Enthalpy and thermochemistry.pdf

8/20/2019 Enthalpy and thermochemistry.pdf

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Copyright©2000 by Houghton

Mifflin Company. All rights reserved.

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Ch 9 Energy, Enthalpy, andThermochemistry


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9.1 The nature of energy

Energy : the capacity to do work or to produce heat.

Law of conservation of energy : the energy can be

converted from one form to another but can be

neither created nor destroyed. That is, the energy

of the universe is constant.


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Energy can be classified as either potential energy

or kinetic energy.

Potential energy is due to position or composition.

I.e., water behind a dam, attractive/repulsive forces…

Kinetic energy of an object is due to the motion &depends on the mass of the object (m) & its velocity

(ν) : KE = 1/2m 2 .


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Figure 9.1: Initial positions ball A has a

higher potential energy than ball B.


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From Figure 9.1 :

1. Frictional heating

2. Work (ball A has done a work on B) : defined as

a force acting over a distance. Work is required to

raise B from its original position to a higher one.Thus, there are two ways to transfer energy :

through work & through heat.


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Regardless of the condition of the hill’s surface,

the total energy transferred will be constant. The

amounts of heat & work will differ. Energy

change is independent of the pathway; however,work & heat are both dependent on the pathway.


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A state function (property) refers to a property of

the system that depends only on its present state.

A state function (property) is that a change in this

function (property) in going from one state to

another is independent of the particular pathway taken between the two states.

Energy is a state function, but work & heat are not

state functions.


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Chemical energy :

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

The universe can be divided into two parts :

the system (reactants/products of the reaction)

& surroundings (everything else).


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When a reaction results in the evolution of heat, it

is said to be exothermic.Reactions that absorb energy from the surroundings

are said to be endothermic.

The energy gained by the surroundings must be

equal to the energy lost by the system.


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In combustion of methane (CH4), the heat flow into

the surroundings results in a lowering of the

potential energy of the reaction system ( this always

holds true).

In any exothermic reaction, the potential energystored in the chemical bonds is being converted to

thermal energy (random kinetic energy) via heat.


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Figure 9.2: Combustion of methane releasesthe quantity of energy


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Figure 9.3: Energy diagram for the

reaction of nitrogen and oxygen


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The study of energy & its interconversion is

called thermodynamics. The law of

conservation of energy is often called the first

law of thermodynamics & is stated as :

the energy of the universe is constant.


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Internal energy (E) is defined as the sum of the

kinetic & potential energies of all the “particles” in

the system.

E = q + w; q : heat, w : work

The meaning of a number (magnitude of the change)& a sign (the direction of the flow & this reflects the

system’s point of view) should be remembered.


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Interconversion between system & surroundings for energy


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Figure 9.4: The piston moving a distance


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Expanded : w = -p V (0)

In dealing with “PV work” keep in mind that the

P in PΔV always refers to the external pressure -

the pressure that causes a compression or that resists an expansion.


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R = 0.08206 Latm/Kmol

= 8.3145 J/Kmol

< 0


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Copyright©2000 by HoughtonMifflin Company. All rights reserved.

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9.2 Enthalpy (焓)

H = E + PV

H : enthalpy, E : internal energy

P, V : properties of the systemH, E, P, & V are all state functions.


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What is exactly enthalpy (H) ?Consider a process carried out at constant pressure :


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Thus, for a process carried out at constant pressure,

H = qP.


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At constant pressure (where only PV work is allow-

ed) the change in enthalpy ( H) of the system is

equal to the energy flow as heat (qP) from the above.

H = Hproducts – Hreactants (for a chemical reaction

the enthalpy change)

H > 0 – endothermic; H < 0 – exothermic.


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9.3 Thermodynamics of ideal gases

Ideal gases : the high temp & low pressure for

real gases (PV = nRT).

(KE)avg = 3/2 RT (Ch. 5 for an ideal gas) represents

average, random, translational energy for 1 moleof gas at a given temp. Energy (“heat”) required

= 3/2 R T.


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The molar heat capacity of a substance is defined

as the energy required to raise the temp. of 1 mole

of that substance by 1 K. Thus, the molar heat

capacity of an ideal gas is 3/2 R.


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Heating an ideal gas at constant volume

CV = 3/2 R = “heat” required to change the tempof 1 mole of gas by 1 K at constant volume ( V=0,

no PV work & all the energy that flows into the gas

is used to increase the translational energies of the

gas molecules).


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Heating an ideal gas at constant pressure

In this case, its volume increases & PV work occurs.

Energy required = “heat”

= energy needed to change the translational energy+ energy needed to do the PV work


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The quantity of work done as the gas expanded by

ΔV is PΔV.

P V = nR T = R T

Thus for a 1K change (ΔT=1K) the work is R, &

heat required to increase the temp of 1 mole ofgas by 1 K (constant P) = 3/2 R (CV) + R = 5/2 R

= CV + R = CP


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Heating a polyatomic gas : as a polyatomic gas is

heated, the gas molecules absorb energy to increase

their rotational & vibrational motions as well as to

move through space (translate) at higher speeds

(previous discussion : the temperature of a mono-atomic ideal gas is an index of the average random

translational energy of the gas).


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Indeed, when a gas is heated, the temperature only

increases to the extent that the translational energies

of the molecules increase. Any energy that isabsorbed to increase the vibrational & rotational

energies does not contribute directly to the

translational kinetic energy. Thus, its heat capacity is

larger than 3/2 R. But, finally the CP – CV = R is also

applicable in all cases, as expected for gases that closely obey the ideal gas law.


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Table 9.1: Molar Heat Capacities of VariousGases at 298 K

R=8.3145 J/KmolCO2 v.s. N2


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Heating a gas : energy and enthalpy

E = 3/2 RT (per mole) (the average translational

energy of an ideal gas - a monoatomic ideal gas)The energy (temp dependence) of an ideal gas :

E = 3/2 R

T (per mole) Note that this expression corresponds to :

E = Cv T (per mole) or E = nC

v T (n moles)


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The CV appears because when a gas is heated at

constant volume, all the input energy (heat) goes toward increasing E (no heat is needed to do

work).


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On the other hand, when a gas is heated at constant

pressure, the volume changes & work occurs :

“heat” required = qP= nCP T = n(CV + R) T

= nCV T + nR T

- E - P V = work required

(E for an ideal gas depends only on T )


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Consider the change in enthalpy when a gas is heated

H = E + PV

H = E + (PV) = E + (nRT)= E + nR T

For a sample of ideal gas containing n moles,substituting E = nCV T

H = nCV

T + nR T = n(CV

+ R) T

= nCP T (CP = CV + R)


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Thus, we can use H = nCP T to calculate the

change in enthalpy when n moles of an ideal gas

is heated, regardless of any conditions on pressureor volume.

E T (E = 3/2 RT ); H T (PV = nRT)Proportional constant Proportional constant

CV CP


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Table 9.2:Thermodynamic properties of an ideal gas


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Figure 9.5: Summary of the two pathways

discussed in example 9.2.


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(constant P)


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H = nCP T = qP

(constant V)


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(constant V)

E = qV = nCV T


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(constant V)


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(constant P)


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E1 = E2 & H1 = H2

Fi 9 6 M i d f h k f h ( )


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Figure 9.6: Magnitude of the work for pathway (a)

and (b) is shown by the colored areas : |w| = |PΔV|

E = q + w; q, w : related to the pathway

E is a state function.


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9.4 Calorimetry

Calorimeter : a device to determine the heat

associated with a chemical reaction experimentally

( temp change & energy absorbed ).

C (heat capacity) = heat absorbed/increase in temp


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Specific heat capacity (JK -1g-1 or J°C-1g-1) : the

energy required to raise the temp of 1 g of a

substance by 1 °C.

Molar heat capacity (JK -1mol-1 or J°C-1mol-1) :

the energy required to raise the temp of 1 mol ofa substance by 1 °C.

Figure 9 7: Coffee cup calorimeter made of two


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Figure 9.7: Coffee cup calorimeter made of two

Styrofoam cups


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For example, suppose we mix 50.0 mL of 1.0 M

HCl with 50.0 mL of 1.0 M NaOH at 25°C in a

calorimeter.H+(aq) + OH

-(aq) → H2O(l)

Energy released by reaction= energy absorbed by the solution

= specific heat capacity x mass of solution x ΔT,

where ΔT = 31.9°C → 25.0° = 6.9 °C


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Mass of solution = 100.0 mLx1.0 g/mL = 1.0 x 102 g

Energy released = (4.18J/°Cg)(1.0 x 102g)(6.9°C)

= 2.9 x 103 J

2.9 x 103 / 5.0 x 10-2 mol (mol of H+) = 5.8 x 104 J

ΔH = -58 KJ/mol


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M


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Copyright©2000 by HoughtonMifflin Company. All rights reserved. 58

Calculation of H & E for cases in which PV

work occurs

(1) ΔE = q (P) + W = ΔH + 0

(at constant pressure, where ΔV=0, W = 0 & thus

ΔE = ΔH = q P / solution reaction)

(2) 2SO2(g) + O2(g) → 2SO3(g) ( gas reaction)

ΔH = q P (at constant pressure)

W = -PΔV (ΔV


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Figure 9.8: Schematic to show the change

in volume for the reaction


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H = qP= -198 kJ


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Calorimetry experiments can also be performed

at constant volume

ΔE = q V + W (=0) = q V (constant volume)

To measure the energy of combustion of octane

(C8H18). A 0.5269 g of octane is placed in a bomb

calorimeter (heat capacity = 11.3 kJ/°C)


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Figure 9 9: Schematic of a bomb calorimeter


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Figure 9.9: Schematic of a bomb calorimeter


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9.5 Hess’s law

In going from a particular set of reactants to products, the change in enthalpy is the same

whether the reaction takes place in one step

or in a series of steps (Hess’s law).


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Figure 9.10: Principle of Hess's law


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Characteristics of enthalpy changes

Two characteristics of H for a reaction :

1. If a reaction is reversed, the sign of H is also

reversed.

2. H is directly proportional to the quantities(coefficients) of reactants & products in a reaction.


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(a) - (b) + 3(c) + 3(d)

過


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Explanations :


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Explanations :

1. Large heat capacity of feet because of water.

2. Although the surface of the coals has a very high

temp, the red-hot layer is very thin. Temp reflects

the intensity of the random kinetic energy in a

given sample of matter. The amount of energy available for heat flow, depends on the quantity

of matter at a given temp.

3. Vaporization of the moisture consumes some ofthe energy from the hot coals.


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9.6 Standard enthalpies of formationSummary : definitions of standard states

• For a gas the standard state is a pressure ofexactly 1 atm.

• For a solution the standard state is a conc of exactly 1 M (1 atm).


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• For a liquid & solid ( condensed state), the

standard state is the pure liquid or solid.

• For an element the standard state is the form

in which the element exists (is more stable)

under conditions of 1 atm & 25°C.


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The importance of tabulated ΔH°f values is thatenthalpies for many reactions can be calculated.

Fig. 9.11: Schematic diagram of the energy changes


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CH4(g) + 2O2(g) → CO2(g) + 2H2O(l )


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Summary : key concepts for doing enthalpy calculations

• When a reaction is reversed, the magnitude ofH

remains the same, but the sign changes.

• When the balanced eq for a reaction is multiplied

by an integer, the value of H must be multiplied by the same integer.


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• The change in enthalpy for a given reaction can

be calculated from the enthalpies of formation of

the reactants & products :

• Elements in their standard states are not included in the ΔHreaction calculation. That is, ΔH°f for an

element in its standard state is zero.


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Figure 9.12: Pathway for the combustion

f i


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of ammonia


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Th th l f b ti f t i


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The enthalpy of combustion per gram of octane isabout twice that per gram of methanol. On this

basis, gasoline appears to be superior to methanol

for use in a racing car, where weight considerations

are usually very important. Why, then, is methanol

used in racing cars? The answer is that methanol burns much more smoothly than gasoline in high-

performance engines, and this advantage more

than compensates for its weight disadvantage.


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(煤

)

(噴射機燃料)

(潤滑 )(瀝青)

(柴

)

9 8 New energy resources


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9.8 New energy resources

• Coal conversion

• Hydrogen as a fuel

• Other energy alternatives - ethanol & methanol

Figure 9.15: Coal gasification


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Possible solution:

H2(g) + M(s) → MH2(s)

MH2(s) → M(s) + H2(g)

Enthalpy and thermochemistry.pdf

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