2013 General Chemistry I 1 Chapter 7. THERMODYNAMICS: THE FIRST LAW 2012 General Chemistry I...

2013 General Chemistry I 1

Chapter 7. THERMODYNAMICS: THE FIRST LAW

2012 General Chemistry I

SYSTEMS, STATES, AND ENERGY

7.1 Systems7.2 Work and Energy7.3 Expansion Work7.4 Heat7.5 The Measurement of Heat7.6 The First Law7.7 A Molecular Interlude: The Origin of Internal Energy


SYSTEMS, STATES, AND ENERGY (Sections 7.1-7.7)

7.1 Systems7.1 Systems

Thermodynamics deals with transformation (from one form to another) and transfer (from one place to another) of energy.

- System means the region in which we are interested

- Surroundings: everything else

- Universe (the system and the surroundings


- Open system: exchanging both matter andenergy with the surroundings

- Closed system: a fixed amount of matter,but exchanging energy with the surroundings

- Isolated system: no contact with its surroundings

Types of systems


235s

Identify the following systems as open, closed, or isolated:

(a) Coffee in a very high quality thermos bottle

(b) Coolant in a refrigerator coil

(c) A bomb calorimeter in which benzene is burned

(d) Gasoline burning in an automobile engine

(e) Mercury in a thermometer

(f) A living plant

Solutions

Example 7.1


7.2 Work and Energy7.2 Work and Energy

Work: the process of achieving motion against an opposing force.

- unit: joule (J), 1 J = 1 kg·m2·s-2 = 1 N·m

Energy: the capacity of a system to do work: Internal energy (U) is the total store of energy in a system. In thermodynamics, "changes" in energy (ΔU) are dealt with.

U = Ufinal – Uinitial

-Symbol w: the energy transferred to a system by doing work; in the absence of other changes, U = w


7.3 Expansion Work7.3 Expansion Work

-Expansion work: the work arising from a change in thevolume of a system. Table 7.1 shows expansion and nonexpansion work.


Only when the external pressure is constant during the expansion.

Note if Pex = 0, a vacuum, w = 0; free expansion (expansion against vacuum)

Units of work

Irreversible expansion work

The negative sign relates work to the system.


Self-Test 7.1B

The gases in the four cylinders of an automobile engineexpand from 0.22 L to 2.2 L during one ignition cycle.Assuming the gear train maintains a steady pressure of9.60 atm on the gases, how much work can the engine doin one cycle?

Using w = -PexV,

w = (9.60 atm) x (1.98 L) = 19.0 L atm

= 19.0 L atm101.325 J1 L atm

= 1.93 kJx

Solution

_

_

_

_


Air in a bicycle pump is compressed by pushing in the handle. If the

inner diameter of the pump is 3.0 cm and the pump is depressed 20 cm

with a pressure of 2.00 atm,

(a) How much work is done in the compression?

(b) Is the work positive or negative with respect to the air in the pump?

238s

Example 7.3


Solution


Reversible process can be reversed by an

infinitesimal change in a variable.

E.g. reversible, isothermal expansion of an ideal gas

PexV = nRTconst = constant

- Work done is the area beneath the ideal gas isothermlying between the initial and the final volumes.


EXAMPLE 7.2

A piston confines 0.100 mol Ar(g) in a volume of 1.00 L at 25 oC. Two

experiments are performed. (a) The gas is allowed to expand through

an additional 1.00 L against a constant pressure of 1.00 atm. (b) The

gas is allowed to expand reversibly and isothermally to the same

final volume. Which process does more work?

(a) Irreversible path:

(b) Reversible path:


Self-Test 7.2A

A cylinder of volume 2.00 L contains 0.100 mol He(g) at 30 oC.Which process does more work on the system, compressingthe gas isothermally to 1.00 L with a constant pressure of 5.00atm, or compressing it reversibly and isothermally to the samefinal volume?

Solution

Isothermal irreversible compression

w = PexV = (5.00 atm) x ( 1.00 L) = +5.00 L atm_ _

= +507 J (1 L atm = 101.325 J)

Isothermal reversible compression

w = nRT ln(Vfinal/Vintitial)_

= (0.100 mol) (8.315 J K-1 mol-1) x (303 K) x lnx1.00L

2.00 L= +175 J.Thus the irreversible compression does more work.

_


7.4 Heat7.4 Heat

Heat is the energy transferred as a result of a temperature difference.

- Thermal energy of a system: the sum of the potential and

kinetic energies arising from the chaotic thermal motion of

atoms, ions, and molecules

-The energy transferred to a system as heat is q

Internal energy of the system changed by transferring energy as heat

U = q

- Unit: cal, the energy needed to raise T of 1 g of water by 1 oC.

1 cal = 4.184 J (exactly)

1 Cal (nutritional calorie) = 1 kcal


Exothermic process: releasing heat into the surroundings

- Thermite reaction:

Endothermic process: absorbing heat from the surroundings


7.5 The Measurement of Heat7.5 The Measurement of Heat

-Adiabatic Process: one performed in isolated system; noheat exchanged with surroundings

Heat capacity, C is the ratio of the heat supplied to the rise in temperature observed.

Heat capacity = heat supplied

temperature rise produced


Some Specific and Molar Heat Capacities (Table 7.2)


Self-Test 7.3A

Potassium perchlorate, KClO4, is used as an oxidizerin fireworks. Calculate the heat required to raise thetemperature of 10.0 g of KClO4 from 25 oC to an ignitiontemperature of 900. oC. The specific heat capacity of KClO4

is 0.8111 J K-1 g-1.

Solution

From q = mCsT,

Heat required = (10.0 g) x (0.8111 J K-1 g-1) x (875 K)

= 7097 J or 7.10 kJ


- Bomb calorimeter: q at constant volume (qV)

- Styrofoam calorimeter: q at constant pressure (qp)

Types of Calorimeters

- Calorimeter: a device in which heat transfer is monitored by recording the change in temperature that it produces


Self-Test 7.4A

A small piece of calcium carbonate was placed in abomb calorimeter of 488 J (oC-1) heat capacity (Ccal) and0.100 L of dilute hydrochloric acid was poured over it.The temperature of the calorimeter rose by 3.57 oC.What is the value of U for the reaction of hydrochloricacid with calcium carbonate?

Solution

Using qcal = CcalT = (488 J (oC-1)) x (3.57 oC)

= 1742 J = 1.74 kJ

Hence U = 1.74 kJ_


(a) Calculate the heat that must be supplied to a 500.0 g copper kettle

containing 400.0 g of water to raise its temperature from 22.0 oC to the

boiling point of water, 100.0 oC (See Table 7.2).

(b) What percentage of the heat is used to raise the temperature of the

water?

246s

Solution

Example 7.17


7.6 The First Law7.6 The First Law

The first law of thermodynamics (closed system) states that the change in internal energy (U) is the sum of the work and heat changes: it is applicable to any process that begins and ends in equilibrium states.

All the energies received are turned into the energy of the system: this is a form of the energy conservation law.


U is a state function: a property that depends only on the current state of the system and is independent of how that state was prepared.

Other state functions are P, V, T, H, S, G


- For any ideal gas, ΔU = 0 for an isothermal process.

No changes in the kinetic energy of an ideal gas if ΔT = 0 No intermolecular forces for ideal gas molecules

No changes in potential energy during expansion or compression

No changes in the total energy; ΔU = 0


Self-Test 7.5B

A system was heated by using 300. J of heat, yet it wasfound that its internal energy decreased by 150. J (soU = -150. J). Calculate w. Was work done on the systemor did the system do work?

Solution

Since U = q + w

( 150 J) = (+300 J) w_ +

w = 450 J Work was done by the system_


EXAMPLE 7.5

Suppose that 1.00 mol of ideal gas molecules maintained at 292 K and

3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm

by two different paths. (a) Path A is an isothermal, reversible expansion.

(b) Path B has two parts. In step 1, the gas is cooled at constant volume

until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed

to expand against a constant pressure of 1.20 atm until its volume is

20.00 L and T = 292 K. Determine for each path the work done, the heat

transferred, and the change in internal energy (w, q, and U).


(a) From ,


(b) Step 1, w = 0

Step 2, from

Less work is done in the irreversible path andless energy has to enter the system as heatto maintain its temperature.


Self-Test 7.6A

Suppose that 2.00 mol CO2 at 2.00 atm and 300. K iscompressed isothermally and reversibly to half itsoriginal volume before being used to produce sodawater. Calculate w, q, and U by treating CO2 as anideal gas.

Solution

Since the process is isothermal, U = 0 and w = q._

From w = nRT ln(Vfinal/Vinitial)_

w = (2.00 mol) x (8.315 J K-1 mol-1) x (300. K) x ln 12

_

= +3.46 kJHence, q = 3.46 kJ_


250s

In an adiabatic process, no energy is transferred as heat. Indicate

whether each of the following statements about an adiabatic process in a

closed system is always true, always false, or true in certain conditions

(specify the conditions):

(a) ΔU = 0

(b) Q = 0

(c) q < 0

(d) ΔU = q

(e) ΔU = w

Solution

Example 7.13


7.7 A Molecular Interlude: The Origin of Internal 7.7 A Molecular Interlude: The Origin of Internal EnergyEnergy

U = K + EP

- A system at high temperature has a greater internal energy than the same system at a lower temperature. (Greater kinetic energy)

Equipartition theorem: The average value of each quadratic contribution to the energy of a molecule in a sample at a temperature T is equal to .

- kB = 1.381×10–23 J·K-1; Boltzmann’s constant

- R = kBNA; RT = 2.48 kJ·mol-1 at 25 oC

-Can be used for vibrational motion only when quantum effects can be neglected, at high temperatures. (kBT >> ΔE)


- Molecules of N atoms; 3N degrees of freedom

(a) Center of mass

3 translational degrees of freedom

(b) Linear molecule

2 rotational degrees of freedom

3N – 5 vibrational degrees of freedom

(c) Nonlinear molecule

3 rotational degreesof freedom

3N – 6 vibrationaldegrees of freedom


- At room temperature, the equipartition theorem holds for translational and rotational motions.

(a) Monatomic ideal gas: translational kinetic energy only (3 modes)

(b) Diatomic or linear ideal gas: translational energy (3 modes) + rotational energy (2 modes)

(c) Nonlinear ideal gas: translational energy (3 modes) + rotational energy (3 modes)


252s

7.25 Which molecular substance do you expect to have the

higher molar heat capacity, NO or NO2? Why?

Solution

Example 7.25




ENTHALPY

7.8 Heat Transfers at Constant Pressure7.9 Heat Capacities at Constant Volume and Constant Pressure7.10 A Molecular Interlude: The Origin of the Heat Capacities of Gases7.11 The Enthalpy of Physical Change7.12 Heating Curves


ENTHALPY (Sections 7.8-7.12)

7.8 Heat Transfers at Constant Pressure7.8 Heat Transfers at Constant Pressure

- At constant volume and no nonexpansion work:

w = -PexV = 0; U = w + q = q

- However, most chemical reactions take place at a constant pressure of about 1 atm.

- At constant pressure Pex, if no work other than

pressure-volume work is done, then


Enthalpy is defined as:

(constant pressure, pressure-volume work only)

Since U, P, and V are state functions, H is also a state function!


-Exothermicreaction (ΔH < 0)

-Endothermicreaction (ΔH > 0)

H = -208 kJ

Exothermic and endothermic reactions


Self-Test 7.7A

In an exothermic reaction at constant pressure, 50. kJ ofenergy left the system as heat and 20. kJ of energy leftthe system as expansion work. What are the values of(a) H and (b) U for this process?

Solution

(a) Since the reaction occurs at constant P, and isexothermic, H = 50. kJ

(b) Since H = U + PV, and PV = +20. kJ,U = ( 50. kJ) (+20. kJ) = 70. kJ

_

_ _ _


7.9 Heat Capacities at Constant Volume and 7.9 Heat Capacities at Constant Volume and Constant PressureConstant Pressure

for any ideal gas


- If CV and CP do not change with temperature,

qV = nCV,m ΔT qP = nCP,m ΔT

qV < qP


7.10 A Molecular Interlude: The Origin of the 7.10 A Molecular Interlude: The Origin of the Heat Capacities of GasesHeat Capacities of Gases

- For monatomic ideal gases (see 7.7),

- For linear molecules (see 7.7),


Variation of molar heat capacity of iodine vaporat constant volume(Fig. 7.20)


Predict the contribution to the heat capacity CV,m made by molecular

motions for each of the following atoms and molecules:

(a) HCN; (b) C2H6; (c) Ar; (d) HBr

255s

Solutions

Example 7.29


EXAMPLE 7.6

Calculate the final temperature and the change in internal energy

when 500 J of energy is transferred as heat to 0.900 mol O2(g) at

298 K and 1.00 atm at (a) constant volume; (b) constant pressure.

Treat the gas as ideal.

(a)

(b)

= +26.7 K


Self-Test 7.8A

Calculate the final temperature and the change in internalenergy when 500. J of energy is transferred as heat to 0.900mol Ne(g) at 298 K and 1.00 atm (a) at constant volume and(b) at constant pressure. Treat the gas as ideal.

Solution

(a) CV,m = 3/2R = 32

(8.315 J K-1 mol-1) = 12.47 J K-1 mol-1

T =q

nCV,m

=(500. J)

(0.900 mol) x (12.47 J K-1 mol-1)

= 44.6 K

Hence, final temperature is 342.6 K or 343 K

U = q at constant volume = 500. J


(b) CP,m = 5/2R = 52

(8.315 J K-1 mol-1) = 20.79 J K-1 mol-1

T =q

nCV,m

=(500. J)

(0.900 mol) x (20.79 J K-1 mol-1)

= 26.7 K

Hence, final temperature is 324.7 K or 325 K

U = q = nCV,mT = (0.900 mol) x (12.47 J K-1 mol-1) x (26.7 K)

= 300. J


7.11 The Enthalpy of Physical Change7.11 The Enthalpy of Physical Change

Enthalpy of vaporization is the difference in molar enthalpy between the vapor and the liquid states (> 0, always).

- Energy needed to separate liquid molecules

- Temperature dependence:


Enthalpy of fusion is the molar enthalpy change that accompanies melting (fusion).

Enthalpy of freezing is the change in molar enthalpy change of a liquid when it solidifies.

Hfreez = Hm(solid) – Hm(liquid)

In general:


Self-Test 7.9A

A sample of benzene, C6H6, was heated to 80 oC, itsnormal boiling point. The heating was continued until15.4 kJ had been supplied; as a result, 39.1 g of boilingbenzene was vaporized. What is the enthalpy ofvaporization of benzene at its boiling point?

Solution

It takes 15.4 kJ of heat to vaporize 0.50 mol of benzene,hence Hvap = (15.4 kJ) / (0.50 mol) = 30.8 kJ mol-1

The molar mass of benzene is 78.2 g/mol, hence 39.1 gcorresponds to 0.50 mol


Enthalpy of sublimation is the molar enthalpy change when a solid sublimes


Standard Enthalpies of Physical Change*Table 7.3)


Self-Test 7.10B

The enthalpy of vaporization of methanol is 38 kJ mol-1 at25 oC and the enthapy of fusion is 3 kJ mol-1 at the sametemperature. What is the enthalpy of sublimation of methanolat this temperature?

Solution

Since Hsub = Hvap + Hfus,Hsub = (38 kJ mol-1) + (3 kJ mol-1) = 41 kJ mol-1


7.12 Heating Curves7.12 Heating Curves

- Heating curve is the graph showing the variation in the temperature of a sample as it is heated at a constant rate at constant pressure and therefore at a constant rate of increase in enthalpy.

-The steeper the slope of a heating curve, the lower the heat capacity.

-The horizontal sections correspond to phase changes:melting and boiling.


- In water, the slope for liquid < those for solid or vapor,

the high heat capacity of the liquid is due largely to the

extensive hydrogen bonding network.

Heating curve of water (Fig. 7.26)


The following data were collected for a new compound used in

cosmetics: Hfus = 10.0 kJ·mol-1, Hvap = 20.0 kJ·mol-1; heat

capacities: 30 J·mol-1 for the solid; 60 J·mol-1 for the liquid; 30 J·mol-1

for the gas. Which heating curve below best matches the data for

this compound?

259s

Solution: (c)

Example 7.41




THE ENTHALPY OF CHEMICAL CHANGE

7.13 Reaction Enthalpies7.14 The Relation Between H and U7.15 Standard Reaction Enthalpies7.16 Combining Reaction Enthalpies: Hess’s Law7.17 The Heat Output of Reactions7.18 Standard Reaction Enthalpies7.19 The Born-Haber Cycle7.20 Bond Enthalpies7.21 The Variation of Reaction Enthalpy with Temperature


THE ENTHALPY OF CHEMICAL CHANGE (Sections 7.13-7.21)

7.13 Reaction Enthalpies7.13 Reaction Enthalpies

-Thermochemical equation: consisting of a chemical equation together with a statement of the reaction enthalpy, the corresponding enthalpy change.


Self-Test 7.11A

When 0.231 g of phosphorus reacts with chlorine to formphosphorus trichloride, PCl3, in a constant pressurecalorimeter of heat capacity 216 J (oC)-1, the temperatureof the calorimeter rises by 11.06 oC. What is the thermo--chemical equation for the reaction?

Solution

The molar mass of P is 30.97 g mol-1, hence 0.231 gcorresponds to 7.46 x 10-3 mol.Heat transferred to calorimeter, qcal = CcalT= (216 J (oC)-1) x (11.06 oC) = 2.39 kJIf the equation for the reaction is2P(s) + 3Cl2(g) 2PCl3(l), then H = 2 (mol) x qcal

no. mol

=(2 mol) x 2.39 kJ

7.46 x 10-3 mol= 641 kJ

_

_ _

Therefore 2P(s) + 3Cl2(g) 2PCl3(l), H = 641 kJ_


7.43 Carbon disulfide can be prepared from coke (an impure form of

carbon) and elemental sulfur:

(a) How much heat is absorbed in the reaction of 1.25 mol S8?

(b) Calculate the heat absorbed in the reaction of 197 g of carbon

with an excess of sulfur.

(c) If the heat absorbed in the reaction was 415 kJ, how much CS2

was produced?

4 C(s) + S8(s) → 4 CS2(l) Ho = +358.8 kJ

261sExample 7.43


7.14 The Relation Between 7.14 The Relation Between HH and and UU

H = Hfinal – Hinitial = U + (nfinal – ninitial)RT = U + ngasRT

- If a gas is formed in the reaction,

- For reactions in liquids and solids only,


Oxygen difluoride is a colorless, very poisonous gas that reacts

rapidly with water vapor to produce O2, HF, and heat:

261s

What is the change in internal energy for the reaction of 1.00 mol

OF2?

OF2(g) + H2O(g) → O2(g) + 2HF(g) H = -318 kJ

Example 7.51

Solution


Self-Test 7.12A

The thermochemical equation for the combustion ofcyclohexane, C6H12, isC6H12(l) + 9O2(g) 6CO2(g) + 6H2O(l), H = -3920 kJat 298 K. What is the change in internal energy for thecombustion of 1.00 mol C6H12((l) at 298 K?

Solution

ngas = nfinal ninitial = 3_ _

Since H = U + ngasRT

( 3920 kJ) = U 3 x (8.315 J K-1 mol-1) x (298 K)_

U = 3910 kJ or 3.91 x 103 kJ

_

_ _


7.15 Standard Reaction Enthalpies7.15 Standard Reaction Enthalpies

- Reaction enthalpies depend on the physical states of the reactants and products

88 kJ = enthalpy of vaporizationof water (44 kJ·mol-1) × 2

It is therefore useful if enthalpies (Ho) can be referred to a standard state.


or for a solute in a liquid solution: concentration of 1 mol·L-1.

Standard reaction enthalpy, Ho is the reaction enthalpy when reactants in their standard states change into products in their standard states.

-Most thermochemical data is reported for 25 oC (298.15 K) but the temperature is not part of standard states.

Standard state: refers to a pure form at exactly 1 bar


7.16 Combining Reaction Enthalpies: Hess’s 7.16 Combining Reaction Enthalpies: Hess’s LawLaw

- Enthalpy is a state function; H is independent of the path.

Hess’s law: the overall reaction enthalpy is the sum of the reaction enthalpies of the steps into which the reaction can be divided.

Ho = -393.5 kJ


EXAMPLE 7.9Consider the synthesis of propane, C3H8, a gas use as camping fuel:

It is difficult to measure the enthalpy change of this reaction directly.However, standard enthalpies of combustion reactions are easy tomeasure. Calculate the standard enthalpy of this reaction from thefollowing experimental data:


Self-Test 7.13B

Methanol is a clean-burning liquid fuel proposed as areplacement for gasoline. Suppose it could be producedby the controlled reaction of oxygen with methane. Findthe standard reaction enthalpy for the formation of 1 molCH3OH(l) from methane and oxygen, given the followinginformation.

(1) CH4(g) + H2O(g) CO(g) + 3H2(g) Ho = +206.10 kJ(2) 2H2(g) + CO(g) CH3OH(l) Ho = 128.33 kJ(3) 2H2(g) + O2(g) 2H2O(g) Ho = 483.64 kJ

__

Solution

The required equation is 2CH4(g) + O2(g) 2CH3OH(l)

Multiply equation (1) and (2) by 2, and add both to equation (3):this gives the required equation.Ho = 2Ho(1) + 2Ho(2) + Ho(3) = 328.10 kJ

_


Calculate the reaction enthalpy for the synthesis of hydrogen

chloride gas, H2(g) + Cl2(g) → 2 HCl(g), from the following data:

NH3(g) + HCl(g) → NH4Cl(s) Ho = -176.0 kJ

N2(g) + 3 H2(g) → 2 NH3(g) Ho = -92.22 kJ

N2(g) + 4 H2(g) + Cl2(g) → 2 NH4Cl(s) Ho = -628.86 kJ

266s

Example 7.67

Solution


7.17 The Heat Output of Reactions7.17 The Heat Output of Reactions

the standard enthalpy of combustion, Hco

is defined as the change in enthalpy per mole of a substance that is burned in a combustion reaction under standard conditions.

There is also the specific enthalpy of combustion: the enthalpy of combustion per gram

Combustion reactions are important throughout chemistry - they are always exothermic.


Standard Enthalpies of Combustion at 25 oC(Table 7.4)*


Self-Test 7.14A

The thermochemical equation for the combustion ofpropane is:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = 2220. kJ

What mass of propane must be burned to supply 350. kJas heat? Would it be easier to pack propane rather thanbutane?

Solution

_

The above equation gives the heat supplied by burning1 mol of propane, hence 350. kJ of heat can be obtainedfrom

3502220

= 0.1577 mol of propane

Since the molar mass of propane is 44.07 g mol-1, themass of propane needed to supply 350. kJ of heat is(44.07 g mol-1) x (0.1577 mol) = 6.95 g

Yes, propane would be slightly easier to pack, becausethe liquid is slightly less dense than butane.


7.18 Standard Enthalpies of Formation7.18 Standard Enthalpies of Formation

Standard enthalpy of formation, Hfo is defined as

the standard reaction enthalpy per mole of formula units for the formation of a substance from its elements in their most stable form.


Standard Enthalpies of Formation at 25 oC (kJ mol-1)(Table 7.5)


Standard reaction enthalpy can be calculated from standard enthalpies of formation of reactants and products.


Self-Test 7.15A

Calculate the standard enthalpy of combustion ofglucose from the standard enthalpies of formationin Table 7.5 and Appendix 2A.

Solution

The combustion equation is

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

nHfo(products) = {(6 mol) x ( 393.51 kJ mol-1)}

+ {(6 mol) x ( 285.83 kJ mol-1)}

=

_

_

4076.04 kJ_

nHfo(reactants) = 1268 kJ + 0 kJ = 1268 kJ

Ho (combustion) =

nHfo(products)

nHfo(reactants)_

_

= 2808 kJ mol-1_

_


EXAMPLE 7.12

Use the information of standard enthalpies of formation and theenthalpy of combustion of propane gas to calculate the enthalpyof formation of propane, a gas commonly used for camping stovesand outdoor barbecues.

= -2323.85 kJ

= -2220 kJ


Self-Test 7.16A

Calculate the standard enthalpy of formation ofethyne, the fuel used in oxyacetylene weldingtorches, from the information in Tables 7.4 and 7.5.

Solution

nHfo(products) nHf

o(reactants)

The thermochemical equation for the combustionof ethyne is:C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(l) Ho = 1300 kJ mol-1

Hor = _

1300 kJ mol-1 = [(2 mol) x ( 393.51 kJ mol-1) + ( 285.83 kJ mol-1)[(Hf

o(ethyne)]__ __

Hfo(ethyne) = + 227 kJ mol-1

_


Calculate the standard enthalpy of formation of PCl5(s) from the

standard enthalpy of formation of PCl3(l) (Hof(PCl3,l) = -319.7 kJ·mol-1)

and PCl3(l) + Cl2(g) → PCl5(s), Ho = -124 kJ.

273sExample 7.73

Solution


7.19 The Born-Haber Cycle7.19 The Born-Haber Cycle

Lattice enthalpy of the solid, HL

- Lattice enthalpies at 25oC (kJ·mol-1)(Table 7.)


Born-Haber cycle: a closed path of steps, one of which is the formation of a solid lattice from the gaseous ions


EXAMPLE 7.13 Lattice enthalpy of KCl

Hf (K, atoms)

Hf (Cl, atoms)

I(K)

-Eea(Cl)

-Hf(KCl)

K+Cl-(s) K+(g) + Cl-(g)


Self-Test 7.17A

Calculate the lattice enthalpy of calcium chloride, CaCl2,by using data in Appendices 2A and 2D.

Ca2+(g) + 2e-(g) + 2Cl(g)

Ca2+(g) 2Cl-(g)+

Ca(g) + 2Cl(g)

Ca(s) + Cl2(g)

Ca2+(Cl-)2(s)

-Hfo = +795.8

kJ mol-1 +178.2 + 244

Solution

The Born-Haber cycle is:

H

+1735

698_(not to scale)

-HL

HL = 795.8 + 178.2+ 244 + 1735 698= 2255 kJ

_


7.20 Bond Enthalpies7.20 Bond Enthalpies

Bond enthalpy, HB is the difference between the standard molar enthalpies of a molecule, X-Y, and its fragments X and Y in the gas phase.

E.g.

- Bond breaking is always endothermic and bond formation is always exothermic.


Mean bond enthalpy: small variations of a specific bond enthalpy in polyatomic molecules give a guide to the average bond strength.

- The enthalpy change from a liquid (or solid) sample,


EXAMPLE 7.14

Estimate the enthalpy of the reaction between gaseous iodoethane

and water vapor:

HBo(C-I) + HB

o(O-H) =

- Breaking the bonds (reactants)

- Forming the bonds (products)

HBo(C-O) + HB

o(H-I) =

- The overall enthalpy change:


Self-Test 17.18B

Use bond enthalpies to estimate the standard enthalpy ofreaction in which 1.00 mol of gaseous CH4 reacts withgaseous F2 to form gaseous CH2F2 and HF.

Solution

The relevant equation isCH4(g) + 2F2(gas) CH2F2(g) + 2HF(g)

In this reaction, 2 mol of F-F bonds and 2 mol of C-H bondsare broken (reactants), whereas 2 mol of C-F bonds and 2 molof H-F bonds are formed.

Ho = mean HB(reactants) mean HB(products)_

= [(2 x +412 kJ mol-1) + (2 x +158 kJ mol-1)][(2 x +484 kJ mol-1) + (2 x +565 kJ mol-1)

= 958 kJ mol-1_

_


277s

7.81 Use the bond enthalpies in the table to estimate the reaction enthalpy

for 3 C2H2(g) → C6H6(g)

Example 7.81

Solution


7.21 The Variation of Reaction Enthalpy with 7.21 The Variation of Reaction Enthalpy with TemperatureTemperature

- The enthalpies of both reactants and products increase with temperature.


Kirchhoff’s law


EXAMPLE 7.15

The standard enthalpy of reaction of N2(g) + 3H2(g) → 2NH3(g) is

-92.22 kJ·mol-1 at 298 K. The industrial synthesis takes place at 450 oC.

What is the standard reaction enthalpy at the latter temperature?


Self-Test 17.19B

The standard enthalpy of formation of ammonium nitrateis 365.56 kJ mol-1 at 298 K. Estimate its value at 250. oC.

Solution

Use Kirchhoff's law: Ho(T2) = Ho(T1) + (T2 T1)Cp_

_

The equation of formation isN2(g) + 3/2O2(g) + 2H2(g) NH4NO3(s)

250. oC = 523 K

Values of Cp (J K-1 mol-1): NH4NO3(s) 84.1 (product);

N2(g) 29.12; 3/2O2(g) 44.04; 2H2(g) 57.64 (reactants)

Ho(T2) = ( 365.56 kJ mol-1) + (225 K)( 46.7 J K-1 mol-1)

= 376 kJ mol-1_

__


(a) Calculate the enthalpy of vaporization of benzene (C6H6) at

298.2 K. The standard enthalpy of formation of gaseous benzene

is +82.93 kJ·mol-1, and that of liquid benzene is +49.0 kJ·mol-1.

278sExample 7.87

Solution


278s

(b) Given that, for liquid benzene, CP,m = 136.1 J·mol-1·K-1 and

that,

for gaseous benzene, CP,m = 81.67 J·mol-1·K-1, calculate the

enthalpy of vaporization of benzene at its boiling point (353.2 K).


278s

(c) Compare the value obtained in part (b) with the real value of

30.8 kJ·mol-1. What is the source of difference between these

numbers?

2013 General Chemistry I 1 Chapter 7. THERMODYNAMICS: THE FIRST LAW 2012 General Chemistry I...

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