Download Schedule and sample problems

Thermodynamics- Chapter 19 Schedule and Notes
Date
Topics
Video cast DUE
One
Review of
thermodynamics
1_thermo_review
AND
Assignment during
class time
Review of thermo
Wksheet
2.1ch19_intro
Optional: 1sc_thermo
Two
19.1-4; state function
demo
2.2ch19_qualitative_two
AND
1, 2, 7, 11, 21, 29a, 30a,
31
3.1ch19_qualitative_three
Optional: 3_sc_thermo
Three
19.4-5
3.2video_thermo
AND
5video_thermo
39, 40, 47, 51, 52
Optional: 4sc_thermo
Four
19.5-6
6sc_thermo
53, 55, 57, 60, 64
Five
19.7
7video_thermo
71, 74, 75, 79
Six
Enthalpy, Entropy and
Gibbs Lab
Review
Seven
Eight
Nine
Practice Test Review
and in-class quiz
Test
Practice Tests
Sample Problems and Notes for Review of Thermodynamics VideoCast 2008
Energy and physical or chemical processes.
Measured in ________________________
1st law of Thermodynamics: ____________________________
Endo vs Exo:
Heat at constant pressure is called:
Diagrams:
Ways to find Enthalpy of a Process:
1) Using info from an equation
2) Calorimetry
3) Hess Law
4) Enthalpy of Formation
5) Bond Enthalpies
1) Information from an Equation
2H2O2(l)
2H2O(l) + O2(g) H= -196 kJ
What is the heat (q) [or H, since it is at constant pressure)] when 5.0g of H2O2
decomposes?
Think of the reverse of this question too….
2) Calorimetry
At constant pressure, two equations:
How much heat is needed to warm 250g of H2O from 22C to 98C. cH2O= 4.184 J/gK
What is the molar heat capacity (Cp) of H2O?
More calorimetry Examples:
When a student mixes 50mL of 1.0M HCl and 50mL of 1.0M NaOH in a coffee-cup
calorimeter, the temperature of the resultant solution increases from 21.0˚C to 27.5˚C.
Calculate the enthalpy change for the reaction, assuming that the calorimeter loses only a
negligible quantity of heat, that the total volume of the solution is 100.0mL, that its
density is 1.0g/mL, and that its specific heat is 4.184J/g˚C.
Constant Volume Calorimetry (Bomb Calorimetry):
Methylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. The combustion of
methylhydrazine with oxygen produces nitrogen, carbon dioxide and water.
2CH6N2 + 5O2  2N2 + 2CO2 + 6H2O
When 4.00g of methylhydrazine is combusted in a bomb calorimeter, the temperature of
the calorimeter increases from 25.00˚C to 39.50˚C. In a separate experiment the heat
capacity of the calorimeter is measured to be 7.794 kJ/˚C. What is the heat of reaction for
the combustion of a mole of CH6N2?
3) Hess Law
Calculate H for the reaction: 2C(s) + H2(g)  C2H2(g)
Given the following reactions and their respective enthalpy changes:
C2H2(g) + 5/2O2(g)  2CO2 + H2O H = -1299.6kJ
C(s) + O2(g)  CO2(g) H = -393.5kJ
H2(g) + 1/2O2(g)  H2O H = -285.8kJ
4) Standard Enthalpies of Formation, p. 191ff…
Use values from appendix C, pg 1123, to determine the enthalpy of reaction for the
combustion of pentane gas to form carbon dioxide gas and liquid water. (p193)
5) Using Bond Enthalpies, p.332
Bond enthalpy is the enthalpy change, H, for the breaking of a particular bond in
one mole of a gaseous substance. Use bond enthalpies from p.330 in your text to
estimate the enthalpy of reaction for the combustion of gaseous ethane, producing
carbon dioxide gas and water vapor. p. 332.
Thermochemistry Review Worksheet
1)
The SI unit for energy is the Joule. Show how the units of meter, second and kilogram can be
used to define Joule.
2)
What is the first law of thermodynamics?
3)
Thermodynamic quantities are always identified by a number and a sign (+ or -). What does each
tell us about a system?
4)
How much heat is released when 4.50g of methane gas is burned in a constant–pressure system?
(the H for the combustion of methane is –890kJ/mol)
5)
Large beds of rocks are used in some solar-heated homes to store heat.
a) calculate the quantity of heat absorbed by 50.0kg of rocks if their temperature increases
by 12.0˚C. (Assume the specific heat of the rocks is 0.82 J/gK)
b) What temperature change would these rocks undergo if they absorbed 450kJ of heat?
6)
When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates.
a) write the thermochemical equation for this process assuming H = –65.5kJ/mol
b) the energy for a reaction such as this is sometimes called the “energy of precipitation”
or “heat of precipitation” or “H of formation.” Calculate the H of formation of 0.200
mol of AgCl.
c)
Calculate the H when 0.350 mmol (yes, milli-moles!) of AgCl dissolves in water.
7)
A coffee-cup calorimeter of the type shown in figure 6.5 on page 185 of your text contains 150.0
g of water at 25.1˚C. A 121.0g block of Cu metal is heated to 100.4˚C by putting it in a beaker of
boiling water. The specific heat of Cu(s) is 0.385 J/gK. The Cu is added to the calorimeter and
after a time the contents of the cup reach a constant temperature of 30.1˚C.
a)
Determine the amount of heat, in Joules, lost by the copper block.
b) Determine the amount of heat gained by the water (the specific heat of water is 4.184
J/gK).
c)
The difference between your answers for (a) and (b) is due to the heat loss though the
styrofoam cups and the heat necessary to raise the temperature of the inner wall of the
apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise
the temperature of the apparatus (the cups and the cover) by 1K. Calculate the heat
capacity of the calorimeter in J/K.
d) If only one Styrofoam cup were used in the apparatus, rather than two, would you
expect the heat capacity of the calorimeter to increase, decrease or stay the same?
Explain.
e)
What would be the final temperature of the system if all the heat lost by the Cu block
were absorbed by the water in the calorimeter?
8)
Calculate H for the reaction NO(g) + O(g)  NO2(g)
Given the following:
NO(g) + O3(g)  NO2(g) + O2(g)
H=-198.9kJ
O2(g)  2O(g)
H=495.0kJ
O3(g)  3/2O2(g)
H=-142.3kJ
9)
What is the “standard enthalpy of formation?”
What does the little ˚ mean in the term H˚?
How do you use “standard enthalpies of formation” to find Hrxn?
10)
Use values from Appendix 4 (page A22) in your text to calculate
a) the enthalpy change for the combustion of 1 mol of ethanol
b) the combustion of 4.5g of ethanol.
c)
11)
Now compare the energy in (b) with the combustion of 4.5g of methane gas (#3
above)…which is greater? Why?
Use bond enthalpies from p.330 in your text to estimate the bond enthalpy of a nitrogen-nitrogen
triple bond, when it is produced in the following reaction:
N2H4(g)
N2(g) + 2H2(g) H = -86kJ
Videos 2.1 and 2.2 and 3.1 qualitative aspects of thermodynamics video notes on this
page. Write neat and large and use lots of white space on this and other pages provided
by you.
Sample Problems, Chapter 19 –These note pages start with video 3.2video_thermo
Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction,
or in equilibrium:
a) when a piece of metal heated to 150˚C is added to water at 40˚C, 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 1atm condenses to liquid benzene at the normal boiling
point of benzene, 80.1˚C.
If a process is nonspontaneous, does that mean that it cannot occur under any circumstances?
The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is
–38.9˚C, and its molar enthalpy of fusion is Hfusion = 2.29kJ/mol. What is the entropy change of the system
when 50.0g of Hg(l) freezes at the normal freezing point?
Extra
The normal boiling point of ethanol is 78.3˚C and its molar enthalpy of vaporization is 38.56 kJ/mol. What
is the change in entropy in the system when 68.3g of C2H5OH(g) at 1atm condenses to liquid at the normal
boiling point?
The second law of thermodynamics says that entropy …..
19.3
Microstates…not super important, but helps in understanding…
S = k lnW
What is W?
Predict whether S is positive or negative for each of the following processes, assuming each occurs at a
constant temperature:
a) H2O(l)  H2O(g)
b) Ag+(aq) + Cl-(aq)  AgCl(s)
c) 4Fe(s) + 3O2(g)  2Fe2O3(s)
d) N2(g) + O2(g)  2NO(g)
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 25˚C
b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25˚C
c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298K
If you are told that the entropy of a certain system is zero, what do you know about the system?
19.4
Calculate the S˚ for the synthesis of ammonia from N2(g) and H2(g) at 298K:
N2(g) + 3H2(g)  2NH3(g)
You try:
Using the standard entropies in Appendix C, calculate the standard entropy change, S˚, for the following
reaction at 298K:
Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)
19.5
What are the criteria for spontaneity in terms of:
a) entropy
b) free energy
Calculate the standard free-energy change for the following reaction at 298K:
P4(g) + 6Cl2(g)  4PCl3(g)
What is G˚ for the reverse of this reaction?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H˚ = -2220 kJ
Without using data from appendix C, predict whether the Gibbs Free Energy for this reaction is more
negative or less negative than H˚.
Use data from appendix C to calculate the standard free energy change for the reaction at 298K. Is your
prediction correct?
19.6
The Haber Process: N2(g) + 3H2(g)  2NH3(g)
Assume that H˚ and S˚ for this reaction do not change with temperature.
a) predict the direction in which G˚ for this reaction changes with increasing temperature.
b) Calculate the values of G˚ for the reaction at 25˚C and 500˚C
Using standard enthalpies of formation and standard entropies in appendix C, calculate H˚ and S˚ at
298K for the following reaction:
2SO2(g) + O2(g)  2SO3(g)
Using the values obtained, estimate G˚ at 400K.
19.7
The Haber Process: N2(g) + 3H2(g)  2NH3(g)
Calculate G at 298K for a reaction mixture that consists of 1.0atm N 2, 3.0 atm H2, and 0.50 atm NH3.
Use standard free energies of formation to calculate the equilibrium constant, K, at 25˚C for the reaction
involved in the Haber process:
The Haber Process: N2(g) + 3H2(g)  2NH3(g)
The standard free-energy change for this reaction was calculated in an earlier practice problem…it was –
33.3kJ/mol = -33300 J/mol.
Use data from appendix C to calculate the standard free-energy change, G˚, and the equilibrium constant,
K, at 298K for the reaction: H2(g) + Br2(l)  2HBr(g)
1991 D (Required)
BCl3(g) + NH3(g)  Cl3BNH3(s)
The reaction represented above is a reversible reaction.
(a) Predict the sign of the entropy change, ∆S, as the reaction proceeds to the right. Explain your prediction.
(b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, ∆H. Explain your prediction.
(c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased
above a specific temperature. Explain.
(d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific
temperature at which the direction of the spontaneous reaction changes? Explain.
Answer:
(a) Because a mixture of 2 gases produces a single pure solid, there is an extremely large decrease in
entropy,  ∆S < 0, i.e. the sign of ∆S is negative.
(b) In order for a spontaneous change to occur in the right direction, the enthalpy change must overcome
the entropy change which favors the reactants (left), since nature favors a lower enthalpy, then the
reaction must be exothermic to the right,  ∆H < 0.
(c)
∆G = ∆H – T∆S, the reaction will change direction when the sign of ∆G changes, since ∆H < 0
and ∆S < 0, then at low temperatures the sign of ∆G is negative and spontaneous to the right. At some
higher T, ∆H = T∆S and ∆G = 0, thereafter, any higher temperature will see ∆G as positive and
spontaneous in the left direction.
(d) At equilibrium, K = e–∆G/RT, where ∆G = 0, K = eo = 1
1993 D
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
The reaction represented above is spontaneous at 25˚C. Assume that all reactants and products are in their
standard state.
(a) Predict the sign of ∆S˚ for the reaction and justify your prediction.
(b) What is the sign of ∆G˚ for the reaction? How would the sign and magnitude of ∆G˚ be affected by an
increase in temperature to 50˚C? Explain your answer.
(c) What must be the sign of ∆H˚ for the reaction at 25˚C? How does the total bond energy of the reactants compare to that of the products?
(d) When the reactants are place together in a container, no change is observed even though the reaction
is known to be spontaneous. Explain this observation.
Answer:
(a) ∆S<0. The number of moles of gaseous products is less than the number of moles of gaseous reactants. OR A liquid is formed from gaseous reactants.
(b) ∆G<0. ∆G becomes less negative as the temperature is increased since ∆S < 0 and ∆G = ∆H –T∆S.
The term “–T∆S” adds a positive number to ∆H.
(c) ∆H<0. The bond energy of the reactants is less than the bond energy of the products.
(d) The reaction has a high activation energy; OR is kinetically slow; OR a specific mention of the need
for a catalyst or spark.
1994 D
2 H2S(g) + SO2(g)  3 S(s) + 2 H2O(g)
At 298 K, the standard enthalpy change, ∆H˚ for the reaction represented above is –145 kilojoules.
(a) Predict the sign of the standard entropy change, ∆S˚, for the reaction. Explain the basis for your
prediction.
(b) At 298 K, the forward reaction (i.e., toward the right) is spontaneous. What change, if any, would
occur in the value of ∆G˚ for this reaction as the temperature is increased? Explain your reasoning
using thermodynamic principles.
(c) What change, if any, would occur in the value of the equilibrium constant, Keq, for the situation described in (b)? Explain your reasoning.
(d) The absolute temperature at which the forward reaction becomes nonspontaneous can be predicted.
Write the equation that is used to make the prediction. Why does this equation predict only an approximate value for the temperature?
Answer:
(a) ∆S˚ is negative (–). A high entropy mixture of two kinds of gases forms into a low entropy solid and a
pure gas; 3 molecules of gas makes 2 molecules of gas, fewer gas molecules is at a lower entropy.
(b) ∆G˚ < 0 if spontaneous. ∆G˚ = ∆H˚ – T∆S˚ Since ∆S˚ is neg. (–), as T gets larger, –T∆S˚ will become
larger than +145 kJ and the sign of ∆G˚ becomes pos. (+) and the reaction is non–spontanseous.
(c) When –T∆S˚ < +145 kJ, Keq > 1,
when –T∆S˚ = +145 kJ, Keq = 1,
when –T∆S˚ > +145 kJ, Keq < 1, but > 0
∆H˚
(d) ∆G = 0 at this point, the equation is T =
; this assumes that ∆H and/or S do not change with tem∆S˚
perature; not a perfect assumption leading to errors in the calculation.
1996 B
C2H2(g) + 2 H2(g)  C2H6(g)
Information about the substances involved in the reaction represented above is summarized in the following
tables.
S˚ (J/mol•K)
∆H˚f (kJ/mol)
C2H2(g)
200.9
226.7
H2(g)
130.7
0
C2H6(g)
----
–84.7
Substance
Bond
Bond Energy (kJ/mol)
C–C
347
C=C
611
C–H
414
H–H
436
(a)
If the value of the standard entropy change, ∆S˚, for the reaction is –232.7 joules per mole•Kelvin,
calculate the standard molar entropy, S˚, of C2H6 gas.
(b)
Calculate the value of the standard free-energy change, ∆G˚, for the reaction. What does the sign of
∆G˚ indicate about the reaction above?
Calculate the value of the equilibrium constant, K, for the reaction at 298 K.
Calculate the value of the CC bond energy in C2H2 in kilojoules per mole.
(c)
(d)
Answer:
(a) –232.7 J/K = S˚(C2H6) – [2(130.7) + 200.9] J/K
S˚(C2H6) = 229.6 J/K
(b)
(c)
(d)
∆H˚ = ∑ ∆Hƒ˚(products) – ∑ ∆Hƒ˚(reactants)
= –84.7 kJ – [226.7 + 2(0)] kJ = –311.4 kJ
∆G˚ = ∆H˚ – T∆S˚
= –311.4 kJ – (298K)(–0.2327 kJ/K) = –242.1 kJ
A ∆G˚ < 0 (a negative ∆G˚) indicates a spontaneous forward reaction.
Keq = e–∆G/RT = e–(–242100/(8.314)(298))
= 2.741042
∆H˚ = bond energy of products – bond energy of reactants
–311.4 kJ = [(2)(436) + Ecc + (2)(414)] – [347 + (6)(414)] kJ
Ecc = 820 kJ
1997 D
For the gaseous equilibrium represented below, it is observed that greater amounts of PCl 3 and Cl2 are produced as the temperature is increased.
PCl5(g)  PCl3(g) + Cl2(g)
(a)
(b)
(c)
(d)
What is the sign of ∆S˚ for the reaction? Explain.
What change, if any, will occur in ∆G˚ for the reaction as the temperature is increased? Explain your
reasoning in terms of thermodynamic principles.
If He gas is added to the original reaction mixture at constant volume and temperature, what will
happen to the partial pressure of Cl2? Explain.
If the volume of the reaction mixture is decreased at constant temperature to half the original volume,
what will happen to the number of moles of Cl2 in the reaction vessel? Explain.
Answer:
(a) The sign of ∆S˚ is (+). There is an increase in the number of gas molecules as well as a change from a
pure gas to a mixture of gases.
(b) ∆G˚ = ∆H˚ – T∆S˚. Both ∆S˚ and ∆H˚ are (+). As temperature increases, at some point the sign of ∆G˚
will change from (+) to (–), when the system will become spontaneous.
(c) There will be no change in the partial pressure of the chlorine. Without a volume or temperature
change, the pressure is independent of the other gases that are present.
(d) The number of moles of Cl2 will decrease. The decrease in volume will result in an increase in
pressure and, according to LeChatelier’s Principle, the equilibrium system will shift to the left (the
side with fewer gas molecules) to reduce this increase in pressure. This will cause a decrease in the
number of moles of products and an increase in the number of moles of reactant.
THERMODYNAMICS
ENTHALPY, ENTROPY and GIBBS FREE ENERGY
OBJECTIVES
1. To measure enthalpy.
2. To calculate entropy.
3. To calculate Gibbs free energy.
The Gibbs free energy is (delta G) is a measure of the spontaneity of a process and of the useful energy
available from it.
Negative delta G indicates a spontaneous process.
Positive delta G indicates a nonspontaneous process or a process spontaneous in opposite direction
Delta G is calculated from the equation:
Delta G= delta H – T delta S
In today’s experiments we will measure experimentally change in enthalpy, delta H, and obtain change in
entropy, delta S, from the table data.
After completing the calculations answer the following questions separately for each experiment:
1. Is the dissolving exothermic or endothermic?
2. Does the entropy decreases or increases during dissolving?
3. Is the Gibbs free energy positive or negative? Is the process of dissolving spontaneous?
Teacher Notes -EXAMPLES OF CALCULATIONS:
1. Measurement and calculations of delta H.
22.0 g of magnesium sulfate was dissolved in 300. mL of water at 23.3 degree C in a coffee-cup calorimeter.
After dissolving, the temperature increased to 36.6 degree C. Write the thermochemical equation for the
reaction, that is, calculate the enthalpy for the reaction:
MgSO4(s) --> Mg2+(aq) + SO42(aq)
Assume that the calorimeter loses only a negligible quantity of heat, density of the solutions is 1.00 g/mL,
and the specific heat of the solutions is 4.184 J/g K. Since adding the crystals doesn’t change the volume
noticeably, assume the mass of solution to be 300. g.
Step 1. Calculate the heat produced in the reaction:
heat = sp. heat of solution x mass of solution x
or
qsoln =
csoln
x
masssoln
qsoln =4.184 J/g K
delta T
x
delta Tsoln
x 300. g x 13.3 K = 16694 J = 16.7 kJ
Step 2. Calculate how many moles are 22.0 g of magnesium sulfate.
22.0 g x 1 mol/120.4 g = 0.183 mol MgSO4
Step 3. Calculate the heat evolved per one mole of magnesium sulfate. This is the sought delta H
value. Enthalpy will be negative, (), since it is an exothermic reaction (the temperature went up).
1 mol x 16.7 kJ/0.183 mol = 91.3 kJ
Step 4. Rewrite the equation with state symbols and with the delta H value. Enthalpy is expressed
per number of moles in the equation, in this case, per one mole.
MgSO4(s) –>Mg2+(aq) + SO42(aq)
delta H = - 91.3 kJ
2. Delta S from standard tabulated data.
S (J/K mol)
MgSO4(s)
91.6
Mg2+(aq)
- 138.1
SO42(aq)
20.1
Delta S = S products - S reactants
Delta S = [Mg2+(aq) + SO42(aq)] - MgSO4(s)
Delta S = [-138.1 J/K mol + 20.1 J/K mol] - 91.6 J/K mol = - 209.6 J/K mol
This value indicates that entropy decreases during the dissolving. This is a rare case. It is related to
formation of regular structures of water surrounding the ions.
3. Calculation of the Gibbs free energy.
In all calculations we will use the standard temperature 298 K which is close to a room temperature.
Enthalpy and entropy doesn’t change significantly in our range of temperatures. Remember that H is
usually expresses in kJ and S in J. Perform appropriate conversions before calculating G.
Delta G= delta H – T delta S
Delta G = - 91300 J/mol - [298 K x (- 209.6 J/K mol)] = - 28800 J = - 28.8 kJ/mol
Negative value of delta G indicates that dissolving magnesium sulfate is spontaneous.