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Chapter 10
Energy Changes in Chemical Reactions
Thermochemistry
10
10.1 Energy and Energy Changes
10.2 Introduction to Thermodynamics
States and State Functions
The First Law of Thermodynamics
Work and Heat
10.3 Enthalpy
Reactions Carried Out at Constant Volume or at Constant Pressure
Enthalpy and Enthalpy Changes
Thermochemical Equations
10.4 Calorimetry
Specific Heat and Heat Capacity
Constant-Pressure Calorimetry
Constant-Volume Calorimetry
10.5 Hess’s Law
10.6 Standard Enthalpies of Formation
10.7 Bond Enthalpy and the Stability of Covalent Molecules
10.8 Lattice Energy and the Stability of Ionic Compounds
The Born-Haber Cycle
Comparison of Ionic and Covalent Compounds
Energy and Energy Changes
The system is a part of the universe that is of specific interest.
The surroundings constitute the rest of the universe outside the system.
The system is usually defined as the substances involved in chemical and physical changes.
System
Surroundings
Universe = System + Surroundings
10.1
Energy and Energy Changes
Thermochemistry is the study of heat (the transfer of thermal energy) in chemical reactions.
Heat is the transfer of thermal energy.
Heat is either absorbed or released during a process.
Surroundings
heat
Energy and Energy Changes
An exothermic process occurs when heat is transferred from the system to the surroundings.
“Feels hot!”
Surroundings
Universe = System + Surroundings
heat
System
2H2(g) + O2(g)
2H2O(l) + energy
Energy and Energy Changes
An endothermic process occurs when heat is transferred from the surroundings to the system.
“Feels cold”
energy + 2HgO(s)
2Hg(l) + O2(g)
Surroundings
Universe = System + Surroundings
heat
System
Introduction to Thermodynamics
Thermodynamics is the study of the interconversion of heat and other kinds of energy.
In thermodynamics, there are three types
of systems:
An open system can exchange mass and
energy with the surroundings.
A closed system allows the transfer
of energy but not mass.
An isolated system does not exchange
either mass or energy with its
surroundings.
10.2
open
mass & energy
Exchange:
closed
energy
isolated
nothing
Internal energy
Internal energy, also called thermodynamic energy
the energy associated with the random, disordered motion of molecules or particles.
It is separated in scale from the macroscopic ordered energy.
Its absolute value is unknown.
Its symbol and unit are U and joule respectively.
Internal Energy
Internal energy: 2 components
Kinetic energy - molecular motion
Potential energy - attractive/repulsive interactions
Internal Energy
We cannot calculate the exact value of internal energy with any certainty
We can calculate the changes in energy of the system experimentally
U = U(products) - U(reactants)
Introduction to Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
The magnitude of change depends only on the initial and final states of the system.
Energy
Pressure
Volume
Temperature
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
energy
, pressure, volume, temperature
DU = Ufinal - Uinitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
The First Law of Thermodynamics
The first law of thermodynamics states that energy can be converted from one form to another, but cannot be created or destroyed.
ΔU is the change in the internal energy.
“sys” and “surr” denote system and surroundings, respectively.
ΔU = Uf – Ui; the difference in the energies of the initial and final states.
ΔUsys + ΔUsurr = 0
ΔUsys = –ΔUsurr
Work and Heat
The overall change in the system’s internal energy is given by:
q is heat
q is positive for an endothermic process (heat absorbed by the system)
q is negative for an exothermic process (heat released by the system)
w is work
w is positive for work done on the system
w is negative for work done by the system
ΔU = q + w
Work and Heat
ΔU = q + w
The Sign Conventions* for q, w and DU
q
w
+
=
DU
+
+
-
-
-
-
+
+
+
-
depends on sizes of q and w
depends on sizes of q and w
* For q: + means system gains heat; - means system loses heat.
* For w: + means work done on system; - means work done by system.
Worked Example 10.1
Strategy Combine the two contributions to internal energy using ΔU = q + w and the sign conventions for q and w.
Calculate the overall change in internal energy, ΔU, (in joules) for a system that absorbs 188 J of heat and does 141 J of work on its surroundings.
Solution The system absorbs heat, so q is positive. The system does work on the surroundings, so w is negative.
ΔU = q + w = 188 J + (-141 J) = 47 J
Think About It Consult Table 10.1 to make sure you have used the proper sign conventions for q and w.
Enthalpy
Sodium azide detonates to give a large quantity of nitrogen gas.
Under constant volume conditions, pressure increases:
2NaN3(s)
2Na(s) + 3N2(g)
10.3
Enthalpy
Sodium azide detonates to give a large quantity of nitrogen gas.
Under constant volume conditions, pressure increases:
2NaN3(s)
2Na(s) + 3N2(g)
Enthalpy
Pressure-volume, or PV work, is done when there is a volume change under constant pressure.
w = PΔV
P is the external opposing pressure.
ΔV is the change in the volume of the container.
Worked Example 10.2
Strategy Determine change in volume (ΔV), identify the external pressure (P), and use w = PΔV to calculate w. The result will be in L atm; use the equality 1 L atm = 101.3 J to convert to joules.
Determine the work done (in joules) when a sample of gas extends from 552 mL to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm, (b) against a constant pressure of 1.00 atm, and (c) against a vacuum (1 L atm = 101.3 J).
Solution ΔV = (891 – 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm, (c) P = 0 atm.
(a) w = -(1.25 atm)(339 mL)
(b) w = -(1.00 atm)(339 mL)
1 L
1000 mL
101.3 J
1 L atm
= -42.9 J
1 L
1000 mL
101.3 J
1 L atm
= -34.3 J
Worked Example 10.2 (cont.)
Solution
(c) w = -(0 atm)(339 mL)
1 L
1000 mL
101.3 J
1 L atm
= 0 J
Think About It Remember that the negative sign in the answers to part (a) and (b) indicate that the system does work on the surroundings. When an expansion happens against a vacuum, no work is done. This example illustrates that work is not a state function. For an equivalent change in volume, the work varies depending on external pressure against which the expansion must occur.
Enthalpy
Pressure-volume, or PV work, is done when there is a volume change under constant pressure.
w = PΔV
When a change occurs at constant volume, ΔV = 0 and no work is done.
ΔU = q + w
qV = ΔU
ΔU = q PΔV
substitute
Enthalpy
Under conditions of constant pressure:
ΔU = q + w
qP = ΔU + PΔV
ΔU = q PΔV
Enthalpy
The thermodynamic function of a system called enthalpy (H) is defined by the equation:
H = U + PV
A note about SI units:
Pressure: pascal; 1Pa = 1 kg/(m . s2)
Volume: cubic meters; m3
PV: 1kg/(m . s2) x m3 = 1(kg . m2)/s2 = 1 J
Enthalpy: joules
U, P, V, and H are all state functions.
Enthalpy and Enthalpy Changes
For any process, the change in enthalpy is:
ΔH = ΔU + Δ(PV)
(1)
ΔH = ΔU + PΔV
If pressure is constant:
(2)
ΔU = ΔH + PΔV
Rearrange to solve for ΔU:
(3)
qp = ΔU + PΔV
Remember, qp:
(4)
qp = (ΔH PΔV) + PΔV
Substitute equation (3) into equation (4) and solve:
(5)
qp = ΔH
for a constant-pressure process
Enthalpy and Enthalpy Changes
The enthalpy of reaction (ΔHr) is the difference between the enthalpies of the products and the enthalpies of the reactants:
Assumes reactions in the lab occur at constant pressure
ΔH > 0 (positive) endothermic process
ΔH < 0 (negative) exothermic process
ΔHr = H(products) – H(reactants)
Thermochemical Equations
Concepts to consider:
Is this a constant pressure process
What is the system
What are the surroundings
ΔH > 0 endothermic
H2O(s)
H2O(l) ΔH = +6.01 kJ/mol
Thermochemical Equations
Concepts to consider:
Is this a constant pressure process
What is the system
What are the surroundings
ΔH < 0 exothermic
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l) ΔH = 890.4 kJ/mol
Thermochemical Equations
Enthalpy is an extensive property.
Extensive properties are dependent on the amount of matter involved.
H2O(l) → H2O(g) ΔH = +44 kJ/mol
2H2O(l) → 2H2O(g) ΔH = +88 kJ/mol
Units refer to
mole of reaction as written
Double the amount of matter
Double the enthalpy
Thermochemical Equations
The following guidelines are useful when considering thermochemical equations:
1) Always specify the physical states of reactants and products because they help determine the actual enthapy changes.
CH4(g) + 2O2(g)
ΔH = 802.4 kJ/mol
CO2(g) + 2H2O(g)
CH4(g) + 2O2(g)
ΔH = +890.4 kJ/mol
CO2(g) + 2H2O(l)
different states
different enthalpies
Thermochemical Equations
The following guidelines are useful when considering thermochemical equations:
2) When multiplying an equation by a factor (n), multiply the ΔH value by same factor.
CH4(g) + 2O2(g)
ΔH = 802.4 kJ/mol
CO2(g) + 2H2O(g)
2CH4(g) + 4O2(g)
ΔH = 1604.8 kJ/mol
2CO2(g) + 4H2O(g)
3) Reversing an equation changes the sign but not the magnitude of ΔH.
CH4(g) + 2O2(g)
ΔH = 802.4 kJ/mol
CO2(g) + 2H2O(g)
CO2(g) + 2H2O(g)
ΔH = +802.4 kJ/mol
CH4(g) + 2O2(g)
Worked Example 10.3
Strategy The thermochemical equation shows that for every mole of C6H12O6 produced, 2803 kJ is absorbed. We need to find out how much energy is absorbed for the production of 75.0 g of C6H12O6. We must first find out how many moles there are in 75.0 g of C6H12O6.
The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is
75.0 g C6H12O6 ×
We will multiply the thermochemical equation, including the enthalpy change, by 0.416, in order to write the equation in terms of the appropriate amount of C6H12O6.
Given the thermochemical equation for photosynthesis,
6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g) ΔH = +2803 kJ/mol
calculate the solar energy required to produce 75.0 g of C6H12O6.
1 mol C6H12O6
180.2 g C6H12O6
= 0.416 mol C6H12O6
Worked Example 10.3 (cont.)
Solution
(0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)]
and (0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives
2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g) ΔH = +1.17×103 kJ
Therefore, 1.17×103 kJ of energy in the form of sunlight is consumed in the production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are canceled when we multiply the thermochemical equation by the number of moles of C6H12O6.
Think About It The specified amount of C6H12O6 is less than half a mole. Therefore, we should expect the associated enthalpy change to be less than half that specified in the thermochemical equation for the production of 1 mole of C6H12O6.
Calorimetry
Calorimetry is the measurement of heat changes.
Heat changes are measured in a device called a calorimeter.
The specific heat (s) of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1°C.
10.4
Specific Heat and Heat Capacity
The heat capacity (C) is the amount of heat required to raise the temperature of an object by 1°C.
The “object” may be a given quantity of a particular substance.
Specific heat capacity has units of J/(g °C)
Heat capacity has units of J/°C
Specific heat capacity of water
heat capacity of
1 kg water
Specific Heat and Heat Capacity
The heat associated with a temperature change may be calculated:
m is the mass.
s is the specific heat.
ΔT is the change in temperature (ΔT = Tfinal – Tinitial).
C is the heat capacity.
q = msΔT
q = CΔT
Worked Example 10.4
Strategy Use q = msΔT to calculate q. s = 4.184 J/g °C, m = 255 g, ΔT = 90.5°C – 25.2°C = 65.3°C.
Calculate the amount of heat (in kJ) required to heat 255 g of water from 25.2°C to 90.5°C.
Solution
q =
Think About It Look carefully at the cancellation of units and make sure that the number of kilojoules is smaller than the number of joules. It is a common error to multiply by 1000 instead of dividing in conversions of this kind.
4.184 J
g °C
× 255 g × 65.3°C = 6.97×104 J or 69.7 kJ
Calorimetry
Calculate the amount of heat required to heat 1.01 kg of water from 0.05°C to 35.81°C.
Solution:
Step 1: Use the equation q = msΔT to calculate q.
Calorimetry
A coffee-cup calorimeter may be used to measure the heat exchange for a variety of reactions at constant pressure:
Heat of neutralization:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Heat of ionization:
H2O(l) → H+(aq) + OH (aq)
Heat of fusion:
H2O(s) → H2O(l)
Heat of vaporization:
H2O(l) → H2O(g)
Calorimetry
Concepts to consider for coffee-cup calorimetry:
qP = ΔH
System: reactants and products (the reaction)
Surroundings: water in the calorimeter
For an exothermic reaction:
the system loses heat
the surroundings gain (absorb) heat
qsys = msΔT
The minus sign is used to keep sign conventions consistent.
qsurr = msΔT
qsys = qsurr
Worked Example 10.5
Strategy Water constitutes the surroundings; the pellet is the system. Use qsurr = msΔT to determine the heat absorbed by the water; then use q = CΔT to determine the heat capacity of the metal pellet.
mwater = 125 g, swater = 4.184 J/g °C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C. The heat absorbed by the water must be released by the pellet: qwater = -qpellet, mpellet = 100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.
A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125 g of water originally at 25.1°C. The final temperature of both pellet and the water is 31.3°C. Calculate the heat capacity C (in J/°C) of the pellet.
Worked Example 10.5 (cont.)
Solution
qwater =
Thus,
qpellet = -3242.6 J
From q = CΔT we have
-3242.6 J = Cpellet × (-57.1°C)
Thus,
Cpellet = 57 J/°C
4.184 J
g °C
× 125 g × 6.2°C = 3242.6 J
Think About It The units cancel properly to give appropriate units for heat capacity. Moreover, ΔTpellet is a negative number because the temperature of the pellet decreases.
Constant-Volume Calorimetry
Constant volume calorimetry is carried out in a device known as a constant-volume bomb.
qcal = qrxn
A constant-volume calorimeter is an isolated system.
Bomb calorimeters are typically used to determine heats of combustion.
Constant-Volume Calorimetry
qrxn = CcalΔT
qcal = CcalΔT
qrxn = qcal
To calculate qcal, the heat capacity of the calorimeter must be known.
Worked Example 10.6
Strategy Use qrxn = -CcalΔT to calculate the heat released by the combustion of the cookie. Divide the heat released by the mass of the cookie to determine its energy content per gram Ccal = 39.97 kJ/°C and ΔT = 3.90°C.
A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a bomb calorimeter to determine its energy content. The heat capacity of the calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of the cookie.
Solution
qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ
Because energy content is a positive quantity, we write
energy content per gram =
1.559×102 kJ
7.25 g
= 21.5 kJ/g
Think About It According to the label on the cookie package, a service size is four cookies, or 29 g, and each serving contains 150 Cal. Convert the energy per gram to Calories per serving to verify the result.
21.5 kJ
g
×
1 Cal
4.184 kJ
×
29 g
serving
= 1.5×102 Cal/serving
Hess’s Law
Hess’s law states that the change in enthalpy for a stepwise process is the sum of the enthalpy changes for each of the steps.
CH4(g) + 2O2(g)
ΔH = 890.4 kJ/mol
CO2(g) + 2H2O(l)
CH4(g) + 2O2(g)
ΔH = 802.4 kJ/mol
CO2(g) + 2H2O(g)
2H2O(l)
ΔH = +88.0 kJ/mol
2H2O(g)
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = 890.4 kJ
ΔH = +88.0 kJ
CO2(g) + 2H2O(g)
ΔH = 802.4 kJ
10.5
Hess’s Law
When applying Hess’s Law:
1) Manipulate thermochemical equations in a manner that gives the overall desired equation.
2) Remember the rules for manipulating thermochemical equations:
3) Add the ΔH for each step after proper manipulation.
4) Process is useful for calculating enthalpies that cannot be found directly.
Always specify the physical states of reactants and products because they help determine the actual enthalpy changes.
When multiplying an equation by a factor (n), multiply the ΔH value by same factor.
Reversing an equation changes the sign but not the magnitude of ΔH.
Worked Example 10.7
Strategy Arrange the given thermochemical equations so that they sum to the desired equation. Make the corresponding changes to the enthalpy changes, and add them to get the desired enthalpy change.
Given the following thermochemical equations,
determine the enthalpy change for the reaction
NO(g) + O(g) → NO2(g)
NO(g) + O3(g) → NO2(g) + O2(g)
ΔH = –198.9 kJ/mol
O3(g) → O2(g)
ΔH = –142.3 kJ/mol
O2(g) → 2O(g)
ΔH = +495 kJ/mol
3
2
Worked Example 10.7 (cont.)
Solution The first equation has NO as a reactant with the correct coefficients, so we will use it as is.
The second equation must be reversed so that the O3 introduced by the first equation will cancel (O3 is not part of the overall chemical equation). We also must change the sign on the corresponding ΔH value.
These two steps sum to give:
NO(g) + O3(g) → NO2(g) + O2(g)
ΔH = –198.9 kJ/mol
O2(g) → O3(g)
ΔH = +142.3 kJ/mol
3
2
NO(g) + O3(g) → NO2(g) + O2(g)
O2(g) → O3(g)
3
2
ΔH = –198.9 kJ/mol
ΔH = +142.3 kJ/mol
NO(g) + O2(g) → NO2(g)
ΔH = –56.6 kJ/mol
1
2
+
O2(g)
1
2
Worked Example 10.7 (cont.)
Solution We then replace O2 on the left with O by incorporating the last equation. To do so, we divide the third equation by 2 and reverse its direction. As a result, we must also divide ΔH value by 2 and change its sign.
Finally, we sum all the steps and add their enthalpy changes.
NO(g) + O3(g) → NO2(g) + O2(g)
ΔH = –198.9 kJ/mol
O2(g) → O3(g)
ΔH = +142.3 kJ/mol
3
2
+
O(g) → O2(g)
ΔH = –247.5 kJ/mol
1
2
O(g) → O2(g)
ΔH = –247.5 kJ/mol
1
2
NO(g) + O(g) → NO2(g)
ΔH = –304 kJ/mol
Think About It Double-check the cancellation of identical items–especially where fractions are involved.
1
2
Standard Enthalpies of Formation
The standard enthalpy of formation (Δ Hf) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states.
Elements in standard states
C(graphite) + O2(g)
CO2(g)
ΔH f° = 393.5 kJ/mol
1 mole of product
10.6
Standard Enthalpies of Formation
The standard enthalpy of formation (ΔfH ) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states.
The superscripted degree sign denotes standard conditions.
1 atm pressure for gases
1 M concentration for solutions
“f” stands for formation.
ΔfH for an element in its most stable form is zero.
ΔfH for many substances are tabulated in Appendix 2 of the textbook.
Standard Enthalpies of Formation
The standard enthalpy of reaction (ΔrxnH ) is defined as the enthalpy of a reaction carried out under standard conditions.
aA + bB → cC + dD
n and m are the stoichiometric coefficients for the reactants and products.
Δ rxnH = Σn ΔfH (products) – Σm ΔfH (reactants)
ΔrxnH = [cΔfH (C) + dΔfH (D) ] – [aΔfH (A) + b ΔfH (B)]
Worked Example 10.8
Strategy Use Δ rxnH = Σn ΔfH (products) – Σm ΔfH (reactants) and ΔfH values from Appendix 2 to calculate ΔrxnH . The ΔfH values for Ag+(aq), Cl-(aq), and AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively.
Using data from Appendix 2, calculate ΔrxnH for Ag+(aq) + Cl-(aq) → AgCl(s).
Solution
ΔrxnH = ΔfH (AgCl) – [ΔfH (Ag+) + ΔfH (Cl-)]
= –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2 kJ/mol)]
= –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/mol
Think About It Watch out for misplaced or missing minus signs. This is an easy place to lose track of them.
Worked Example 10.9
Strategy Arrange the equations that are provided so that they will sum to the desired equation. This may require reversing or multiplying one or more of the equations. For any such change, the corresponding change must also be made to the Δ rxnH value. The desired equation, corresponding to the standard enthalpy of formation of acetylene, is
2C(graphite) + H2(g) → C2H2(g)
Given the following information, calculate the standard enthalpy of formation of acetylene (C2H2) from its constituent elements:
C(graphite) + O2(g) → CO2(g)
Δ rxnH = –393.5 kJ/mol
H2(g) + O2(g) → H2O(l)
Δ rxnH = –285.5 kJ/mol
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
Δ rxnH = –2598.8 kJ/mol
1
2
(1)
(2)
(3)
Worked Example 10.9 (cont.)
Solution We multiply Equation (1) and its ΔH °rxn value by 2:
We include Equation (2) and its ΔH °rxn value as is:
We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):
Summing the resulting equations and the corresponding ΔrH values:
2C(graphite) + 2O2(g) → 2CO2(g)
ΔrH = –787.0 kJ/mol
H2(g) + O2(g) → H2O(l)
ΔrH = –285.5 kJ/mol
2CO2(g) + H2O(l) → C2H2(g) + O2(g)
ΔrH = +1299.4 kJ/mol
1
2
5
2
2C(graphite) + 2O2(g) → 2CO2(g)
ΔrH = –787.0 kJ/mol
H2(g) + O2(g) → H2O(l)
ΔrH = –285.5 kJ/mol
2CO2(g) + H2O(l) → C2H2(g) + O2(g)
ΔrH = +1299.4 kJ/mol
1
2
5
2
2C(graphite) + H2(g) → C2H2(g)
ΔrH = +226.6 kJ/mol
Think About It Remember that a Δ rxnH is only a ΔfH when there is just one product, just one mole produced, and all the reactants are elements in their standard states.
Bond Enthalpy and the Stability of Covalent
Molecules
The bond enthalpy is the enthalpy change associated with breaking a bond in 1 mole of gaseous molecule.
The enthalpy for a gas phase reaction is given by:
H2(g) → H(g) + H(g) ΔH° = 436.4 kJ/mol
ΔH = total energy input – total energy released
bonds broken
bonds formed
10.7
ΔH = ΣBE(reactants) – ΣBE(products)
Bond Enthalpy and the Stability of Covalent Molecules
Bond enthalpy change in an exothermic reaction.:
Bond Enthalpy and the Stability of Covalent Molecules
Bond enthalpy change in an endothermic reaction:
Worked Example 10.10
Strategy Draw Lewis structures to determine what bonds are to be broken and what bonds are to be formed.
Bonds to break: 4 C–H and 2 O=O
Bonds to form: 2 C=O and 4 H–O
Bond enthalpies from Table 10.4: 414 kJ/mol (C–H), 498.7 kJ/mol (O=O), 799 kJ/mol (C=O in CO2), and 460 kJ/mol (H–O).
Use bond enthalpies from Table 10.4 to estimate the enthalpy of reaction for the combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Solution
[4(414 kJ/mol) + 2(498.7 kJ/mol)] – [2(799 kJ/mol) + 4(460 kJ/mol)] = –785 kJ/mol
Think About It Use equation ΔrH = Σn ΔfH (products) – Σm ΔfH (reactants) and data from Appendix 2 to calculate this enthalpy of reaction again; then compare your results using the two approaches. The difference in this case is due to two things: most tabulated bond enthalpies are averages and, by convention, we show the product of combustion as liquid water–but average bond enthalpies apply to species in the gas phase, where there is little or no influence exerted by neighboring molecules.
Lattice Energy and the Stability of Ionic Compounds
A Born-Haber cycle is a cycle that relates the lattice energy of an ionic compound to quantities that can be measured.
Na(s) + Cl2(g) → Na+(g) + Cl-(g)
10.8
1
2
Lattice Energy and the Stability of Ionic Compounds
A Born-Haber cycle is a cycle that relates the lattice energy of an ionic compound to quantities that can be measured.
Na(s) + Cl2(g) → Na+(g) + Cl-(g)
1
2
Lattice Energy and the Stability of Ionic Compounds
Worked Example 10.11
Strategy Using the Born-Haber Cycle Figure as a guide, combine pertinent thermodynamic data and Hess’s law to calculate the lattice energy.
From Figure 4.8, IE1(Cs) = 376 kJ/mol. From Figure 4.10, EA1(Cl) = 349.0 kJ/mol. From Appendix 2, ΔH °f [Cs(g)] = 76.50 kJ/mol, ΔH °f [Cl(g)] = 121.7 kJ/mol, ΔH °f [CsCl(s)] = – 422.8 kJ/mol. Because we are interested in magnitudes only, we can use the absolute values of the thermodynamic data. And, because only the standard heat of formation of CsCl(s) is a negative number, it is only one for which the sign changes.
Using data from Figure 4.8 and 4.10 and Appendix 2, calculate the lattice energy of cesium chloride (CsCl).
Solution
{Δf H [Cs(g)] + Δf H [Cl(g)] + IE1(Cs) + Δf H [CsCl(s)]} – EA1(Cl) = lattice energy
= (76.50 kJ/mol + 121.7 kJ/mol + 376 kJ/mol + 422.8 kJ/mol) – 349.0 kJ/mol
= 648 kJ/mol
Think About It Compare this value to that for NaCl in Figure 10.14 (787 kJ/mol). Both compounds contain the same anion (Cl-) and both have cations with the same charge (+1), so the relative sizes of the cations will determine the relative strengths of their lattic energies. Because Cs+ is larger than Na+, the lattice energy of CsCl is smaller than the lattice energy of NaCl.
Chapter Summary: Key Points
10
Energy and Energy Changes
Forms of Energy
Energy Changes in Chemical Reactions
Units of Energy
Introduction to Thermodynamics
States and State Functions
The First Law of Thermodynamics
Work and Heat
Enthalpy
Reactions Carried Out at Constant Volume or at Constant Pressure
Enthalpy and Enthalpy Changes
Thermochemical Equations
Calorimetry Specific Heat and Heat Capacity
Constant-Pressure Calorimetry
Constant-Volume Calorimetry
Hess’s Law
Standard Enthalpies of Formation
System
Surroundings
Universe = System + Surroundings
Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 18
Entropy, Free Energy, and Equilibrium
Entropy, Free Energy, and Equilibrium
18
18.1 Spontaneous Processes
18.2 Entropy
A Qualitative Description of Entropy
A Quantitative Description of Entropy
18.3 Entropy Changes in a System
Calculating ΔSsys
Standard S
Qualitatively Predicting ΔsysS
18.4 Entropy Changes in the Universe
Calculating ΔSsurr
The Second Law of Thermodynamics
The Third Law of Thermodynamics
18.5 Predicting Spontaneity
Gibbs Free-Energy Change, ΔG
Standard Free-Energy Changes, ΔG
Using ΔG and ΔG to Solve Problems
18.6 Free Energy and Chemical Equilibrium
Relationship Between ΔG and ΔG
Relationship Between ΔG and K
18.7 Thermodynamics in Living Systems
Spontaneous Processes
A process that does occur under a specific set of conditions is called a spontaneous process.
A process that does not occur under a specific set of conditions is called nonspontaneous.
18.1
Spontaneous Processes
A process that results in a decrease in the energy of a system often is spontaneous:
The sign of ΔH alone is insufficient to predict spontaneity in every circumstance:
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -890.4 kJ/mol
H2O(l) H2O(s) T > 0°C; ΔH = -6.01 kJ/mol
Entropy
18.2
To predict spontaneity, both the enthalpy and entropy must be known.
Entropy (S) of a system is a measure of how spread out or how dispersed the system’s energy is.
Entropy
Spontaneity is favored by an increase in entropy.
k is the Boltzmann constant (1.38 x 10–23 J/K)
W is the number of different arrangements
S = k ln W
The number of arrangements possible is given by:
X is the number of cells in a volume
N is the number of molecules
W = X N
Entropy
Entropy
There are three possible states for this system:
1) One molecule on each side (eight possible arrangements)
2) Both molecules on the left (four possible arrangements)
3) Both molecules on the right (four possible arrangements)
The most probable state has the largest number of arrangements.
Entropy Changes in a System
The change in entropy for a system is the difference in entropy of the final state and the entropy of the initial state.
Alternatively:
18.3
ΔSsys = Sfinal – Sinitial
Worked Example 18.1
Strategy This is the isothermal expansion of an ideal gas. Because the molecules spread out to occupy a greater volume, we expect there to be an increase in the entropy of the system. Use ΔSsys = Sfinal – Sinitial to solve for ΔSsys. R = 8.314 J/K mol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L.
Determine the change in entropy for 1.0 mole of an ideal gas originally confined to one-half of a 5.0-L container when the gas is allowed to expand to fill the entire container at constant temperature.
Solution
ΔSsys = nR ln
Vfinal
Vinitial
= 1.0 mol ×
8.314 J
K mol
×ln
5.0 L
2.5 L
= 5.8 J/K
Think About It Remember that for a process to be spontaneous, something must favor spontaneity. If the process is spontaneous but not exothermic (in this case, there is no enthalpy change), then we should expect ΔSsys to be positive.
Worked Example 18.1 (cont.)
Solution These equilibrium concentrations are then substituted into the equilibrium expression to give
Because we expect x to be very small (even smaller than 1.34×10-3 M–see above), because the ionization of CH3COOH is suppressed by the presence of CH3COO-, we assume
(0.10 – x) M ≈ 0.10 M and (0.050 + x) M ≈ 0.050 M
Therefore, the equilibrium expression simplifies to
and x = 3.6×10-5 M. According to the equilibrium table, [H+] = x, so pH = –log(3.6×10-5) = 4.44.
1.8×10-5 =
(x)(0.050 + x)
0.010 – x
1.8×10-5 =
(x)(0.050)
0.010
Entropy Changes in a System
The standard entropy (S ) is the absolute entropy of a substance at 1 atm.
Temperature is not part of the standard state definition and must be specified.
Entropy Changes in a System
There are several important trends in entropy:
S liquid > S solid
S gas > S liquid
S increases with molar mass
S increases with molecular complexity
S increases with the mobility of a phase (for an element with two or more allotropes)
Entropy Changes in a System
In addition to translational motion, molecules exhibit vibrations and rotations.
For a chemical reaction
aA + bB → cC + dD
Alternatively,
Entropy Changes in a System
ΔrxnS = [cS (C) + dS (D)] – [aS (A) + bS (B)]
ΔrxnS = ΣnS (products) – ΣmS (reactants)
Worked Example 18.2
Strategy Look up standard enthalpy values and calculate ΔrxnS . Just as we did when we calculated standard enthalpies of reaction, we consider stoichiometric coefficients to be dimensionless–giving ΔrxnS units of J/K mol.
From Appendix 2, S [CaCO3(s)] = 92.9 J/K mol, S [CaO(s)] = 39.8 J/K mol, S [CO2(g)] = 213.6 J/K mol, S [N2(g)] = 191.5 J/K mol, S [H2(g)] = 131.0 J/K mol, S [NH3(g)] = 193.0 J/K mol, S [Cl2(g)] = 223.0 J/K mol, and S [HCl(g)] = 187.0 J/K mol.
From the standard enthalpy values in Appendix 2, calculate the standard entropy changes for the following reactions at 25°C.
(a) CaCO3(s) → CaO(s) + CO2(g) (b) N2(g) + 3H2(g) → 2NH3(g)
(c) H2(g) + Cl2(g) → 2HCl(g)
Worked Example 18.2 (cont.)
Solution
(a) S rxn = [S (CaO) + S (CO2)] – [S (CaCO3)]
= [(39.8 J/K mol) + (213.6 J/K mol)] – (92.9 J/K mol)
= 160.5 J/K mol
(b) S rxn = [2S (NH3)] – [S (N2) + 3S (H2)]
=(2)(193.0 J/K mol) – [(191.5 J/K mol) + (3)(131.0 J/K mol)]
= –198.5 J/K mol
(c) S rxn = [2S (HCl)] – [S (H2) + S°(Cl2)]
= (2)(187.0 J/K mol) – [(131.0 J/K mol) + (223.0 J/K mol)]
= 20.0 J/K mol
Think About It Remember to multiply each standard entropy value by the correct stoichiometric coefficient. Like Equation 10.18, Equation 18.5 can only be used with a balanced chemical equation.
Entropy Changes in a System
Several processes that lead to an increase in entropy are:
Melting
Vaporization or sublimation
Temperature increase
Reaction resulting in a greater number of gas molecules
Entropy Changes in a System
The process of dissolving a substance can lead to either an increase or a decrease in entropy, depending on the nature of the solute.
Molecular solutes (i.e. sugar): entropy increases
Ionic compounds: entropy could decrease or increase
Worked Example 18.3
Strategy Consider the change in energy/mobility of atoms and the resulting change in number of possible positions that each particle can occupy in each case. An increase in the number of arrangements corresponds to an increase in entropy and therefore a positive ΔS.
For each process, determine the sign of ΔS for the system: (a) decomposition of CaCO3(s) to give CaO(s) and CO2(g), (b) heating bromine vapor from 45°C to 80°C, (c) condensation of water vapor on a cold surface, (d) reaction of NH3(g) and HCl(g) to give NH4Cl(s), and (e) dissolution of sugar in water.
Solution Increases in entropy generally accompany solid-to-liquid, liquid-to-gas, and solid-to-gas transitions; the dissolving of one substance in another; a temperature increase; and reactions that increase the net number of moles of gas.
ΔS is (a) positive
(b) positive
(c) negative
(d) negative
(e) positive
Think About It For reactions involving only liquids and solids, predicting the sign of ΔS can be more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase of entropy.
Entropy Changes in the Universe
Correctly predicting the spontaneity of a process requires us to consider entropy changes in both the system and the surroundings.
An ice cube spontaneously melts in a room at 25°C.
A cup of hot water spontaneously cools to room temperature.
The entropy of both the system AND surroundings are important!
18.4
Perspective Components ΔS
System ice positive
Surroundings everything else negative
Perspective Components ΔS
System hot water negative
Surroundings everything else positive
Entropy Changes in the Universe
The change in entropy of the surroundings is directly proportional to the enthalpy of the system.
The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive.
ΔSuniverse = ΔSsys + ΔSsurr
Entropy Changes in the Universe
The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive.
ΔSuniverse > 0 for a spontaneous process
ΔSuniverse < 0 for a nonspontaneous process
ΔSuniverse = 0 for an equilibrium process
ΔSuniverse = ΔSsys + ΔSsurr
Worked Example 18.4
Strategy For each process, use ΔS°rxn = ΣnS°(products) – ΣmS°(reactants) to determine ΔS °sys; ΔSsurr = (–ΔHsys/T)and ΔH °sys = ΣnΔH f°(products) – ΣmΔH f°(reactants) to determine ΔH °sys and ΔS °surr. At the specified temperature, the process is spontaneous if ΔS °sys and ΔS °surr sum to a positive number, nonspontaneous is they sum to a negative number, and an equilibrium process if they sum to zero. Note that because the reaction is the system, ΔSrxn and ΔSsys are used interchangeably.
Determine if each of the following is a spontaneous process, a nonspontaneous process, or an equilibrium process at the specified temperature: (a) H2(g) + I2(g) → 2HI(g) at 0°C, (b) CaCO3(s) → CaO(s) + CO2(g) at 200°C, (c) CaCO3(s) → CaO(s) + CO2(g) at 1000°C, (d) Na(s) → Na(l) at 98°C. (Assume that the thermodynamic data in Appendix 2 do not vary with temperature.)
Worked Example 18.4 (cont.)
Solution (a) S [H2(g)] = 131.0 J/K mol, S [I2(g)] = 260.57 J/K mol, S [HI(g)] = 206.3 J/K mol; ΔfH [H2(g)] = 0 kJ/mol, ΔfH [I2(g)] = 62.25 kJ/mol, ΔfH [HI(g)] = 25.9 kJ/mol.
ΔrxnS = [2S (HI)] – [S (H2) + S (I2)]
= (2)(206.3 J/K mol) – [131.0 J/K mol + 160.57 J/K mol] = 21.03 J/K mol
ΔrxnH = [2 ΔfH (HI)] – [ΔfH (H2) + ΔfH (I2)]
= (2)(25.9 kJ/mol) – [0 kJ/mol + 62.26 kJ/mol] = 10.5 kJ/mol
ΔsurrS = = = 0.0385 kJ/K mol = 38.5 J/K mol
ΔSuniverse = ΔSsys + ΔSsurr = 21.03 J/K mol + 38.5 J/K mol = 59.5 J/K mol
ΔSuniverse is positive; therefore the reaction is spontaneous at 0°C.
( 10.5 kJ/mol)
273 K
ΔrxnH
T
Worked Example 18.4 (cont.)
Solution (b) In Worked Example 18.2(a), we determined that for this reaction, ΔrxnS = 160.5 J/K mol; ΔfH [CaCO3(s)] = 1206.9 kJ/mol, ΔfH [CaO(s)] = 635.6 kJ/mol, Δf H [CO2(g)] = 393.5 kJ/mol.
(b), (c)
ΔrxnS = 160.5 J/K mol
ΔrxnH = [ΔfH (CaO) + ΔfH (CO2)] – [ΔfH (CaCO3)]
= [-635.6 kJ/mol + (–393.5 kJ/mol)] – (–1206.9 kJ/mol) = 177.8 kJ/mol
(b) T = 200°C and
ΔSsurr = = = 0.376 kJ/K mol = 376 J/K mol
ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K mol + ( 376 J/K mol) = 216 J/K mol
Δsuniverse is negative, therefore the reaction is nonspontaneous at 200°C.
(177.8 kJ/mol)
473 K
ΔHsys
T
Worked Example 18.4 (cont.)
Solution (c) T = 1000°C and
ΔSsurr = = = 0.1397 kJ/K mol = 139.7 J/K mol
ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K mol + ( 139.7 J/K mol) = 20.8 J/K mol
In this case, ΔSuniverse is positive; therefore, the reaction is spontaneous at 1000°C.
(177.8 kJ/mol)
473 K
ΔHsys
T
Worked Example 18.4 (cont.)
Solution (d) S [Na(s)] = 51.05 J/K mol, S [Na(l)] = 57.56 J/K mol; ΔfH [Na(s)] = 0 kJ/mol, ΔfH [Na(l)] = 2.41 kJ/mol.
ΔrxnS = S [Na(l)] – S [Na(s)]
= 57.56 J/K mol – 51.05 J/K mol = 6.51 J/K mol
ΔrxnH = ΔfH [Na(l)] – ΔfH [Na(s)]
= 2.41 kJ/mol – 0 kJ/mol = 2.41 kJ/mol
ΔSsurr = = = 0.0650 kJ/K mol = 6.50 J/K mol
ΔSuniverse = ΔSsys + ΔSsurr = 6.51 J/K mol 6.50 J/K mol = 0.01 J/K mol ≈ 0
ΔSuniverse is zero; therefore, the reaction is an equilibrium process at 98°C. In fact, this is the melting point of sodium.
(2.41 kJ/mol)
371 K
ΔHrxn
T
Think About It Remember that standard enthalpies of formation have units of kJ/mol, whereas standard absolute entropies have units of J/K mol. Make sure that you convert kilojoules to joules, or vice versa, before combining the terms.
Entropy Changes in the Universe
The third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at absolute zero.
Entropy increases in a
substance as temperature
increases from absolute
zero.
Predicting Spontaneity
Measurements on the surroundings are seldom made, limiting the use of the second law of thermodynamics.
Gibbs free energy (G) or simply free energy can be used to express spontaneity more directly.
The change in free energy for a system is:
18.5
G = H – TS
ΔG = ΔH – TΔS
Predicting Spontaneity
Using the Gibbs free energy, it is possible to make predictions on spontaneity.
ΔG < 0 The reaction is spontaneous in the forward direction.
ΔG > 0 The reaction is nonspontaneous in the forward direction.
ΔG = 0 The system is at equilibrium
ΔG = ΔH – TΔS
Worked Example 18.5
Strategy The temperature that divides high from low is the temperature at which ΔH = TΔS (ΔG = 0). Therefore, we use ΔG = ΔH – TΔS, substituting 0 for ΔG and solving for T to determine temperature in kelvins; we then convert to degrees Celsius.
According to Table 18.4, a reaction will be spontaneous only at high temperatures if both ΔH and ΔS are positive. For a reaction in which ΔH = 199.5 kJ/mol and ΔS = 476 J/K mol, determine the temperature (in °C) above which the reaction is spontaneous.
Solution
ΔS =
476 J
K mol
= 0.476 kJ/K mol
1 kJ
1000 J
T =
ΔH
ΔS
= 419 K
199.5 kJ/mol
0.476 kJ/K mol
=
= (419 – 273) = 146°C
Think About It Spontaneity is favored by a release of energy (ΔH being negative) and by an increase in entropy (ΔS being positive). When both quantities are positive, as in this case, only the entropy change favor spontaneity. For an endothermic process such as this, which requires the input of heat, it should make sense that adding more heat by increasing the temperature will shift the equilibrium to the right, thus making it “more spontaneous.”
Predicting Spontaneity
The standard free energy of reaction (ΔrxnG ) is free-energy change for a reaction when it occurs under standard-state conditions.
The following conditions define the standard states of pure substances and solutions are:
Gases 1 atm pressure
Liquids pure liquid
Solids pure solid
Elements the most stable allotropic form at 1 atm and 25°C
Solutions 1 molar concentration
Entropy Changes in a System
For a chemical reaction
aA + bB → cC + dD
Alternatively,
Δf G for any element in its most stable allotropic form at 1 atm is defined as zero.
ΔrxnG = [cΔfG (C) + dΔfG (D)] – [aΔfG (A) + bΔfG (B)]
ΔrxnG = ΣnΔfG (products) – ΣmΔfG (reactants)
Worked Example 18.6
Strategy Look up the ΔfG values for the reactants and products in each equation, and use ΔrxnG = ΣnΔfG (products) – ΣmΔfG (reactants) to solve for ΔrxnG .
Calculate the standard free-energy changes for the following reactions at 25°C:
(a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
(b) 2MgO(s) → 2Mg(s) + O2(g)
Solution From Appendix 2, we have the following values: ΔfG [CH4(g)] = 50.8 kJ/mol, ΔfG [CO2(g)] = 394.4 kJ/mol, Δf G [H2O(l)] = 237.2 kJ/mol, and ΔfG [MgO(s)] = 569.6 kJ/mol. All the other substances are elements in their standard states and have, by definition, ΔfG = 0.
(a) ΔrxnG = (ΔfG [CO2(g)] + 2ΔfG [H2O(l)]) – (ΔfG [CH4(g)] + ΔfG [O2(g)])
= [( 394.4 kJ/mol) + (2)( 237.2 kJ/mol)] [( 50.8 kJ/mol) + (2)(0 kJ/mol)]
= 818.0 kJ/mol
Worked Example 18.6 (cont.)
Solution From Appendix 2, we have the following values: Δ [CH4(g)] = 50.8 kJ/mol, Δf G [CO2(g)] = 394.4 kJ/mol, Δf G [H2O(l)] = 237.2 kJ/mol, and ΔfG [MgO(s)] = 596.6 kJ/mol. All the other substances are elements in their standard states and have, by definition, ΔfG = 0.
(b) ΔG °rxn = (2ΔfG [Mg(s)] + ΔfG [O2(g)]) – (2Δf G [MgO(s)])
= [(2)(0 kJ/mol) + (0 kJ/mol)] [(2)( 569.6 kJ/mol)]
= 1139 kJ/mol
Think About It Note that, like standard enthalpies of formation (ΔfH ), standard free energies of formation (ΔfG ) depend on the state of matter. Using water as an example, ΔfG [H2O(l)] = 237.2 kJ/mol and ΔfG [H2O(g)] = 228.6 kJ/mol. Always double-check to make sure you have selected the right value from the table.
Worked Example 18.7
Strategy The solid-liquid transition at the melting point and the liquid-vapor transition at the boiling points are equilibrium processes. Therefore, because ΔG is zero at equilibrium, in each case we can use ΔG = ΔH – TΔS, substituting 0 for ΔG and solving for ΔS, to determine the entropy change associated with the process.
The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid-to-liquid and liquid-to-vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
Solution
= 0.0391 kJ/K mol or 39.1 J/K mol
ΔSfus =
10.9 kJ/mol
278.7 K
ΔHfus
Tmelting
=
= 0.0877 kJ/K mol or 87.7 J/K mol
ΔSvap =
31.0 kJ/mol
353.3 K
ΔHvap
Tboiling
=
Think About It For the same substance, ΔSvap is always significantly larger than ΔSfus. The change in number of arrangements is always bigger in a liquid-to-gas transition than in a solid-to-liquid transition.
Free Energy and Chemical Equilibrium
It is the sign of ΔG (not ΔG ) that determines spontaneity.
The relationship between ΔG and ΔG° is:
R is the gas constant (8.314 J/K·mol).
T is the kelvin temperature.
Q is the reaction quotient.
18.6
ΔG = ΔG + RT lnQ
Consider the following equilibrium:
H2(g) + I2(g) 2HI(g)
ΔG at 25°C = 2.60 kJ/mol
ΔG depends on the partial pressures of each chemical species.
If PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 3.0 atm:
Then:
Free Energy and Chemical Equilibrium
The spontaneity can be manipulated by changing the partial pressures of the reaction components:
H2(g) + I2(g) 2HI(g)
ΔG at 25°C = 2.60 kJ/mol
If PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 1.0 atm:
Then:
Free Energy and Chemical Equilibrium
Worked Example 18.8
Strategy Use the partial pressures of N2O4 and NO2 to calculate the reaction quotient QP, and then use ΔG = ΔG + RT lnQ to calculate ΔG.
The reaction quotient expression is
The equilibrium constant, KP, for the reaction
N2O4(g) 2NO2(g)
is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.4 kJ/mol. In a certain experiment, the initial pressures are PN2O4 = 0.453 atm and PNO2 = 0.122 atm. Calculate ΔG for the reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously to establish equilibrium.
QP =
(0.122)2
0.453
(PNO2)2
PN2O4
=
= 0.0329
Worked Example 18.8 (cont.)
Solution
Because ΔG is negative, the reaction proceeds spontaneously from left to right to reach equilibrium.
(298 K)(ln 0.0329)
=
8.314×10-3 kJ
K mol
5.4 kJ
mol
+
= 5.4 kJ/mol – 8.46 kJ/mol
ΔG = ΔG + RT lnQ
= –3.1 kJ/mol
Think About It Remember, a reaction with a positive ΔG value can be spontaneous if the starting concentrations of reactants and products are such that Q < K.
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG + RT ln K
Free Energy and Chemical Equilibrium
ΔG = –RT ln K
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG + RT ln K
Free Energy and Chemical Equilibrium
ΔG = –RT ln K
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG + RT ln K
Free Energy and Chemical Equilibrium
ΔG = –RT ln K
Worked Example 18.9
Strategy Use data from Appendix 2 and ΔrxnG = ΣnΔf G (products) – ΣmΔfG (reactants) to calculate ΔG for the reaction. Then use ΔG = RT lnK to solve for KP.
Using data from Appendix 2, calculate the equilibrium constant, KP, for the following reaction at 25°C:
2H2O(l) 2H2(g) + O2(g)
Solution
ΔG ° = (2ΔfG [H2(g)] + ΔfG [O2(g)]) – (2ΔfG [H2O(l)])
= [2(0 kJ/mol) + (0 kJ/mol)] [(2)( 237.2 kJ/mol)]
= 474.4 kJ/mol
474.4 kJ
mol
(298 K) ln KP
=
8.314×10-3 kJ
K mol
191.5 = ln KP
KP = e 191.5 = 7×10-84
ΔG = RT ln KP
Think About It This is an extremely small equilibrium constant, which is consistent with the large, positive value of ΔG . We know from everyday experience that water does not decompose spontaneously into its constituent elements at 25°C.
Worked Example 18.10
Strategy Use ΔG = RT lnK to calculate ΔG°.
The equilibrium constant, Ksp, for the dissolution of silver chloride in water at 25°C:
AgCl(s) Ag+(aq) + Cl-(aq)
is 1.6×10-10. Calculate ΔG° for the process.
Solution
R = 8.314×10-3 kJ/K mol and T = (25 + 273) = 298 K.
(298 K) ln (1.6×10-10)
=
8.314×10-3 kJ
K mol
= 55.9 kJ/mol
ΔG = RT ln Ksp
Think About It The relatively large, positive ΔG , like the very small K value, corresponds to a process that lies very far to the left. Note that the K in ΔG = RT lnK can be any type of Kc (Ka, Kb, Ksp, etc.) or KP.
Thermodynamics of Living Systems
Many biological reactions have positive ΔG value, making the reaction nonspontaneous.
None spontaneous reactions can be coupled with spontaneous reactions in order to drive a process forward:
alanine + glycine → alanylglycine ΔG = 29 kJ/mol
ATP + H2O → ADP + H3PO4 ΔG = –31 kJ/mol
ATP + H2O + alanine + glycine → ADP + H3PO4 + alanylglycine
ΔG = 29 kJ/mol + –31 kJ/mol = –2 kJ/mol
18.7
Thermodynamics of Living Systems
Many biological reactions have positive ΔG value, making the reaction nonspontaneous.
Key Concepts
18
A Qualitative Description of Entropy
A Quantitative Description of Entropy
Calculating ΔsysS
Standard S
Qualitatively Predicting ΔsysS
Calculating ΔsurrS
The Second Law of Thermodynamics
The Third Law of Thermodynamics
Gibbs Free-Energy Change, ΔG
Standard Free-Energy Changes, ΔG
Using ΔG and ΔG to Solve Problems
Relationship Between ΔG and ΔG
Relationship Between ΔG and K

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