ap chemistry chapter 19 chemical...

AP Chemistry Chapter 19 Chemical Thermodynamics

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Chapter 19. Chemical Thermodynamics 19.1 Spontaneous Processes

Chemical thermodynamics is concerned with energy relationships in chemical reactions, i.e. Will a reaction occur? Two components: • enthalpy (H) • entropy (S) (randomness or disorder): note that in nature things tend towards disorder, much easier to go from low S high S, but natural tendency to disorder can be overcome by enthalpy. First law of thermodynamics: energy is conserved: ∆E= q + w ∆E = change in internal energy, q = heat absorbed by the system from the surroundings w = work done. Spontaneous process = any process that occurs without outside intervention. • Two eggs are dropped spontaneously break. • Reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous. a spontaneous process has a direction. • A process that is spontaneous in one direction is not spontaneous in the opposite direction. • Temperature may also affect the spontaneity of a process.

Seeking a Criterion for Spontaneity

• To understand why some processes are spontaneous we must look at the ways in which the state of a system might change. • Temperature, internal energy, and enthalpy are state functions. • Heat transferred between a system and the surroundings, as well as work done on or by a system, are not state functions.

Spontaneous expansion of an ideal gas into an evacuated space. In (a) flask B holds an ideal gas at 1 atm pressure and flask A is evacuated. In (b) the stopcock connecting the flasks has been opened. The ideal gas expands to occupy both flasks A and B at a pressure of 0.5 atm. The reverse process—all the gas molecules moving back into flask B—is not spontaneous.


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Reversible and Irreversible Processes

• A reversible process = one that can go back and forth between states along the same path. • Reverse process restores the system to its original state. • Path taken back to the original state is exactly the reverse of the forward process. • There is no net change in the system or the surroundings when this cycle is completed. • Completely reversible processes are too slow to be attained in practice. e.g. the interconversion of ice and water at 1 atm, 0oC. • Ice and water are in equilibrium. • Add heat to the system from the surroundings. 1 mole of ice is melted to form 1 mole of liquid water. q=∆Hfus • To return to the original state we reverse the procedure. • We remove the same amount of heat from the system to the surroundings.

Spontaneity can depend on the temperature. At T > 0 °C ice melts spontaneously to liquid water. At T < 0 °C the reverse process, water freezing to ice, is spontaneous. At T = 0 °C the two states are in equilibrium.


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Sample Exercise 19.1 (p. 803)

Predict whether the following processes are spontaneous as described, are spontaneous in the reverse direction,

or are in equilibrium:

a) When a piece of metal heated to 150oC is added to water at 40oC, the water gets hotter.

b) Water at room temperature decomposes into H2(g) and O2(g).

c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point

of benzene, 80.1oC.

Practice Exercise 19.1

Under 1 atm pressure CO2(s) (“dry ice”) sublimes at -78oC. Is the transformation of CO2(s) to CO2(g) a spontaneous process at -100oC and 1 atm pressure?

• An irreversible process cannot be reversed to restore the system and surroundings back to their original

state. • A different path (with different values of q and w) must be taken. • Consider a gas in a cylinder with a piston. • Remove the partition and the gas expands to fill the space. • No P-V work is done on the surroundings w = 0. • Now use the piston to compress the gas back to the original state. • The surroundings must do work on the system w > 0. • A different path is required to get the system back to its original state. • Note that the surroundings are NOT returned to their original conditions.

• For a system at equilibrium, reactants and products can interconvert reversibly. • For a spontaneous process, the path between reactants and products is irreversible.


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An irreversible process. Restoring the system to its original state after an irreversible process changes the surroundings. In (a) the gas is confined to the right half of the cylinder by a partition. When the partition is removed (b), the gas spontaneously (irreversibly) expands to fill the whole cylinder. No work is done by the system during this expansion. In (c) we can use the piston to compress the gas back to its original state. Doing so requires that the surroundings do work on the system, which changes the surroundings forever. • Consider the expansion of an ideal gas. • Consider an initial state: two 1-liter flasks connected by a closed stopcock. • One flask is evacuated and the other contains 1 atm of gas. • We open the stopcock while maintaining the system at constant temperature. • Initial state: an ideal gas confined to a cylinder kept at constant temperature in a water bath. • The process is isothermal at constant temperature. • ∆E = 0 for an isothermal process. • Thus, q = –w. • Allow the gas to expand from V1 to V2. • Pressure ↓ from P1 to P2. • The final state: two flasks connected by an open stopcock. • Each flask contains gas at 0.5 atm. the gas does no work and heat is not transferred. • Why does the gas expand? • Why is the process spontaneous? • Why is the reverse process nonspontaneous? • When the gas molecules spread out into the 2 L system there is an ↑ in the randomness or disorder. • Processes in which the disorder or entropy of the system ↑ tend to be spontaneous.

When there are many molecules, it is much more probable that the molecules will distribute among to the two flasks than all remain in only one flask.


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19.2 Entropy and the Second Law of Thermodynamics Entropy Change

• Entropy, S, is a thermodynamic term that reflects the disorder, or randomness, of the system. • The more disordered, or random, the system is, the larger the value of S. • Entropy is a state function. • independent of path. • For a system, ∆S = Sfinal – Sinitial. • If ∆S > 0 the randomness ↑, if ∆S < 0 the order ↑. • Suppose a system changes reversibly between state 1 and state 2. • Then, the change in entropy is given by,

• Where qrev is the amount of heat added reversibly to the system. • The subscript “rev” reminds us that the path between states is reversible. • Example: A phase change occurs at constant T with the reversible addition of heat.

• with the units J• K-1 or J• (mol-1K-1)

∆S for Phase Changes

• Phase changes (such as melting a substance at its melting point) are isothermal processes.

Sample Exercise 19.2 (p. 807) The element, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9oC, and its molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?

(∆Ssys = -2.44 J/K)

TqS rev

=∆

∆Sfusion =qrev

T=

∆Hfusion

T


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The normal boiling point of ethanol, C2H5OH, is 78.3oC, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point?

(-163 J/K)

The Second Law of Thermodynamics

• The second law of thermodynamics explains why spontaneous processes have a direction. • In any spontaneous process, the total entropy of the universe will increase . • The change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings.

∆Suniv = ∆Ssys + ∆Ssurr • For a reversible process:

∆Suniv = ∆Ssys + ∆Ssurr = 0 • For a spontaneous process (i.e., irreversible):

∆Suniv = ∆Ssys + ∆Ssurr > 0 • Entropy is not conserved: ∆Suniv is continually ↑. • Note: The second law states that the entropy of the universe must ↑ in a spontaneous process. • It is possible for the entropy of a system to ↓ as long as the entropy of the surroundings ↑.


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19.3 The Molecular Interpretation of Entropy

• The entropy of a system indicates its disorder. • A gas is less ordered than a liquid which is less ordered than a solid. • Any process that ↑ the number of gas molecules ↑ in entropy. • When NO(g) reacts with O2(g) to form NO2(g), the total number of gas molecules ↓.

2NO(g) + O2(g) 2NO2(g) the entropy ↓.

• How can we relate changes in entropy to changes at the molecular level? • Formation of the new N-O bonds “ties up” more of the atoms in the products than in the reactants. • The degrees of freedom associated with the atoms have changed. • ↑ freedom of movement and degrees of freedom ↑ the entropy of the system. • Individual molecules have degrees of freedom associated with motions within the molecule. • There are three atomic modes of motion: • Translational motion. • The moving of a molecule from one point in space to another. • Vibrational motion. • The shortening and lengthening of bonds, including the change in bond angles. • Rotational motion. • The spinning of a molecule about some axis.


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• Energy is required to get a molecule to translate, vibrate or rotate. • These forms of motion are ways molecules can store energy. • ↑ energy stored in translation, vibration and rotation ↑ entropy.

Boltzmann’s Equation and Microstates

• Statistical thermodynamics is a field that uses statistics and probability to link the microscopic and macroscopic worlds.

• Entropy may be connected to the behavior of atoms and molecules. • Envision a microstate: a snapshot of the positions and speeds of all molecules in a sample of a

particular macroscopic state at a given point in time. • Consider a molecule of ideal gas at a given temperature and volume. • A microstate is a single possible arrangement of the positions and kinetic energies of the gas

molecules. • Other snapshots are possible (different microstates). • Each thermodynamic state has a characteristic number of microstates (W). • The Boltzmann equation shows how entropy (S) relates to W.

S = k lnW, where k is Boltzmann’s constant (1.38 × 10–23 J/K). • Entropy is thus a measure of how many microstates are associated with a particular macroscopic state. • Any change in the system that increases the number of microstates gives a positive value of ∆S and vice

versa. • In general, the number of microstates will increase with an increase in volume, an increase in

temperature, or an increase in the number of molecules because any of these changes increases the possible positions and energies of the molecules.

Probability and the locations of gas molecules.

The two molecules are colored red and blue to keep track of them. (a) Before the stopcock is opened, both molecules are in the right-hand flask. (b) After the stopcock is opened, there are four possible arrangements of the two molecules. Only one of the four arrangements corresponds to both molecules being in the right-hand flask. The greater number of possible arrangements corresponds to greater disorder in the system. In general, the probability that the molecules will stay in the original flask is (½)n, where n is the number of molecules.


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Making Qualitative Predictions About ∆S

• In most cases an increase in the number of microstates (and thus entropy) parallels an increase in: • temperature • volume • number of independently moving particles.

• Consider the melting of ice. • In the ice, the molecules are held rigidly in a lattice.

• The intermolecular attractions in the three dimensional lattice restrict the molecules to vibrational motion only.

• When it melts, the molecules have more freedom to move (↑ degrees of freedom). • The molecules are more randomly distributed.

• Consider a KCl crystal dissolving in water. • The solid KCl has ions in a highly ordered arrangement. • crystal dissolves the ions have more freedom more randomly distributed. • However, now the water molecules are more ordered. • Some must be used to hydrate the ions. • Thus this example involves both ordering and disordering. • The disordering usually predominates (for most salts).


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• In general, entropy will increase when: • liquids or solutions are formed from solids, • gases are formed from solids or liquids, or • the number of gas molecules increases.


Predict whether ∆S is positive or negative for each of the following processes, assuming each occurs at constant

temperature:

a) H2O(l) H2O(g)

b) Ag+(aq) + Cl-

(aq) AgCl(s)

c) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

d) N2(g) + O2(g) 2 NO(g)


Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system:

a) CO2(s) CO2(g)

b) CaO(s) + CO2(g) CaCO3(s)

c) HCl(g) + NH3(g) NH4Cl(s)

d) 2 SO2(g) + O2(g) 2 SO3(g)


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Sample Exercise 19.4 (p. 815) Choose the sample of matter that has greater entropy in each pair, and explain your choice:

a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25oC. b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25oC

c) 1 mol of HCl(g) or 1 mol or Ar(g) at 298 K.

Practice Exercise 19.4 Choose the substance with the greater entropy in each case: a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100oC and 0.5 atm b) 1 mol of H2O(s) at 0oC or 1 mol of H2O(l) at 25oC c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.

The Third Law of Thermodynamics • In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules = state of perfect order • Entropy will ↑ as we ↑ the temperature of the perfect crystal. • Third law of thermodynamics: The entropy of a perfect pure crystal at 0 K = 0. • Entropy will increase as we increase the temperature of the perfect crystal. • Molecules gain vibrational motion ↑ degrees of freedom.


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• As we heat a substance from 0 K the entropy must ↑. • The entropy changes dramatically at a phase change. • When a solid melts, the molecules and atoms have a large ↑ in freedom of movement. • Boiling corresponds to a much greater change in entropy than melting. • In general, entropy will ↑ when liquids or solutions are formed from solids.

• Gases are formed from solids or liquids. • The number of gas molecules increase.

19.4 Entropy Changes in Chemical Reactions

• Absolute entropy can be determined from complicated measurements. • Values are based on a reference point of zero for a perfect crystalline solid at 0 K (the 3rd law). • Standard molar entropy, So: molar entropy of a substance in its standard state. • Similar in concept to ∆Ho. • Units: J/mol-K. • Note: the units of ∆H are kJ/mol.

In general, the more complex a molecule (that is, the greater the number of atoms present), the greater the molar entropy of the substance, as illustrated by the molar entropies of these three simple hydrocarbons:


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( ) ( )∑∑ °−°=°∆ reactantsproducts mSnSS

• Some observations about So values: • Standard molar entropies of elements ≠ zero. • So

gas > Soliquid or So

solid. • So tends to ↑ with ↑ molar mass of the substance. • So tends to ↑ with the number of atoms in the formula of the substance. • For a chemical reaction which produces n products from m reactants:

• Example: Consider the reaction:

N2(g) + 3H2(g) 2NH3(g)

∆So = {2So(NH3) – [So(N2) + 3So(H2)]}


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Sample Exercise 19.5 (p. 818) Calculate ∆So for the synthesis of ammonia from N2(g) and H2(g) at 298 K. N2(g) + 3 H2(g) 2 NH3(g) (-198.3 J/K) Practice Exercise 19.5 Using the standard entropies in Appendix C, calculate the standard entropy change, ∆So for the following reaction at 298 K: Al2O3(s) + 3 H2(g) 2 Al(s) + 3 H2O(g) (180.39 J/K)

Entropy Changes in the Surroundings

• For an isothermal process, • ∆Ssurr = –qsys / T • For a reaction at constant pressure, • qsys = ∆H • Example: consider the reaction:

N2(g) + 3H2(g) 2NH3(g) • The entropy gained by the surroundings is > the entropy lost by the system. • This is the sign of a spontaneous reaction: the overall entropy change of the universe is positive. • ∆Suniv > 0


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19.5 Gibbs Free Energy

• For a spontaneous reaction the entropy of the universe must ↑. • Reactions with large negative ∆H values tend to be spontaneous. • How can we use ∆S and ∆H to predict whether a reaction is spontaneous? • The Gibbs free energy, (free energy), G, of a state is:

G = H–TS • Free energy is a state function. • For a process occurring at constant temperature, the free energy change is: ∆G = ∆H–T∆S • Recall: • ∆Suniv = ∆Ssys + ∆Ssurr = ∆Ssys + [–∆Hsys / T] –T∆Suniv = ∆Hsys – T∆Ssys

• The sign of ∆G is important in predicting the spontaneity of the reaction. • If ∆G < 0 then the forward reaction is spontaneous. • If ∆G = 0 then the reaction is at equilibrium and no net reaction will occur. • If ∆G > 0 then the forward reaction is not spontaneous. • However, the reverse reaction is spontaneous. • If ∆G > 0, work must be supplied from the surroundings to drive the reaction. • The equilibrium position in a spontaneous process is given by the minimum free energy available to the

system. • The free energy ↓ until it reaches this minimum value.


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Standard Free-Energy Changes • We can tabulate standard free energies of formation, ∆Go

f . • Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and ∆Go = 0 for

elements. • We most often use 25oC (or 298 K) as the temperature. • The standard free-energy change for a process is given by:

• The quantity ∆Go for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (∆Go > 0) or products (∆Go < 0).

Free energy and equilibrium.

In the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g), if the reaction mixture has too much N2 and H2 relative to NH3 (left), the equilibrium lies too far to the left (Q < K) and NH3 forms spontaneously. If there is too much NH3 in the mixture (right), the equilibrium lies too far to the right (Q > K) and the NH3 decomposes spontaneously into N2 and H2. Both of these spontaneous processes are "downhill" in free energy. At equilibrium (center), Q = K and the free energy is at a minimum (ΔG = 0).

( ) ( )∑∑ °∆−°∆=°∆ reactantsproducts ff GmGnG


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Calculate the standard free energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K:

N2(g) + O2(g) 2 NO(g)

Given that ∆Ho = 180.7 kJ and ∆So = 24.7 J/K. Is the reaction spontaneous under these circumstances?


A particular reaction has ∆Ho = 24.6 kJ and ∆So = 132 J/K at 298 K. Calculate ∆Go. Is the reaction spontaneous under these conditions?


a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K:

P4(g) + 6 Cl2(g) 4 PCl3(g) (-1102.8 kJ)

b) What is ∆Go for the reverse of the above reaction? (+1102.8 kJ)


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Practice Exercise 19.7 By using the data from Appendix C, calculate ∆Go at 298 K for the combustion of methane: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) (-800.7 kJ) Sample Exercise 19.8 (p. 823) In Section 5.7 we used Hess’s law to calculate ∆Ho for the combustion of propane gas at 298 K:

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

a) Without using data from Appendix C, predict whether ∆Go for this reaction is more negative or less negative than ∆Ho.

b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is

your prediction from part (a) correct? (-2108 kJ) Practice Exercise 19.8 Consider the combustion of propane to form CO2(g) and H2O(g) at 298 K: C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) ∆Ho = -2220 kJ Would you expect ∆Go to be more negative or less negative than ∆Ho?


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19.6 Free Energy and Temperature

• The sign of ∆G tells us if the reaction is spontaneous, at a given T and P condition. • Focus on ∆G = ∆H – T∆S. • If ∆H <0 and –T∆S <0: • ∆G will always be <0. • Thus the reaction will be spontaneous. • If ∆H >0 and –T∆S >0: • ∆G will always be >0. • Thus the reaction will not be spontaneous. • If ∆H and –T∆S have different signs: • The sign of ∆G will depend on the sign and magnitudes of the other terms. • Temperature will be an important factor. • For example, consider the following reaction:

H2O(s) H2O(l) ∆H >0, ∆S > 0 • At a T < 0oC: • ∆H > T∆S • ∆G> 0 • The melting of ice is not spontaneous when the temperature is < 0oC. • At a T > 0oC: • ∆H < T∆S • ∆G < 0 • The melting of ice is spontaneous when the temperature is > 0oC. • At 0oC: • ∆H = T∆S • ∆G = 0 • Ice and water are in equilibrium at 0oC. • Note that we have assumed that both ∆H° and ∆S° were independent of temperature; they aren’t, but the

changes are negligible. • Even though a reaction has a negative ∆G it may occur too slowly to be observed. • Thermodynamics gives us the direction of a spontaneous process; it does not give us the rate of the

process.


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Sample Exercise 19.9 (p. 826) The Haber process for the production of ammonia involves the following equilibrium: N2(g) + 3 H2(g) 2 NH3(g) Assume that ∆Ho and ∆So for this reaction do not change with temperature.

a) Predict the direction in which ∆Go for this reaction changes with increasing temperature.

b) Calculate the values of ∆Go for the reaction at 25oC and 500oC. (-33.3 kJ, 61 kJ)


a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ∆Ho and ∆So at 298 K for the following reaction:

2 SO2(g) + O2(g) 2 SO3(g).

(∆Ho = -196.6 kJ, ∆So = -189.6 J/K)

b) Using the values obtained in part (a), estimate ∆Go at 400 K. (∆Go = -120.8 kJ)


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19.7 Free Energy and the Equilibrium Constant • Recall that ∆Go and Keq (equilibrium constant) apply to standard conditions. • Recall that ∆G and Q (equilibrium quotient) apply to any conditions. • It is useful to determine whether substances will react under specific conditions:

∆G = ∆Go + RTlnQ Sample Exercise 19.10 (p. 827) As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm.

a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l).

b) What is the value of ∆Go for the equilibrium in part (a)?

c) Use thermodynamic data in Appendix C and ∆Go = ∆Ho – T∆So to estimate the normal boiling point of CCl4. (70oC)

Practice Exercise 19.10 Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3). (330 K)


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Sample Exercise 19.11 (p. 828) We will continue to explore the Haber process for the synthesis of ammonia:

N2(g) + 3 H2(g) 2 NH3(g) Calculate ∆G at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. (-44.9 kJ/mol) Practice Exercise 19.11 Calculate ∆G at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. (-26.0 kJ/mol)


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• At equilibrium, Q = Keq and ∆G = 0, so: ∆G = ∆Go + RTlnQ 0 = ∆Go + RTlnKeq ∴ ∆Go = – RTlnKeq • From the above we can conclude: • If ∆Go < 0, then Keq > 1. • If ∆Go = 0, then Keq = 1. • If ∆Go > 0, then Keq < 1.

Sample Exercise 19.12 (p. 829) Use standard free energies of formation to calculate the equilibrium constant Keq at 25oC for the reaction involved in the Haber process:

N2(g) + 3 H2(g) 2 NH3(g) (7 x 105) Practice Exercise 19.12 Use data from Appendix C to calculate the standard free-energy change, ∆Go, and the equilibrium constant, K, at 298 K for the following reaction: H2(g) + Br2(l) 2 HBr(g) (-106.4 kJ/mol; 4 x 1018)


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Driving Nonspontaneous Reactions

• If ∆G > 0, work must be supplied from the surroundings to drive the reaction. • Biological systems often use one spontaneous reaction to drive another nonspontaneous reaction. • These reactions are coupled reactions. • The energy required to drive most nonspontaneous reactions comes from the metabolism of foods. • Example: Consider the oxidation of glucose:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) ∆Go = –2880 kJ. • The free energy released by glucose oxidation is used to convert low energy adenosine diphosphate

(ADP) and inorganic phosphate into high energy adenosine triphosphate (ATP). • When ATP is converted back to ADP the energy released may be used to “drive” other reactions. Sample Integrative Exercise 19 (p. 831) Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions: NaCl(s) Na+

(aq) + Cl-(aq)

AgCl(s) Ag+(aq) + Cl-

(aq)

a) Calculate the value of ∆Go at 298 K for each of the preceding reactions. b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or

the entropy term of the standard free-energy change?

c) Use the values of ∆Go to calculate Ksp values for the two salts at 298 K.

d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)?

e) How will ∆Go for the solution process of these salts change with increasing T? What effect should this

change have on the solubility of the salts?

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