PMU Distinguish Between Spontaneous and Nonspontaneous Processes Exam i have chem exam tommrow same stuff as always will post quiesion and same time same l

PMU Distinguish Between Spontaneous and Nonspontaneous Processes Exam i have chem exam tommrow same stuff as always will post quiesion and same time same length of time. CHEMISTRY 2e
Chapter 16 THERMODYNAMICS
PowerPoint Image Slideshow
Chapter Outline
• 16.1 Spontaneity
• 16.2 Entropy
• 16.3 The Second and Third Laws of Thermodynamics
• 16.4 Free Energy
Figure 16.1
Geysers are a dramatic display of thermodynamic principles in nature. As water inside the
earth heats up, it rises to the surface through small channels. Pressure builds up until the water
turns to steam, and steam is expelled forcefully through a hole at the surface. (credit:
modification of work by Yellowstone National Park)
Learning Objectives
• 16.1 Spontaneity
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain
spontaneous processes
Spontaneous Process
• Will a reaction occur “naturally” at a given temperature and pressure,
without the exertion of any outside force?
• In other words, is a reaction spontaneous?
• A process that does occur naturally under a specific set of conditions
is called a spontaneous process.
• A process that does not occur naturally under a specific set of
conditions is called nonspontaneous.
Spontaneous Process
• A spontaneous process will “just happen.”
• A spontaneous process occurs naturally at a given set of conditions,
without outside forces.
• Example: A mixture of hydrogen and oxygen will react in the
presence of a spark.
2H2(g) + O2(g) + spark → 2H2O(g)
Spontaneity
• It is important to not confuse spontaneous with fast.
• The rate of a reaction (Chapter 12) and spontaneity are not
necessarily connected.
Figure 16.2
Both U-238 and Tc-99m undergo spontaneous radioactive decay, but at drastically different
rates. Over the course of one week, essentially all of a Tc-99m sample and none of a U-238
sample will have decayed.
Figure 16.3
The conversion of carbon from the diamond allotrope to the graphite allotrope is spontaneous at ambient
pressure, but its rate is immeasurably slow at low to moderate temperatures. This process is known as
graphitization, and its rate can be increased to easily measurable values at temperatures in the 1000–2000 K
range. (credit “diamond” photo: modification of work by “Fancy Diamonds”/Flickr; credit “graphite” photo:
modificaton of work by images-of-elements.com/carbon.php)
Spontaneity
• If a reaction is spontaneous in one direction, it will be
nonspontaneous in the reverse direction under the same conditions.
2H2(g) + O2(g) → 2H2O(l)
Spontaneous
2H2O(l) → 2H2(g) + O2(g)
Nonspontaneous
• A nonspontaneous reaction is still possible with the continual input
of energy.
Energy and Spontaneity
• Many spontaneous processes proceed with a decrease in energy.
• Recall that exothermic reactions also proceed with a decrease in
energy.
• Spontaneous reactions are often exothermic, but not always.
2H2(g) + O2(g) + spark → 2H2O(g)
Exceptions
• Which is spontaneous at room temperature and a pressure of 1 atm?
H2O(s) → H2O(l)
DH = +6.01 kJ/mol
H2O(l) → H2O(s)
DH = –6.01 kJ/mol
Dispersal of Matter
• When the valve opens, the gas spontaneously expands to fill both
containers.
• With an ideal gas, this process results in no change in energy.
Dispersal of Energy
• Two objects at different temperature are placed in contact.
• Heat spontaneously flows from the hotter object to the colder object.
• But overall, there is no change in energy.
Figure 16.5
When two objects at different temperatures come in contact, heat spontaneously flows from
the hotter to the colder object.
Figure 16.6
(credit a: modification of work by Jenny Downing; credit b: modification of work by
“Fuzzy Gerdes”/Flickr; credit c: modification of work by Sahar Atwa)
KNOWLEDGE CHECK 6
Difference between a spontaneous and non-spontaneous
process?
Which of these processes are spontaneous?
a) A bicyle going uphill.
b) Wind scattering leaves in a pile.
c) A drop of ink dispersing in water.
Some Other Factor
• Some factor other than energy is important to the spontaneity of a
process.
• This other factor is entropy.
Learning Objectives
• 16.2 Entropy
• Define entropy
• Explain the relationship between entropy and the number of microstates
• Predict the sign of the entropy change for chemical and physical processes
Entropy
• Spontaneity is favored by an increase in entropy (S).
S = k ln W
• k is the Boltzmann constant (1.38 × 10–23 J/K).
• W is the number of microstates possible
• Microstate: A specific configuration of the locations and energies of the
particles in a system.
• The number of microstates possible is given by:
W = nN
• n is the number of boxes
• N is the number of particles
Entropy and Microstates
• Consider two particles distributed between two boxes.
Entropy and Microstates
• Now consider four particles distributed between two boxes.
• Microstates with equivalent particle arrangements are grouped
together and called distributions.
Figure 16.8
The sixteen microstates associated with placing four particles in two boxes are shown. The
microstates are collected into five distributions—(a), (b), (c), (d), and (e)—based on the
numbers of particles in each box.
Entropy and Microstates
• There are five possible distributions for this system.
• Which distribution is most probable?
Entropy and Microstates
• The most probable distribution has the largest number of
microstates.
• The most probable distribution, therefore, is the one of greatest
entropy.
S = k ln W
• States of high entropy are favored because they are the most
probable.
KNOWLEDGE CHECK 7
• (Refer to Figure 16.8 to answer this question). The number of
microstates possible is given by:
W = nN
How many microstates are possible if instead of 2 boxes, there were 3
boxes in Figure 16.8?
Entropy and Microstates
• The most probable distribution has the largest number of
microstates.
• This same principle applies to all systems, including those with larger
numbers of particles.
• The most probable state will be the one in which the particles are
divided evenly throughout the container.
Entropy Changes
• Entropy is another state function.
• The change in entropy for a process is the difference in entropy
between the final state and the initial state.
ΔSsys = Sfinal – Sinitial
Factors That Influence Entropy
1) The phase of the substance.
2) The temperature of the substance.
• Temperature is proportional to the average kinetic energy of the particles.
• With higher temperature, the particles have greater freedom to move
around.
Figure 16.10 (Phase of the substance and entropy)
The entropy of a substance increases (ΔS > 0) as it transforms from a relatively ordered solid, to
a less-ordered liquid, and then to a still less-ordered gas. The entropy decreases (ΔS < 0) as the substance transforms from a gas to a liquid and then to a solid. ALEKS (Predicting qualitatively how entropy changes with temperature and volume) Figure 16.11 Entropy increases as the temperature of a substance is raised, which corresponds to the greater spread of kinetic energies. When a substance melts or vaporizes, it experiences a significant increase in entropy. Factors that Influence Entropy 3) The type and number of particles that make up the substance. 4) Variations in the type of particles. • Pure substances vs. mixture? ALEKS (How entropy changes with mixing) KNOWLEDGE CHECK 8 • Predict whether ΔS is positive or negative for these processes: Process 1: Taking dry ice from a freezer where its temperature is -80oC and allowing it to warm to room temperature. Process 2: dissolving salt in water. Process 3: condensing gaseous bromine to liquid bromine. Learning Objectives • 16.3 The Second and Third Laws of Thermodynamics • State and explain the second and third laws of thermodynamics • Calculate entropy changes for phase transitions and chemical reactions under standard conditions The Second and Third Laws of Thermodynamics • The system is the part of the universe that is of specific interest. • The surroundings constitute the rest of the universe outside the system. Generally, chemists just focus on the immediate surroundings. • Correctly predicting the spontaneity of a process requires us to consider entropy changes in both the system and the surroundings. The Second Law of Thermodynamics • The entropy change of the universe is the sum of the entropy changes for the system and surroundings. DSuniverse = DSsystem + DSsurroundings • The second law of thermodynamics states that all spontaneous changes cause an increase in the entropy of the universe. • For a spontaneous process, ΔSuniverse must be positive. • A process with –ΔSsystem can still be spontaneous if … The Second Law of Thermodynamics DS universe = DSsystem + DSsurroundings • ΔSuniverse > 0 for a spontaneous process
• ΔSuniverse < 0 for a nonspontaneous process (spontaneous in the reverse direction). • ΔSuniverse = 0 for a process at equilibrium KNOWLEDGE CHECK 9 • How can the sign of ΔSuniverse be used to predict the spontaneity of the reaction? ALEKS (Using the Second Law to predict spontaneous change) The Third Law of Thermodynamics • Third law of thermodynamics: The entropy of a pure perfect crystalline substance at zero Kelvin is zero. • • • • Zero Kelvin is called absolute zero. There is no lower temperature than zero Kelvin. At zero Kelvin, all molecular movement completely stops. There is only one possible way to arrange the molecules. Standard Entropies • It is possible to determine the absolute entropy of a substance. • Standard Entropies, S° • These values are for 1 mole of a substance at a pressure of 1 bar and a temperature of 298 K. • Aqueous species at 1 M concentration. • Standard entropy values can be used to calculate the standard entropy change (DS°) for a process. Table 16.2 Substance S°298 (J mol–1 K–1) carbon Substance S°298 (J mol–1 K–1) hydrogen C(s, graphite) 5.740 H2(g) 130.57 C(s, diamond) 2.38 H(g) 114.6 CO(g) 197.7 H2O(g) 188.71 CO2(g) 213.8 H2O(l) 69.91 CH4(g) 186.3 HCl(g) 186.8 C2H4(g) 219.5 H2S(g) 205.7 C2H6(g) 229.5 oxygen CH3OH(I) 126.8 O2(g) C2H5OH(I) 160.7 205.03 Standard entropies for selected substances measured at 1 atm and 298.15 K. (Values are approximately equal to those measured at 1 bar, the currently accepted standard state pressure.) Knowledge Check 10 • (Table 16.2) Look at the standard entropy values of gases CH4, C2H4, and C2H6. What trend do you notice? • Also, look at the general trend in the standard entropy values of solid, liquid, and gas. What do you notice? ΔS° for Reactions • The equation for calculating DS° is similar to that for DH°: DH =  DH f products − DH f reactants DS =  S products − S reactants • When calculating DS° and DH°, remember to multiply the standard entropies and standard enthalpies of formation by the coefficients of the balanced equation. ALEKS (Calculating reaction entropy using the standard molar entropies of reactants) Learning Objectives • 16.4 Free Energy • Define Gibbs free energy, and describe its relation to spontaneity • Calculate free energy change for a process using free energies of formation for its reactants and products • Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products • Explain how temperature affects the spontaneity of some processes • Relate standard free energy changes to equilibrium constants Free Energy, G • The second law of thermodynamics can be used to predict spontaneity. • But measurements on the surroundings are seldom made. • This limits the use of the second law of thermodynamics. • It is convenient to have a thermodynamic function that focuses on just the system and predicts spontaneity. Gibbs Free Energy Change, ΔG • Second Law of Thermodynamics: DS univ = DSsys + DSsurr  0 Gibbs Free Energy Change, ΔG • The changes in Gibbs free energy (DG) or simply change in free energy allow us to predict spontaneity by focusing on the system only. DG = DH – TDS ΔG and Spontaneity DG = DH – TDS • The sign of DG indicates if a reaction will be spontaneous or not. • If DG < 0, the reaction is spontaneous in the forward direction. • If DG > 0, the reaction is nonspontaneous in the forward direction
• If DG = 0, the system is at equilibrium
KNOWLEDGE CHECK 11
• How does the sign of delta G help predict the spontaneity of a
reaction?
Relationship among ΔG, ΔH, and ΔS
DG = DH – TDS
• Spontaneous reactions, those with –ΔG, generally have
• ΔH < 0 • Exothermic reaction. • A negative ΔH will contribute to a negative ΔG. • ΔS > 0
• A positive ΔS will contribute to a negative ΔG.
• Note that a reaction can still be spontaneous (have a –ΔG) when ΔH
is positive or ΔS is negative, but not both.
• Also note that there is a temperature dependence.
ΔG = ΔH – TΔS
ΔH
ΔS
–
+
+
–
–
–
ΔG
Spontaneous?
Figure 16.12
There are four possibilities regarding the signs of enthalpy and entropy changes.
ALEKS (Using the conditions of spontaneity to deduce
the signs of ΔH and ΔS
Direction of Spontaneity Change
• To calculate the temperature at which the spontaneity of a reaction
changes from …
• Spontaneous to nonspontaneous
• Or nonspontaneous to spontaneous
… find the temperature at which ΔG = 0
ΔG = 0 = ΔH – TΔS
T = ΔH / ΔS
• This is the temperature at which ΔG = 0 and, by definition, the system
is at equilibrium.
The Standard Free Energy Change, ΔG°
• Although the Change in Gibbs Free Energy equation is valid under all
conditions, we will most often apply it at standard conditions.
• Standard conditions:
• Under standard conditions, ΔG° = ΔH° – TΔS°
• Pay attention to J vs. kJ in calculations!
Standard Free Energy of Formation, ∆G°f
• The standard free energy of formation (∆G°f) for a compound is
defined as the free energy change for the formation of one mole of a
substance from its elements in their standard state at 1 bar and 25
°C.
• Analogous to the ∆H°f discussed in Chapter 5.
• Example:
H2(g) + ½O2(g) → H2O(l)
∆G°f = –237.2 kJ/mol
∆G°f Values Can Be Used to Calculate ΔG°
• For a chemical reaction:
mA + nB → xC + yD
ΔG° = [xΔG°f (C) + yΔG°f (D)] – [mΔG°f(A) + nΔG°f (B)]
ΔG°= ΣnΔG°f(products) – ΣnΔG°f(reactants)
∆G°f Values Can Be Used to Calculate ΔG°
ΔG° = ΣnΔG°f (products) – ΣnΔG°f (reactants)
• This equation only works for calculating ∆G° of a reaction at the
temperature for which the values of ∆G°f are tabulated, which is 298
K.
• ΔG°f for any element in its most stable form at standard conditions is
defined as zero.
ALEKS (Calculating standard reaction free energy from standard
free energies of formation)
Pressure and Concentration Effects
• Most of our discussion on free energy to this point has involved the
standard free energy change ΔG°.
• All species are at 1 bar of partial pressure or 1 M concentration.
• There is a general equation that enables you to calculate ΔG under
non-standard conditions.
DG = DG + RT ln Q
• T is temperature in K
• R = 0.008314 kJ/mol . K
• Q is the reaction quotient
ΔG and the Equilibrium Constant
• Thus far in this chapter, we have focused heavily on the relationship
between the free energy change and the spontaneity of a reaction.
• For a reaction to be spontaneous, ΔG° must be negative.
• Another measure of reaction spontaneity is the equilibrium constant,
K.
• For a reaction to be spontaneous, K must be greater than 1.
• This should make sense because we have discussed that if K is greater than 1,
then the reaction is product favored.
ΔG and the Equilibrium Constant
• The relationship between ΔG° and K can be found starting with this
general equation.
• Remember that at equilibrium, ΔG = 0 and Q = K. So,
• Therefore,
ALEKS (Calculating reaction free energy under nonstandard
conditions)
ΔG and the Equilibrium Constant
• This relationship between ΔG° and K holds for all equilibrium
constants we have discussed in this course.
Kc, Kp, Ka, Kb, Kw, Ksp, Kf, Kd
• We can now relate the standard free energy change of a reaction to the
extent of a reaction.
ALEKS (Using thermodynamic data to calculate K)
KNOWLEDGE CHECK 12
• Calculate the Ksp for PbCl2 at 25oC. (The free energies of formation of
Pb2+, Cl-, and PbCl2 are -24.4, -131.2, and -314.1 kJ/mol respectively.)
Remaining topics from acidbase equilibria
Week 4_Part 1
Polyprotic Acids and Bases
• Polyprotic acids such as H2SO4, H3PO4, and H2CO3 contain more than one
ionizable proton
• The protons are lost in a stepwise manner
• The fully protonated species is always the strongest acid
• It is easier to remove a proton from a neutral molecule than from a negatively charged ion
• H2SO4 is more acidic than HSO4-
16-2
ALEKS problem on polyprotic acids
Sulfuric acid H2SO4 is a polyprotic acid. Write balanced chemical
equations for the sequence of reactions that sulfuric acid can undergo
when it’s dissolved in water.
Knowledge Check 1
• Write balanced chemical equations for the 2nd sequence of reaction
of phosphoric acid when it’s dissolved in water. (Hint: Use the 2nd
sequence in the previous slide as a template to guide you.)
ALEKS (Solving a polyprotic acid equilibrium
composition problem)
Knowledge Check 2
• What will be the equilibrium molarity of the sulfate ion if the
prepared concentration of sulfuric acid was 0.015 M instead of 0.031
M. (Hint: You will have to use the quadratic formula. Also, use the
same Ka1 and Ka2 values from the previous problem.)
Molecular Structure and Acid–Base Strength
• Bond Strengths: The stronger the A–H or B–H+ bond
• The less likely the bond is to break to form H+ ions and thus the less acidic the substance
• The trend in bond energies is due to a steady decrease in overlap between the
orbital of hydrogen and the valence orbital of the halogen atom as the size of
the halogen increases
16-7
Knowledge Check 3
• Here are some hypothetical acids and their bond strength in kJ/mol.
Rank these acids from the lowest to the highest pKa. (Assume the
element A, B, C, X, and Y are in the same group in the periodic table.)
H-A = 412 kJ/mol
H-B = 345 kJ/mol
H-C = 555 kJ/mol
H-X = 320 kJ/mol
H-Y = 575 kJ/mol
Acid Base Titrations
An acid–base titration is a method of quantitative analysis for determining the
concentration of an acid or base by exactly neutralizing it with a standard solution
of base or acid having known concentration.
End Point
• The point at which the
indicator changes its color
during titration
Equivalence Point
• The point in a titration where
a stoichiometric amount of
the titrant has been added
End point and choice of an indicator
Properties of a good indicator:
•
The color change must be easily detected
•
The color change must be rapid
•
The indicator molecule must not react with the substance being
titrated
•
The indicator must change color within a specific narrow pH range.
Figure: The Titration of (a) a Strong Acid with a Strong
Base and (b) a Strong Base with a Strong Acid
For the titration of a monoprotic
strong acid (HCl) with a monobasic
strong base (NaOH),
How much 0.20 M NaOH is required to completely neutralize 50.0 mL of a 0.10 M
solution of HCl?
Ans: 25 mL
16-11
Figure: The Titration of (a) a Weak Acid with a Strong Base and (b) a
Weak Base with a Strong Acid
ALEKS (Calculating the pH of a weak acid titrated with a strong base)
In a titration experiment, 5.00 mL of 0.200 M NaOH was added to 50.00 mL of 0.100 M CH3C…
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