## Chapter 7. THERMODYNAMICS: THE FIRST LAW

### SYSTEMS, STATES, AND ENERGY

7.1 Systems 7.2 Work and Energy 7.3 Expansion Work 7.4 Heat 7.5 The Measurement of Heat 7.6 The First Law 7.7 A Molecular Interlude: The Origin of Internal Energy

1

### SYSTEMS, STATES, AND ENERGY (Sections 7.1-7.7) 7.1 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 2012 General Chemistry I

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Types of systems - Open system: exchanging both matter and energy with the surroundings - Closed system: a fixed amount of matter, but exchanging energy with the surroundings - Isolated system: no contact with its surroundings 2012 General Chemistry I

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### 7.2 Work and Energy

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

unit: joule (J), 1 J = 1 kg·m 2 ·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.

D

final

### – U

initial

-

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

D

U = w 2012 General Chemistry I

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### 7.3 Expansion Work

-

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

2012 General Chemistry I

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Units of work Irreversible expansion work Only when the external pressure is constant during the expansion. The negative sign relates work to the system.

Note if P ex = 0, a vacuum, w = 0; free expansion (expansion against vacuum) 2012 General Chemistry I

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Self-Test 7.1B

The gases in the four cylinders of an automobile engine expand from 0.22 L to 2.2 L during one ignition cycle.

Assuming the gear train maintains a steady pressure of 9.60 atm on the gases, how much work can the engine do in one cycle?

Solution Using w = -P ex

D

V, = _ 19.0 L atm x 101.325 J 1 L atm _ 19.0 L atm = _ 1.93 kJ 2012 General Chemistry I

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Reversible process can be reversed by an infinitesimal change in a variable.

E.g. reversible, isothermal expansion of an ideal gas P ex V = nRT const = constant - Work done is the area beneath the ideal gas isotherm lying between the initial and the final volumes .

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EXAMPLE 7.2

A piston confines 0.100 mol Ar(g) in a volume of 1.00 L at 25 o C. 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: 2012 General Chemistry I

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Self-Test 7.2A

A cylinder of volume 2.00 L contains 0.100 mol He(g) at 30 o C.

Which process does more work on the system, compressing the gas isothermally to 1.00 L with a constant pressure of 5.00

atm, or compressing it reversibly and isothermally to the same final volume?

Solution Isothermal irreversible compression w = P ex

D

V = _ = +507 J (1 L atm = 101.325 J) Isothermal reversible compression w = _ nRT ln(V final /V intitial ) = _ (0.100 mol) x (8.315 J K -1 mol -1 ) x (303 K) x ln 1.00L

2.00 L = +175 J.

Thus the irreversible compression does more work.

2012 General Chemistry I

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### 7.4 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

D

U = q Unit: cal, the energy needed to raise T of 1 g of water by 1 o C.

1 cal = 4.184 J (exactly) 1 Cal (nutritional calorie) = 1 kcal 2012 General Chemistry I

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Exothermic process: releasing heat into the surroundings - Thermite reaction:

Endothermic process: absorbing heat from the surroundings 2012 General Chemistry I

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### 7.5 The Measurement of Heat

-

Adiabatic Process: one performed in isolated system; no heat 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 2012 General Chemistry I

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Some Specific and Molar Heat Capacities (Table 7.2) 2012 General Chemistry I

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Self-Test 7.3A

Potassium perchlorate, KClO 4 , is used as an oxidizer in fireworks. Calculate the heat required to raise the temperature of 10.0 g of KClO 4 temperature of 900.

o from 25 C. The specific heat capacity of KClO 4 is 0.8111 J K -1 g -1 .

o C to an ignition Solution From q = mC s

D

T, Heat required = (10.0 g) x (0.8111 J K -1 g -1 ) x (875 K) = 7097 J or 7.10 kJ 2012 General Chemistry I

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- Calorimeter: a device in which heat transfer is monitored by recording the change in temperature that it produces - Styrofoam calorimeter: q at constant pressure (q

p

) - Bomb calorimeter: q at constant volume (q

V

) Types of Calorimeters 2012 General Chemistry I

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Self-Test 7.4A

A small piece of calcium carbonate was placed in a bomb calorimeter of 488 J ( o C -1 ) heat capacity (C cal ) and 0.100 L of dilute hydrochloric acid was poured over it.

The temperature of the calorimeter rose by 3.57

What is the value of o C.

D

U for the reaction of hydrochloric acid with calcium carbonate?

Solution Using q cal = C cal

D

T = (488 J ( o C -1 )) x (3.57

o C) = 1742 J = 1.74 kJ Hence

D

U = _ 1.74 kJ 2012 General Chemistry I

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### 7.6 The First Law

The first law of thermodynamics (closed system) states that the change in internal energy (

D

U) is the sum of the work and heat changes: it is a pplicable 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.

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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 2012 General Chemistry I

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- 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 2012 General Chemistry I

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Self-Test 7.5B

A system was heated by using 300. J of heat, yet it was found that its internal energy decreased by 150. J (so

D

U = -150. J). Calculate w. Was work done on the system or did the system do work?

Solution Since

D

U = q + w + w w = _ 450 J Work was done by the system 2012 General Chemistry I

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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

D

U).

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(a) From , 2012 General Chemistry I

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(b) Step 1, w = 0 Step 2, from Less work is done in the irreversible path and less energy has to enter the system as heat to maintain its temperature.

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Self-Test 7.6A

Suppose that 2.00 mol CO 2 at 2.00 atm and 300. K is compressed isothermally and reversibly to half its original volume before being used to produce soda water. Calculate w, q, and

D

U by treating CO 2 as an ideal gas.

Solution Since the process is isothermal,

D

U = 0 and w = From w = _ nRT ln(V final /V initial ) _ q.

w = _ (2.00 mol) x (8.315 J K -1 mol -1 ) x (300. K) x ln 1 2 = +3.46 kJ Hence, q = _ 3.46 kJ 2012 General Chemistry I

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### 7.7 A Molecular Interlude: The Origin of Internal Energy

U = K + E P - 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 .

- k

B = 1.381

×

10 –23 J·K -1 ; Boltzmann’s constant - R = k

B N A

; RT = 2.48 kJ·mol -1 at 25 o C

-

Can be used for vibrational motion only when quantum effects can be neglected, at high temperatures. (k B T >> ΔE) 2012 General Chemistry I

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Molecules of N atoms; 3N degrees of freedom (a) Center of mass 3 translational degrees of freedom (b) Linear molecule (c) Nonlinear molecule 2 rotational degrees of freedom 3N – 5 vibrational degrees of freedom 3 rotational degrees of freedom 3N – 6 vibrational degrees of freedom 2012 General Chemistry I

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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) 2012 General Chemistry I

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## Chapter 7. THERMODYNAMICS: THE FIRST LAW

### ENTHALPY

7.8 Heat Transfers at Constant Pressure 7.9 Heat Capacities at Constant Volume and Constant Pressure 7.10 A Molecular Interlude: The Origin of the Heat Capacities of Gases 7.11 The Enthalpy of Physical Change 7.12 Heating Curves

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### ENTHALPY (Sections 7.8-7.12)

- At constant volume and no nonexpansion work: w = -P

ex

D

V = 0;

D

U = w + q = q - However, most chemical reactions take place at a constant pressure of about 1 atm.

### 7.8 Heat Transfers at Constant Pressure

- At constant pressure P

ex

, if no work other than pressure-volume work is done, then 2012 General Chemistry I

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Enthalpy is defined as: (constant pressure, pressure-volume work only) Since U, P, and V are state functions, H is also a state function!

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Exothermic and endothermic reactions

-

Exothermic reaction ( ΔH < 0)

-

Endothermic reaction ( ΔH > 0)

D

H = -208 kJ 2012 General Chemistry I

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Self-Test 7.7A

In an exothermic reaction at constant pressure, 50. kJ of energy left the system as heat and 20. kJ of energy left the system as expansion work. What are the values of (a)

D

H and (b)

D

U for this process?

Solution (a) Since the reaction occurs at constant P, and is exothermic,

D

H = _ 50. kJ (b) Since

D D

H = _

D

U + P

D

V, and P

D

V = +20. kJ, _ (+20. kJ) = _ 70. kJ 2012 General Chemistry I

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

2012 General Chemistry I for any ideal gas

34

- If C

V

and C

P

do not change with temperature,

q V

= nC V,m ΔT

q P

= nC P,m ΔT

q V

### < q

P

2012 General Chemistry I

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

- For monatomic ideal gases (see 7.7), - For linear molecules (see 7.7), 2012 General Chemistry I

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Variation of molar heat capacity of iodine vapor at constant volume (Fig. 7.20) 2012 General Chemistry I

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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 O 2 (g) at 298 K and 1.00 atm at (a) constant volume; (b) constant pressure.

Treat the gas as ideal.

(a) = +26.7 K (b) 2012 General Chemistry I

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Self-Test 7.8A

Calculate the final temperature and the change in internal energy when 500. J of energy is transferred as heat to 0.900

mol 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) C V,m

D

T = = 3/2R = q nC V,m = 3 2 (8.315 J K -1 mol -1 ) = 12.47 J K (500. J) (0.900 mol) x (12.47 J K -1 mol -1 ) -1 mol -1 = 44.6 K Hence, final temperature is 342.6 K or 343 K

D

U = q at constant volume = 500. J 2012 General Chemistry I

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(b) C P,m = 5/2R =

D

T = q nC V,m = 5 2 (8.315 J K -1 mol -1 ) = 20.79 J K -1 (500. J) (0.900 mol) x (20.79 J K -1 mol -1 ) mol -1 = 26.7 K Hence, final temperature is 324.7 K or 325 K

D

U = q = nC V,m

D

T = (0.900 mol) x (12.47 J K -1 mol -1 ) x (26.7 K) = 300. J 2012 General Chemistry I

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### 7.11 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: 2012 General Chemistry I

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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.

D

H

freez

m (solid)

### – H

m (liquid) In general: 2012 General Chemistry I

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Self-Test 7.9A

A sample of benzene, C 6 H 6 , was heated to 80 o C, its normal boiling point. The heating was continued until 15.4 kJ had been supplied; as a result, 39.1 g of boiling benzene was vaporized. What is the enthalpy of vaporization of benzene at its boiling point?

Solution The molar mass of benzene is 78.2 g/mol, hence 39.1 g corresponds to 0.50 mol It takes 15.4 kJ of heat to vaporize 0.50 mol of benzene, hence

D

H vap = (15.4 kJ) / (0.50 mol) = 30.8 kJ mol -1 2012 General Chemistry I

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Enthalpy of sublimation is the molar enthalpy change when a solid sublimes 2012 General Chemistry I

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Standard Enthalpies of Physical Change* Table 7.3) 2012 General Chemistry I

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Self-Test 7.10B

The enthalpy of vaporization of methanol is 38 kJ mol -1 25 o C and the enthapy of fusion is 3 kJ mol -1 at at the same temperature. What is the enthalpy of sublimation of methanol at this temperature?

Solution Since

D

H sub

D

H sub =

D

H vap +

D

H fus , = (38 kJ mol -1 ) + (3 kJ mol -1 ) = 41 kJ mol -1 2012 General Chemistry I

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### 7.12 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.

2012 General Chemistry I

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Heating curve of water (Fig. 7.26) 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.

2012 General Chemistry I

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## Chapter 7. THERMODYNAMICS: THE FIRST LAW

### THE ENTHALPY OF CHEMICAL CHANGE

7.13 Reaction Enthalpies 7.14 The Relation Between

D

H and

D

U

7.15 Standard Reaction Enthalpies 7.16 Combining Reaction Enthalpies: Hess’s Law 7.17 The Heat Output of Reactions 7.18 Standard Reaction Enthalpies 7.19 The Born-Haber Cycle 7.20 Bond Enthalpies 7.21 The Variation of Reaction Enthalpy with Temperature

49

### THE ENTHALPY OF CHEMICAL CHANGE (Sections 7.13-7.21) 7.13 Reaction Enthalpies

-

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

2012 General Chemistry I

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Self-Test 7.11A

When 0.231 g of phosphorus reacts with chlorine to form phosphorus trichloride, PCl 3 , in a constant pressure calorimeter of heat capacity 216 J ( o C) -1 , the temperature of the calorimeter rises by 11.06

o C. What is the thermo -chemical equation for the reaction?

Solution The molar mass of P is 30.97 g mol -1 , hence 0.231 g corresponds to 7.46 x 10 -3 mol.

Heat transferred to calorimeter, q = (216 J ( o C) -1 ) x (11.06

cal o C) = 2.39 kJ = C cal

D

T If the equation for the reaction is 2P(s) + 3Cl 2 (g) 2PCl 3 (l), then

D

H = _ 2 (mol) x q no. mol cal = _ (2 mol) x 2.39 kJ 7.46 x 10 -3 mol = _ 641 kJ Therefore 2P(s) + 3Cl 2 (g) 2PCl 3 (l),

D

H = 641 kJ 2012 General Chemistry I

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D

D

### U

- For reactions in liquids and solids only, - If a gas is formed in the reaction,

D

H = H final – H initial =

D

U + (n final – n initial )RT =

D

U +

D

n gas RT

2012 General Chemistry I

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Self-Test 7.12A

The thermochemical equation for the combustion of cyclohexane, C 6 H 12 , is C 6 H 12 (l) + 9O 2 (g) 6CO 2 (g) + 6H 2 O(l),

D

H = -3920 kJ at 298 K. What is the change in internal energy for the combustion of 1.00 mol C 6 H 12 ((l) at 298 K?

Solution

D

n gas = n final Since

D

H = _ n initial = _ 3

D

U +

D

n gas RT ( _ 3920 kJ) =

D

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

D

U = _ 3910 kJ or 2012 General Chemistry I _ 3.91 x 10 3 kJ

53

### 7.15 Standard Reaction Enthalpies

- Reaction enthalpies depend on the physical states of the reactants and products 88 kJ = enthalpy of vaporization of water (44 kJ·mol -1 ) × 2 It is therefore useful if enthalpies (

D

H o ) can be referred to a standard state.

2012 General Chemistry I

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Standard state: refers to a pure form at exactly 1 bar or for a solute in a liquid solution: concentration of 1 mol·L -1 .

Standard reaction enthalpy,

D

H

o is the reaction enthalpy when reactants in their standard states change into products in their standard states.

-

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

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### 7.16 Combining Reaction Enthalpies: Hess’s Law

- Enthalpy is a state function;

D

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.

D

H

o = -393.5 kJ 2012 General Chemistry I

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EXAMPLE 7.9

Consider the synthesis of propane, C 3 H 8 , 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 to measure. Calculate the standard enthalpy of this reaction from the following experimental data: 2012 General Chemistry I

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Self-Test 7.13B

Methanol is a clean-burning liquid fuel proposed as a replacement for gasoline. Suppose it could be produced by the controlled reaction of oxygen with methane. Find the standard reaction enthalpy for the formation of 1 mol CH 3 OH(l) from methane and oxygen, given the following information.

(1) CH (2) 2H (3) 2H 2 2 4 (g) + H (g) + CO(g) (g) + O 2 2 O(g) (g) CH CO(g) + 3H 2 (g) 3 OH(l)

D

H o = 2H 2 O(g)

D

H o = _

D

H o = +206.10 kJ 128.33 kJ _ 483.64 kJ Solution The required equation is 2CH 4 (g) + O 2 (g) 2CH 3 OH(l) Multiply equation (1) and (2) by 2, and add both to equation (3): this gives the required equation.

D

H o = 2

D

H o (1) + 2

D

H o (2) +

D

H o (3) = _ 328.10 kJ 2012 General Chemistry I

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### 7.17 The Heat Output of Reactions

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

the standard enthalpy of combustion,

D

H

c o 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 2012 General Chemistry I

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Standard Enthalpies of Combustion at 25 o C (Table 7.4)* 2012 General Chemistry I

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Self-Test 7.14A

The thermochemical equation for the combustion of propane is: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l)

D

H o = _ 2220. kJ What mass of propane must be burned to supply 350. kJ as heat? Would it be easier to pack propane rather than butane?

Solution The above equation gives the heat supplied by burning 1 mol of propane, hence 350. kJ of heat can be obtained from 350 = 0.1577 mol of propane 2220 Since the molar mass of propane is 44.07 g mol -1 , the mass 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, because the liquid is slightly less dense than butane.

2012 General Chemistry I

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### 7.18 Standard Enthalpies of Formation

Standard enthalpy of formation,

D

H

f o 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.

2012 General Chemistry I

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Standard Enthalpies of Formation at 25 o C (kJ mol -1 ) (Table 7.5) 2012 General Chemistry I

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

2012 General Chemistry I

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Self-Test 7.15A

Calculate the standard enthalpy of combustion of glucose from the standard enthalpies of formation in Table 7.5 and Appendix 2A.

Solution The combustion equation is C 6 H 12 O 6 (s) + 6O 2 (g)

6CO 2 (g) + 6H 2 O(l) n

D

H f o (products) = {(6 mol) x ( 393.51 kJ mol -1 )} _ + {(6 mol) x ( 285.83 kJ mol -1 )} = =

n

D

H f o (reactants) =

D

H o (combustion) = _ 4076.04 kJ _ 1268 kJ + 0 kJ = _ 1268 kJ

n

D

H f o (products) _

n

D

H f o (reactants) _ 2808 kJ mol -1 2012 General Chemistry I

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EXAMPLE 7.12

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

= -2323.85 kJ = -2220 kJ 2012 General Chemistry I

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Self-Test 7.16A

Calculate the standard enthalpy of formation of ethyne, the fuel used in oxyacetylene welding torches, from the information in Tables 7.4 and 7.5.

Solution The thermochemical equation for the combustion of ethyne is: C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O(l)

D

H o = _ 1300 kJ mol -1 _

D

H o r =

n

D

H f o (products) 1300 kJ mol -1 _

n

D

H f o (reactants) = [(2 mol) x ( 393.51 kJ mol _ [(

D

H f o _ (ethyne)] -1 _ -1 )

D

H f o (ethyne) = + 227 kJ mol -1 2012 General Chemistry I

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### 7.19 The Born-Haber Cycle

Lattice enthalpy of the solid,

D

H

L Lattice enthalpies at 25 o C (kJ·mol -1 )(Table 7.) 2012 General Chemistry I

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Born-Haber cycle: a closed path of steps, one of which is the formation of a solid lattice from the gaseous ions 2012 General Chemistry I

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EXAMPLE 7.13

Lattice enthalpy of KCl

D

H

f (K, atoms)

D

H

f (Cl, atoms) I(K) -E ea (Cl) -

D

H

f (KCl) K + Cl (s) K + (g) + Cl (g) 2012 General Chemistry I

70

Self-Test 7.17A

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

Solution The Born-Haber cycle is: (not to scale)

D

H L = 795.8 + 178.2

+ 244 + 1735 _ 698 = 2255 kJ H kJ mol -1 Ca +1735 2+ (g) + 2e Ca 2+ (g) (g) + 2Cl(g) + _ 698 2Cl (g) Ca(g) + 2Cl(g) +178.2 + 244 Ca(s) + Cl 2 (g) -

D

H f o = +795.8

Ca 2+ (Cl ) 2 (s) -

D

H L 2012 General Chemistry I

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### 7.20 Bond Enthalpies

Bond enthalpy,

D

H

B 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.

2012 General Chemistry I

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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, 2012 General Chemistry I

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EXAMPLE 7.14

Estimate the enthalpy of the reaction between gaseous iodoethane and water vapor: - Breaking the bonds (reactants)

D

H

B o (C-I) +

D

H

B o (O-H) = - Forming the bonds (products)

D

H

B o (C-O) +

D

H

B o (H-I) = - The overall enthalpy change: 2012 General Chemistry I

74

Self-Test 17.18B

Use bond enthalpies to estimate the standard enthalpy of reaction in which 1.00 mol of gaseous CH 4 gaseous F 2 to form gaseous CH 2 F 2 and HF.

reacts with Solution The relevant equation is CH 4 (g) + 2F 2 (gas) CH 2 F 2 (g) + 2HF(g) In this reaction, 2 mol of F-F bonds and 2 mol of C-H bonds are broken (reactants), whereas 2 mol of C-F bonds and 2 mol of H-F bonds are formed.

D

H o =

mean

D

H B (reactants) _

mean

D

H B (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 2012 General Chemistry I

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### 7.21 The Variation of Reaction Enthalpy with Temperature

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

2012 General Chemistry I

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Kirchhoff’s law 2012 General Chemistry I

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EXAMPLE 7.15

The standard enthalpy of reaction of N 2 (g) + 3H 2 (g) → 2NH 3 (g) is 92.22 kJ·mol -1 at 298 K. The industrial synthesis takes place at 450 o C.

What is the standard reaction enthalpy at the latter temperature?

2012 General Chemistry I

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Self-Test 17.19B

The standard enthalpy of formation of ammonium nitrate is _ 365.56 kJ mol -1 at 298 K. Estimate its value at 250.

o C.

Solution The equation of formation is N 2 (g) + 3/2O 2 (g) + 2H 2 (g) NH 4 NO 3 (s) 250.

o C = 523 K Use Kirchhoff's law: Values of C p (J K -1

D

H o (T 2 ) =

D

H o (T 1 ) + (T 2 _ T 1 )

D

C p mol -1 ): NH 4 NO 3 (s) 84.1 (product); N 2 (g) 29.12; 3/2O 2 (g) 44.04; 2H 2 (g) 57.64 (reactants)

D

H o (T 2 _ -1 ) + (225 K)( 46.7 J K -1 mol -1 ) = _ 376 kJ mol -1 2012 General Chemistry I

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