ap ppt ch 6 ap only copy

Chapter 6

Thermochemistry

AP*

AP Learning Objectives

▪ LO 3.11 The student is able to interpret observations regarding macroscopic energy changes associated with a reaction of process to generate a relevant symbolic and/or graphical representation of the energy changes. (Sec 6.1-6.2)

▪ LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecules collisions. (Sec 6.1-6.2)

▪ LO 5.4 The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow. (Sec 6.1)

▪ LO 5.5 The student is able to use the conservation of energy to relate the magnitudes of the energy changes when two nonreacting substances are mixed or brought into contact with one another. (Sec 6.2)

AP Learning Objectives

▪ LO 5.6 The student is able to use calculations or estimations to relate energy changes associated with heating/cooling a substance to the heat capacity, relate energy changes associated with a phase transition to the enthalpy of fusion/vaporization, relate energy changes associated with a chemical reaction to the enthalpy of the reaction, and relate energy changes to PDV work. (Sec 6.2)

▪ LO 5.7 The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change in enthalpy of a chemical process (heating/cooling, phase transition, or chemical reaction) at constant pressure. (Sec 6.2)

▪ LO 5.8 The student is able to draw qualitative and quantitative connections between the reaction enthalpy and the energies involved in the breaking and formation of chemical bonds. (Sec 6.4)

Section 6.1 The Nature of Energy

AP Learning Objectives, Margin Notes and References

▪ Learning Objectives ▪ LO 3.11 The student is able to interpret observations regarding macroscopic energy changes associated

with a reaction of process to generate a relevant symbolic and/or graphical representation of the energy changes.

▪ LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecules collisions.

▪ LO 5.4 The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow.

▪ Additional AP References ▪ LO 5.3 (see Appendix 7.2, “Thermal Equilibrium, the Kinetic Molecular Theory, and the Process of Heat”)


Copyright © Cengage Learning. All rights reserved 5

▪ Capacity to do work or to produce heat. ▪ Law of conservation of energy – energy

can be converted from one form to another but can be neither created nor destroyed. ▪ The total energy content of the

universe is constant.

Energy

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▪ Potential energy – energy due to position or composition.

▪ Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.

Energy

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▪ In the initial position, ball A has a higher potential energy than ball B.

Initial Position



▪ After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.

Final Position



▪ Heat involves the transfer of energy between two objects due to a temperature difference.

▪ Work – force acting over a distance. ▪ Energy is a state function; work and heat are

not ▪ State Function – property that does not

depend in any way on the system’s past or future (only depends on present state).

Energy



▪ System – part of the universe on which we wish to focus attention.

▪ Surroundings – include everything else in the universe.

Chemical Energy



▪ Endothermic Reaction: ▪ Heat flow is into a system. ▪ Absorb energy from the surroundings.

▪ Exothermic Reaction: ▪ Energy flows out of the system.

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

Chemical Energy

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Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. !

a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it. c) Water boils in a kettle being heated on

a stove. d) Water vapor condenses on a cold pipe. e) Ice cream melts.

!Exo !Endo Endo !Exo Endo

CONCEPT CHECK!


Thermodynamics

▪ The study of energy and its interconversions is called thermodynamics.

▪ Law of conservation of energy is often called the first law of thermodynamics.


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• Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.

▪ To change the internal energy of a system: ΔE = q + w q represents heat w represents work

Internal Energy

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▪ Thermodynamic quantities consist of two parts: ▪ Number gives the magnitude of the

change. ▪ Sign indicates the direction of the flow.

Internal Energy

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Internal Energy

▪ Sign reflects the system’s point of view. ▪ Endothermic Process: ▪ q is positive

▪ Exothermic Process: ▪ q is negative


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▪ Sign reflects the system’s point of view. ▪ System does work on surroundings: ▪ w is negative

▪ Surroundings do work on the system: ▪ w is positive

Internal Energy

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240 Chapter Six Thermochemistry

In this text the same conventions also apply to the flow of work. If the system doeswork on the surroundings (energy flows out of the system), w is negative. If the surround-ings do work on the system (energy flows into the system), w is positive. We define workfrom the system’s point of view to be consistent for all thermodynamic quantities. Thatis, in this convention the signs of both q and w reflect what happens to the system; thuswe use .

In this text we always take the system’s point of view. This convention is not fol-lowed in every area of science. For example, engineers are in the business of designingmachines to do work, that is, to make the system (the machine) transfer energy to its sur-roundings through work. Consequently, engineers define work from the surroundings’point of view. In their convention, work that flows out of the system is treated as positivebecause the energy of the surroundings has increased. The first law of thermodynamics isthen written !E " q # w$, where w$ signifies work from the surroundings’ point of view.

¢E " q % w

Surroundings

∆E < 0

Exothermic Endothermic

System

Surroundings

∆E > 0

System

Energy Energy

The convention in this text is to take thesystem’s point of view; q " #x denotesan exothermic process, and q " %xdenotes an endothermic one.

The joule (J) is the fundamental SI unitfor energy:

One kilojoule (kJ) " 103 J.

J "kg ! m2

s2

S E E E X E R C I S E S 6 . 2 7 T H R O U G H 6 . 3 0

A common type of work associated with chemical processes is work done by a gas(through expansion) or work done to a gas (through compression). For example, in anautomobile engine, the heat from the combustion of the gasoline expands the gases inthe cylinder to push back the piston, and this motion is then translated into the motionof the car.

Internal EnergyE X A M P L E 6 . 1

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

Solution

We use the equation

where q " %15.6 kJ, since the process is endothermic, and w " %1.4 kJ, since work isdone on the system. Thus

The system has gained 17.0 kJ of energy.

¢E " 15.6 kJ % 1.4 kJ " 17.0 kJ

¢E " q % w

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Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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6.1 The Nature of Energy 241

Suppose we have a gas confined to a cylindrical container with a movable piston asshown in Fig. 6.4, where F is the force acting on a piston of area A. Since pressure isdefined as force per unit area, the pressure of the gas is

Work is defined as force applied over a distance, so if the piston moves a distance!h, as shown in Fig. 6.4, then the work done is

Since P " F!A or F " P # A, then

Since the volume of a cylinder equals the area of the piston times the height of the cylinder(see Fig. 6.4), the change in volume !V resulting from the piston moving a distance !h is

Substituting !V " A # !h into the expression for work gives

This gives us the magnitude (size) of the work required to expand a gas !V against a pres-sure P.

What about the sign of the work? The gas (the system) is expanding, moving the pis-ton against the pressure. Thus the system is doing work on the surroundings, so from thesystem’s point of view the sign of the work should be negative.

For an expanding gas, !V is a positive quantity because the volume is increasing.Thus !V and w must have opposite signs, which leads to the equation

Note that for a gas expanding against an external pressure P, w is a negative quantity asrequired, since work flows out of the system. When a gas is compressed, !V is a nega-tive quantity (the volume decreases), which makes w a positive quantity (work flows intothe system).

w " $P¢V

Work " P # A # ¢h " P¢V

¢V " final volume $ initial volume " A # ¢h

Work " F # ¢h " P # A # ¢h

Work " force # distance " F # ¢h

P "FA

∆h

P = FA

Initialstate

P = FA

Finalstate

∆h

Area = A

∆V

(a) (b)

∆V

Figure 6.4(a) The piston, moving a distance !hagainst a pressure P, does work on thesurroundings. (b) Since the volume of acylinder is the area of the base times itsheight, the change in volume of the gasis given by !h # A " !V.

w and P!V have opposite signs becausewhen the gas expands (!V is positive),work flows into the surroundings (w isnegative).

For an ideal gas, work can occur onlywhen its volume changes. Thus, if a gasis heated at constant volume, thepressure increases but no work occurs.


PV WorkE X A M P L E 6 . 2

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a con-stant external pressure of 15 atm.

Solution

For a gas at constant pressure,

In this case P " 15 atm and !V " 64 $ 46 " 18 L. Hence

Note that since the gas expands, it does work on its surroundings.

Reality Check: Energy flows out of the gas, so w is a negative quantity.

w " $15 atm # 18 L " $270 L ! atm

w " $P¢V

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In dealing with “PV work,” keep in mind that the P in P!V always refers to the ex-ternal pressure—the pressure that causes a compression or that resists an expansion.


A propane burner is used to heat the airin a hot-air balloon.

Internal Energy, Heat, and WorkE X A M P L E 6 . 3

A balloon is being inflated to its full extent by heating the air inside it. In the final stagesof this process, the volume of the balloon changes from 4.00 " 106 L to 4.50 " 106 Lby the addition of 1.3 " 108 J of energy as heat. Assuming that the balloon expands againsta constant pressure of 1.0 atm, calculate !E for the process. (To convert between L ! atmand J, use 1 L ! atm # 101.3 J.)

Solution

Where are we going?! To calculate !E

What do we know?

✓

✓

✓

✓

✓

What do we need?✓

How do we get there?What is the work done on the gas?

What is !V?

What is the work?

The negative sign makes sense because the gas is expanding and doing work on the sur-roundings. To calculate !E, we must sum q and w. However, since q is given in units ofJ and w is given in units of L ! atm, we must change the work to units of joules:

Then

!

Reality Check: Since more energy is added through heating than the gas expends do-ing work, there is a net increase in the internal energy of the gas in the balloon. Hence!E is positive.

¢E # q $ w # 1$1.3 " 108 J2 $ 1%5.1 " 107 J2 # 8 " 107 J

w # %5.0 " 105 L ! atm "101.3 JL ! atm

# %5.1 " 107 J

w # %P¢V # %1.0 atm " 5.0 " 105 L # %5.0 " 105 L ! atm

¢V # V2 % V1 # 4.50 " 106 L % 4.00 " 106 L # 5.0 " 105 L

w # %P¢V

¢E # q $ w

V2 # 4.50 " 106 L

1 L ! atm # 101.3 J

P # 1.0 atm

q # $1.3 " 108 J

V1 # 4.00 " 106 L

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delta E= W+QQ= GIVENW= - P*DELTAV

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Section 6.2 Enthalpy and Calorimetry


▪ Learning Objectives ▪ LO 3.11 The student is able to interpret observations regarding macroscopic energy changes associated

with a reaction of process to generate a relevant symbolic and/or graphical representation of the energy changes.

▪ LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecules collisions.

▪ identification of the systems, the type (heat versus work), or the direction of energy flow. ▪ LO 5.5 The student is able to use the conservation of energy to relate the magnitudes of the energy

changes when two nonreacting substances are mixed or brought into contact with one another. ▪ LO 5.6 The student is able to use calculations or estimations to relate energy changes associated with

heating/cooling a substance to the heat capacity, relate energy changes associated with a phase transition to the enthalpy of fusion/vaporization, relate energy changes associated with a chemical reaction to the enthalpy of the reaction, and relate energy changes to PDV work.

▪ LO 5.7 The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change in enthalpy of a chemical process (heating/cooling, phase transition, or chemical reaction) at constant pressure.

▪ Additional AP References ▪ LO 5.3 (see Appendix 7.2, “Thermal Equilibrium, the Kinetic Molecular Theory, and the Process of Heat”) ▪ LO 5.7 (see APEC #12, “Analysis by Calorimetry”)


Change in Enthalpy

▪ State function ▪ ΔH = q at constant pressure

▪ ΔH = Hproducts – Hreactants



Calorimetry

▪ Science of measuring heat ▪ Specific heat capacity: ▪ The energy required to raise the temperature of

one gram of a substance by one degree Celsius. ▪ Molar heat capacity: ▪ The energy required to raise the temperature of

one mole of substance by one degree Celsius.


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Calorimetry

▪ If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.

▪ An endothermic reaction cools the solution.



A Coffee–Cup Calorimeter Made of Two Styrofoam Cups



Calorimetry

▪ Energy released (heat) = s × m × ΔT ! s = specific heat capacity (J/°C·g) m = mass of solution (g) ΔT = change in temperature (°C)


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6.2 Enthalpy and Calorimetry 243

6.2 ! Enthalpy and Calorimetry

EnthalpySo far we have discussed the internal energy of a system. A less familiar property of asystem is its enthalpy H, which is defined as

where E is the internal energy of the system, P is the pressure of the system, and V is thevolume of the system.

Since internal energy, pressure, and volume are all state functions, enthalpy is also astate function. But what exactly is enthalpy? To help answer this question, consider aprocess carried out at constant pressure and where the only work allowed is pressure–volume work (w ! "P#V ). Under these conditions, the expression

becomes

or

where qP is the heat at constant pressure.We will now relate qP to a change in enthalpy. The definition of enthalpy is

H ! E $ PV. Therefore, we can say

or

Since P is constant, the change in PV is due only to a change in volume. Thus

and

This expression is identical to the one we obtained for qP:

Thus, for a process carried out at constant pressure and where the only work allowed isthat from a volume change, we have

At constant pressure (where only PV work is allowed), the change in enthalpy #H of thesystem is equal to the energy flow as heat. This means that for a reaction studied at con-stant pressure, the flow of heat is a measure of the change in enthalpy for the system. Forthis reason, the terms heat of reaction and change in enthalpy are used interchangeablyfor reactions studied at constant pressure.

For a chemical reaction, the enthalpy change is given by the equation

In a case in which the products of a reaction have a greater enthalpy than the reac-tants, #H will be positive. Thus heat will be absorbed by the system, and the reaction is

¢H ! Hproducts " Hreactants

¢H ! qP

qP ! ¢E $ P¢V

¢H ! ¢E $ P¢V

¢ 1PV2 ! P¢V

¢H ! ¢E $ ¢ 1PV2Change in H ! 1change in E2 $ 1change in PV2

qP ! ¢E $ P¢V

¢E ! qP " P¢V

¢E ! qP $ w

H ! E $ PV

Enthalpy is a state function. A change inenthalpy does not depend on the pathwaybetween two states.

Recall from the previous section that wand P#V have opposite signs:

w ! "P¢V

#H ! q only at constant pressure.

The change in enthalpy of a system hasno easily interpreted meaning except atconstant pressure, where #H ! heat.

At constant pressure, exothermic means#H is negative; endothermic means #His positive.

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6.2 Enthalpy and Calorimetry 243

6.2 ! Enthalpy and Calorimetry

EnthalpySo far we have discussed the internal energy of a system. A less familiar property of asystem is its enthalpy H, which is defined as

where E is the internal energy of the system, P is the pressure of the system, and V is thevolume of the system.

Since internal energy, pressure, and volume are all state functions, enthalpy is also astate function. But what exactly is enthalpy? To help answer this question, consider aprocess carried out at constant pressure and where the only work allowed is pressure–volume work (w ! "P#V ). Under these conditions, the expression

becomes

or

where qP is the heat at constant pressure.We will now relate qP to a change in enthalpy. The definition of enthalpy is

H ! E $ PV. Therefore, we can say

or

Since P is constant, the change in PV is due only to a change in volume. Thus

and

This expression is identical to the one we obtained for qP:

Thus, for a process carried out at constant pressure and where the only work allowed isthat from a volume change, we have

At constant pressure (where only PV work is allowed), the change in enthalpy #H of thesystem is equal to the energy flow as heat. This means that for a reaction studied at con-stant pressure, the flow of heat is a measure of the change in enthalpy for the system. Forthis reason, the terms heat of reaction and change in enthalpy are used interchangeablyfor reactions studied at constant pressure.

For a chemical reaction, the enthalpy change is given by the equation

In a case in which the products of a reaction have a greater enthalpy than the reac-tants, #H will be positive. Thus heat will be absorbed by the system, and the reaction is

¢H ! Hproducts " Hreactants

¢H ! qP

qP ! ¢E $ P¢V

¢H ! ¢E $ P¢V

¢ 1PV2 ! P¢V

¢H ! ¢E $ ¢ 1PV2Change in H ! 1change in E2 $ 1change in PV2

qP ! ¢E $ P¢V

¢E ! qP " P¢V

¢E ! qP $ w

H ! E $ PV

Enthalpy is a state function. A change inenthalpy does not depend on the pathwaybetween two states.

Recall from the previous section that wand P#V have opposite signs:

w ! "P¢V

#H ! q only at constant pressure.

The change in enthalpy of a system hasno easily interpreted meaning except atconstant pressure, where #H ! heat.

At constant pressure, exothermic means#H is negative; endothermic means #His positive.

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endothermic. On the other hand, if the enthalpy of the products is less than that of thereactants, !H will be negative. In this case the overall decrease in enthalpy is achievedby the generation of heat, and the reaction is exothermic.


CalorimetryThe device used experimentally to determine the heat associated with a chemical reactionis called a calorimeter. Calorimetry, the science of measuring heat, is based on observ-ing the temperature change when a body absorbs or discharges energy as heat. Substancesrespond differently to being heated. One substance might require a great deal of heat en-ergy to raise its temperature by one degree, whereas another will exhibit the same tem-perature change after absorbing relatively little heat. The heat capacity C of a substance,which is a measure of this property, is defined as

When an element or a compound is heated, the energy required will depend on theamount of the substance present (for example, it takes twice as much energy to raise the

C "heat absorbed

increase in temperature

EnthalpyE X A M P L E 6 . 4

When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is re-leased as heat. Calculate !H for a process in which a 5.8-g sample of methane is burnedat constant pressure.

Solution

Where are we going?! To calculate !H

What do we know?

✓

✓

✓

How do we get there?What are the moles of CH4 burned?

What is !H?

Thus, when a 5.8-g sample of CH4 is burned at constant pressure,

!

Reality Check: In this case, a 5.8-g sample of CH4 is burned. Since this amount issmaller than 1 mol, less than 890 kJ will be released as heat.

¢H " heat flow " #320 kJ

¢H " 0.36 mol CH4 $#890 kJmol CH4

" #320 kJ

5.8 g CH4 $1 mol CH4

16.04 g CH4" 0.36 mol CH4

Molar mass CH4 " 16.04 g

Mass " 5.8 g CH4

qP " ¢H " #890 kJ/mol CH4

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endothermic. On the other hand, if the enthalpy of the products is less than that of thereactants, !H will be negative. In this case the overall decrease in enthalpy is achievedby the generation of heat, and the reaction is exothermic.


CalorimetryThe device used experimentally to determine the heat associated with a chemical reactionis called a calorimeter. Calorimetry, the science of measuring heat, is based on observ-ing the temperature change when a body absorbs or discharges energy as heat. Substancesrespond differently to being heated. One substance might require a great deal of heat en-ergy to raise its temperature by one degree, whereas another will exhibit the same tem-perature change after absorbing relatively little heat. The heat capacity C of a substance,which is a measure of this property, is defined as

When an element or a compound is heated, the energy required will depend on theamount of the substance present (for example, it takes twice as much energy to raise the

C "heat absorbed

increase in temperature

EnthalpyE X A M P L E 6 . 4

When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is re-leased as heat. Calculate !H for a process in which a 5.8-g sample of methane is burnedat constant pressure.

Solution

Where are we going?! To calculate !H

What do we know?

✓

✓

✓

How do we get there?What are the moles of CH4 burned?

What is !H?

Thus, when a 5.8-g sample of CH4 is burned at constant pressure,

!

Reality Check: In this case, a 5.8-g sample of CH4 is burned. Since this amount issmaller than 1 mol, less than 890 kJ will be released as heat.

¢H " heat flow " #320 kJ

¢H " 0.36 mol CH4 $#890 kJmol CH4

" #320 kJ

5.8 g CH4 $1 mol CH4

16.04 g CH4" 0.36 mol CH4

Molar mass CH4 " 16.04 g

Mass " 5.8 g CH4

qP " ¢H " #890 kJ/mol CH4

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Section 6.4 Standard Enthalpies of Formation


▪ Learning Objectives ▪ LO 5.8 The student is able to draw qualitative and quantitative connections between the reaction

enthalpy and the energies involved in the breaking and formation of chemical bonds.


Standard Enthalpy of Formation (ΔHf°)

▪ Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.



Conventional Definitions of Standard States

▪ For a Compound ▪ For a gas, pressure is exactly 1 atm. ▪ For a solution, concentration is exactly 1 M. ▪ Pure substance (liquid or solid)

▪ For an Element

▪ The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.



A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ



Problem-Solving Strategy: Enthalpy Calculations

1. When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes.

2. When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer.



Problem-Solving Strategy: Enthalpy Calculations

3. The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:

!! ΔH°rxn = ΣnpΔHf°(products) - ΣnrΔHf°(reactants)

!4. Elements in their standard states are not included

in the ΔHreaction calculations because ΔHf° for an element in its standard state is zero.



Calculate ΔH° for the following reaction:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

!Given the following information: ΔHf° (kJ/mol)

Na(s) 0 H2O(l) –286

NaOH(aq) –470 H2(g) 0

! ! ! ! ΔH° = –368 kJCopyright © Cengage Learning. All rights reserved 42

EXERCISE!

ap ppt ch 6 ap only copy

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