 2 F e 2 O 3 (s ) [1 ] 4 F e O (s ) + O 2 ( g ) → 2 F e 2 O 3

− 1648 kJ − 560 kJ

Hess’s Law • Example 15-7: Given the following equations and ∆Ho values o ∆H

( kJ )

[1] 2 N 2( g ) + O 2( g ) → 2 N 2 O ( g )

164.1

[2] N 2( g ) + O 2( g ) → 2 NO ( g )

180.5

[3] N 2( g ) + 2 O 2( g ) → 2 NO 2 ( g )

66.4

calculate ∆Ho for the reaction below. N 2O

(g )

+ NO

2 (g

)

→ 3 NO

(g )

∆H

o

= ?

You do it!

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Hess’s Law • Use a little algebra and Hess’s Law to get the appropriate ∆Ho values o (kJ) ∆H∆ ∆oH H(kJ) (kJ)

o

1 O N ×[[−− → −11 1]]] N N222OO O( g( g) )→ →NN N2(2g()g)++ +1212O 2( g2)( g) 2 O

112 × 2

×[ 2]

33 × 2 1 × 2

[ − 3]

( g)

33 2

N2( g) +

2( g)

33 22

2( g)

-82.05 -82.05 -82.05

O22((gg)) →33NO NO((gg)) 270.75 270.75

NO2( g) → 12 N2( g) + O2( g)

N2O( g) + NO2( g) → 3 NO( g)

-33.2 155.5

Hess’s Law • The + sign of the ∆Ho value tells us that the reaction is endothermic. • The reverse reaction is exothermic, i.e.,

3 NO ( g ) → N 2 O ( g ) + NO 2 ( g )

∆H o = - 155.5 kJ

Hess’s Law • Hess’s Law in a more useful form. – For any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants.

∆H 0rxn = ∑ n ∆H f0 products − ∑ n ∆H f0 reactants n

n

n = stoichiometric coefficients

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Hess’s Law

Hess’s Law • Calculate the ∆H o298 for the following reaction from data in Appendix L.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H 2O(l )

Hess’s Law • Calculate ∆Ho298 for the following reaction from data in Appendix L.

C3H8( g) + 5 O2( g) → 3 CO2( g) + 4 H2O( l )

[

][

∆Ho298 = 3∆Hof CO2( g ) + 4∆Hof H2O( l ) − ∆Hof C3H8( g ) + 5∆Hof O2 ( g)

]

= {[ 3(−3935 . ) + 4(−2858 . )] − [ (−1038 . ) + 5(0)]} kJ

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Hess’s Law • Calculate ∆Ho298 for the following reaction from data in Appendix L.

C3H8( g) + 5 O2( g) → 3 CO2( g) + 4 H2O( l )

[

][

∆Ho298 = 3∆Hof CO2( g) + 4∆Hof H2O( l ) − ∆Hof C3H8( g) + 5∆Hof O2 ( g)

]

= {[ 3(−3935 . ) + 4(−2858 . )] − [ (−1038 . ) + 5(0)]} kJ = -2211.9 kJ ∆Ho298 = − 22119 . kJ, and so the reaction is exothermic.

Hess’s Law • Application of Hess’s Law and more algebra allows us to calculate the ∆Hfo for a substance participating in a reaction for which we know ∆Hrxno , if we also know ∆Hfo for all other substances in the reaction. • Example 15-9: Given the following information, calculate ∆Hfo for H2S(g). 2 H2S(g) + 3 O2( g) → 2 SO2( g) + 2 H2O(l ) ∆Hof

?

0

o ∆H298 = -1124 kJ

- 296.8 - 285.8

(kJ / mol)

You do it!

Hess’s Law

[ {

][ [

∆H o298 = 2∆H of SO2 ( g ) + 2∆H of H 2O ( l ) − 2∆H of H2S( g ) + 3∆H of O 2( g )

]}

]

− 1124 kJ = [ 2( −296.8 ) + 2( −285.8] − 2∆H of H 2S( g ) + 3( 0) kJ now we solve for ∆H of H 2S( g ) 2∆H of H 2S( g ) = −412 . kJ ∆H of H 2S( g ) = −20.6 kJ

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Bond Energies • Bond energy is the amount of energy required to break the bond and separate the atoms in the gas phase. – To break a bond always requires an absorption of energy!

A - B ( g ) + bond energy → A ( g ) + B ( g ) H - Cl ( g ) + 432 kJ mol → H ( g ) + Cl ( g )

Bond Energies • Table of average bond energies • Molecule Bond Energy (kJ/mol) • F2 159 243 • Cl2 • Br2 192 • O2 (double bond) 498 946 • N2 (triple bond)

Bond Energies • Bond energies can be calculated from other ∆Ho298 values.

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Bond Energies • Bond energies can be calculated from other ∆Ho298 values • Example 15-9: Calculate the bond energy for hydrogen fluoride, HF. H - F( g ) + BE HF → H ( g ) + F( g ) atoms NOT ions or H - F( g ) → H ( g ) + F( g )

[

∆H o298 = BE HF

] [

∆H o298 = ∆H of H ( g ) + ∆H of F( g ) − ∆H of HF( g )

]

Bond Energies

H - F( g ) + BE HF → H ( g ) + F( g ) atoms NOT ions or H - F( g ) → H ( g ) + F( g )

[

∆ H o298 = BE HF

] [

∆ H o298 = ∆ H of H ( g ) + ∆ H of F( g ) − ∆ H of HF( g ) ∆ H o298

]

= [218 .0 kJ + 78.99 kJ ] − [ − 271 kJ ]

∆ H o298 = 568 .0 kJ

← BE for HF

Bond Energies • Example 15-10: Calculate the average N-H bond energy in ammonia, NH3. You do it!

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Bond Energies NH 3(g ) → N (g ) + 3 H (g )

[

∆H o298 = 3 BE N -H

][

∆H o298 = ∆H fo N ( g ) + 3 ∆H fo H ( g ) − ∆H fo NH 3( g )

]

∆H o298 = {[(472.7) + 3(218)] − [− 46.11]} kJ ∆H o298 = 1173 kJ average BE N-H =

1173 kJ = 391 kJ mol N-H bonds 3

Changes in Internal Energy, ∆E • The internal energy, E, is all of the energy contained within a substance. – This function includes all forms of energy such as kinetic, potential, gravitational, electromagnetic, etc. – The First Law of Thermodynamics states that the change in internal energy, ∆E, is determined by the heat flow, q, and the work, w.

Changes in Internal Energy, ∆E ∆E = E products − E reactants ∆E = q + w q > 0 if heat is absorbed by thesystem. q < 0 if heat is absorbed by thesystem. w > 0 if the surroundings do work on the system. w < 0 if the system does work on the surroundings.

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Changes in Internal Energy, ∆E • ∆E is negative when energy is released by a system undergoing a chemical or physical change. – Energy can be written as a product of the process.

C5H12( l) + 8 O2(g) → 5 CO2(g) + 6 H 2O( l) + 3.516× 103 kJ ∆E = - 3.516× 103 kJ

Changes in Internal Energy, ∆E  ∆E is positive when energy is absorbed by a system undergoing a chemical or physical change. – Energy can be written as a reactant of the process.

5 CO2(g) + 6 H2O(l) + 3.516× 103 kJ → C5H12( l) + 8 O2(g) ∆E = + 3.516× 103 kJ

Changes in Internal Energy, ∆E •

Example 15-12: If 1200 joules of heat are added to a system in energy state E1, and the system does 800 joules of work on the surroundings, what is the : 1. energy change for the system, ∆Esys?

∆E = E 2 - E1 = q + w

∆ E = 1200 J + (-800 J) ∆ E = 400 J ∆ E sys = + 400 J

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Changes in Internal Energy, ∆E 2. energy change of the surroundings, ∆Esurr?

∆E surr = −400 J

Changes in Internal Energy, ∆E 3. energy of the system in the new state, E 2?

∆Esys = E2 - E1

E2 = E1 + ∆Esys = E1 + 400 J

Changes in Internal Energy, ∆E • In most chemical and physical changes, the only kind of work is pressure-volume work. • Pressure is force per unit area.

P =

force F = 2 area d

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Changes in Internal Energy, ∆E • Volume is distance cubed.

V = d3 • P∆V is a work term, i.e., the same units are used for energy and work.

( )

F P∆V =  2  d3 = F × d ← which is work d 

Relationship of ∆H and ∆E • The total amount of heat energy that a system can provide to its surroundings at constant temperature and pressure is given by ∆H= ∆E + P ∆V – which is the relationship between ∆H and ∆E.

• ∆H = change in enthalpy of system • ∆E = change in internal energy of system • P∆V = work done by system

Relationship of ∆H and ∆E • At the start of Chapter 6 we defined ∆H = qP. • Here we define ∆H = ∆E + P∆V. – Are these two definitions compatible?

• Remember ∆E = q + w. • We have also defined w = -P∆V . – Thus ∆E = q + w = q -P∆V

• Consequently, ∆H = q- P∆V + P∆V = q – At constant pressure ∆H = qP.

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Relationship of ∆H and ∆E • For reactions in which the volume change is very small or equal to zero.

For small volume changes, ∆V ≈ 0 and P∆V ≈ 0. Since ∆H = ∆E + P∆V then ∆H ≈ ∆E. For no volume change, ∆H = ∆E.

Relationship of ∆H and ∆E • Change in enthalpy, ∆H, or heat of reaction is amount of heat absorbed or released when a reaction occurs at constant pressure. • The change in energy, ∆E, is the amount of heat absorbed or released when a reaction occurs at constant volume. • How much do the ∆H and ∆E for a reaction differ? – The difference depends on the amount of work performed by the system or the surroundings.

Relationship of ∆H and ∆E  ∆Ho = -3523 kJ/mol for the combustion of n-pentane, n-C5H12. Combustion of one mol of n-pentane at constant pressure releases 3523 kJ of heat. What are the values of the work term and ∆E for this reaction?

C5 H12( l ) + 8 O 2(g) → 5 CO 2(g) + 6 H 2 O ( l )

26

Relationship of ∆H and ∆E • Determine the work done by this reaction. You do it! C5 H12( l ) + 8 O 2(g) → 5 CO 2(g) + 6 H 2O ( l ) 1442443 14442444 3 8 mol gas

5 mol gas

Since ∆H o = - 3523 kJ/mol, we know that T = 298 K. w = - P∆V = - (∆n gas )RT ∆n gas = 5 − 8 mol = - 3 mol w = -(-3 mol)(8.314 J mol K )( 298 K) = 7433 J = 7.433 kJ

Relationship of ∆H and ∆E • Now calculate the ∆E for this reaction from the values of ∆H and w that we have determined.

∆H = ∆E + P∆V ∴ ∆E = ∆H - P∆V since w = - P∆V = 7.433 kJ then P∆V = - 7.433 kJ ∆E = - 3523 kJ - (-7.433 kJ) = -3516 kJ

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