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Homework Chapter 19
Homework Chapter 19
Due: 11:59pm on Wednesday, November 16, 2016
You will receive no credit for items you complete after the assignment is due. Grading Policy
Sample Exercise 19.1 Practice Exercise 1 with feedback
Part A ­ Identifying Spontaneous Processes
The process of iron being oxidized to make iron(III) oxide (rust) is spontaneous. Which of these statements about this process is/are true?
ANSWER:
The reduction of iron(III) oxide to iron is also spontaneous.
The oxidation of iron is endothermic.
Equilibrium is achieved in a closed system when the rate of iron oxidation is equal to the rate of iron(III) oxide reduction.
Because the process is spontaneous, the oxidation of iron must be fast.
The energy of the universe is decreased when iron is oxidized to rust.
Correct
Equilibrium occurs when the rate of the forward reaction is equal to the rate of the reverse reaction. Thus, equilibrium is achieved in a closed system
when the rate of iron oxidation is equal to the rate of iron(III) oxide reduction.
Since the process of iron being oxidized is spontaneous, the reverse reaction would not be considered spontaneous. The energy of the universe will
remain constant regardless, however, following the first law of thermodynamics, which states that energy is conserved.
Two predictions we cannot make are regarding the speed and ethalpy of the reaction. Spontaneous reacts may occur at any rate, and may be
endothermic or exothermic.
Problem 19.11
Part A
Which of the following processes are spontaneous, and which are nonspontaneous? (a) the ripening of a banana, (b) dissolution of sugar in a cup of hot coffee, (c) the reaction of nitrogen atoms to form N2 molecules at 25 ∘ C and 1 atm,
(d) lightning, (e) formation of CH4 and O2 molecules from CO2 and H2 O at room temperature and 1 atm of pressure.
Sort these processes into the proper categories.
ANSWER:
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Sample Exercise 19.2 Practice Exercise 1 with feedback
Part A ­ Calculating ΔS for a Phase Change
Do all exothermic phase changes have a negative value for the entropy change of the system?
ANSWER:
Yes, because the heat transferred from the system has a negative sign.
Yes, because the temperature decreases during the phase transition.
No, because the entropy change depends on the sign of the heat transferred to or from the system.
No, because the heat transferred to the system has a positive sign.
More than one of the previous answers is correct.
Correct
Problem 19.28
Part A
Does the entropy of the surroundings increase for spontaneous processes?
ANSWER:
Yes.
No.
Not necessarily.
Correct
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Homework Chapter 19
Part B
In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of ΔSsurr ?
ANSWER:
ΔSsurr is negative and less than the magnitude of the decrease in ΔSsys .
ΔSsurr is positive and greater than the magnitude of the decrease in
ΔSsys
.
ΔSsurr is positive and less than the magnitude of the decrease in ΔSsys .
ΔSsurr is negative and greater than the magnitude of the decrease in ΔSsys .
Correct
Part C
During a certain reversible process, the surroundings undergo an entropy change, ΔSsurr = ­ 73 What is the entropy change of the system for this process?
. J/K
Express your answer using two significant figures.
ANSWER:
ΔSsys
= 73 J/K Correct
Problem 19.4
Part A
Predict the sign of ΔS accompanying reaction on the figure.
ANSWER:
ΔS > 0
ΔS < 0
ΔS = 0
Correct
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Part B
ΔH
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Homework Chapter 19
Predict the sign of ΔH accompanying this reaction on the figure.
ANSWER:
ΔH
> 0
ΔH
< 0
ΔH
= 0
Correct
Part C
Explain your choice.
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before
submitting your answer.
ANSWER:
Reset
negative
five blue­blue
twenty blue­red
five blue­blue and twenty
blue­red
Help
1. Both ΔH and ΔS for this reaction are positive . The net change is breaking five blue­blue
bonds. Enthalpies for this process are always positive .
In the depicted reaction, both reactants and products are in the gas phase (they are far apart and
randomly placed). There are twice as many molecules (or moles) of gas in the products, so
ΔH
is positive for this reaction.
the same amount of
twice as many
positive
half as many
Correct
Sample Exercise 19.3 Practice Exercise 1 with feedback
Part A ­ Predicting the Sign of ΔS
Select each process that produces an increase in the entropy of the system:
Check all that apply.
ANSWER:
CaO(s) + CO2 (g) → CaCO3 (s)
CO2 (s) → CO2 (g)
2 SO2 (g) + O2 (g) → 2 SO3 (g)
HCl(g) + NH3 (g) → NH4 Cl(s)
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The entropy of a system is expected to increase for processes in which
1. Gases form from either solids or liquids.
2. Liquids or solutions form from solids.
3. The number of gas molecules increases during a chemical reaction.
Problem 19.38
Part A
How does the entropy of the system change when the temperature of the system increases?
ANSWER:
increases
decreases
stays the same
Correct
Part B
How does the entropy of the system change when the volume of a gas increases?
ANSWER:
increases
decreases
stays the same
Correct
Part C
How does the entropy of the system change when equal volumes of ethanol and water are mixed to form a solution?
ANSWER:
increases
decreases
stays the same
Correct
Problem 19.44
Predict the sign of ΔSsys for each of the following processes.
Part A
Molten gold solidifies.
ANSWER:
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Homework Chapter 19
ΔSsys
> 0
ΔSsys
< 0
Correct
Part B
Gaseous Cl2 in the stratosphere to form gaseous Cl atoms.
ANSWER:
ΔSsys
> 0
\Delta{\rm S}_{\rm sys}<0
Correct
Part C
Gaseous \rm CO reacts with gaseous \rm H_2to form liquid methanol, \rm CH_3OH.
ANSWER:
\Delta{\rm S}_{\rm sys}>0
\Delta{\rm S}_{\rm sys}<0
Correct
Part D
Calcium phosphate precipitates upon mixing \rm Ca(NO_3)_2(aq) and \rm (NH_4)_3PO_4(aq).
ANSWER:
\Delta{\rm S}_{\rm sys}>0
\Delta{\rm S}_{\rm sys}<0
Correct
Sample Exercise 19.5 Practice Exercise 1 with feedback
Part A ­ Calculating \Delta S^\circ from Tabulated Entropies
Using the standard molar entropies below, calculate the standard entropy change, \Delta S^\circ , for the “water­splitting” reaction at 298 \rm K:
\rm 2H_2O(l)~\rightarrow~\rm 2H_2(g)~+~O_2(g)
Standard Molar Entropies of Selected Substances at 298 \rm K
Substance
S^\circ (\rm J/mol \cdot K)
\rm H_2(g)
130.6
\rm O_2(g)
205.0
\rm H_2O(g)
188.8
\rm H_2O(l)
69.9
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ANSWER:
326.4 \rm J/K
265.7 \rm J/K
163.2 \rm J/K
88.5 \rm J/K
−326.3 \rm J/K
Correct
The value of \Delta S^\circ is positive, and corresponds to the qualitative prediction that the entropy would increase because of an increase in gas
molecules and change of state from liquid to gas.
Sample Exercise 19.6 Practice Exercise 1 with feedback
Part A ­ Calculating Free­Energy Change from \Delta H^\circ, T, and \Delta S^\circ.
Which of these statements is true?
ANSWER:
All spontaneous reactions have a negative free energy change.
All spontaneous reactions have a negative entropy change.
All spontaneous reactions have a positive entropy change.
All spontaneous reactions have a positive free energy change.
All spontaneous reactions have a negative enthalpy change.
Correct
The change in free energy, \Delta G can be used to determine the spontaneity of a reaction:
If \Delta G~<~0, the reaction is spontaneous in the forward direction.
If \Delta G~=~0, the reaction is at equilibrium.
If \Delta G~>~0, the reaction in the forward direction is nonspontaneous (work must be done to make it occur) but the reverse reaction
is spontaneous.
Sample Exercise 19.8 Practice Exercise 1 with feedback
Part A ­ Predicting and Calculating \Delta G^\circ
If a reaction is exothermic and its entropy change is positive, which statement is true?
ANSWER:
The reaction is spontaneous at all temperatures.
The reaction is spontaneous only at lower temperatures.
The reaction is nonspontaneous at all temperatures.
The reaction is spontaneous only at higher temperatures.
Correct
For an exothermic reaction, the enthalpy change is negative, and since the entropy of the reaction is positive, T\Delta S^\circ is positive. Using the
free­energy equation, it can be deduced that the reaction is spontaneous at all temperatures because regardless the temperature a positive value
(T\Delta S^\circ) is always being subtracted by a negative value (\Delta H^\circ) to give a negative value.
\Delta G^\circ~=~\Delta H^\circ~­~T\Delta S^\circ
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Homework Chapter 19
Sample Exercise 19.9 Practice Exercise 1 with feedback
Part A ­ Determining the Effect of Temperature on Spontaneity
Given that \Delta H^\circ=­92.38~\rm kJ and \Delta S^\circ=­198.3~\rm J/K, what is the temperature above which the Haber ammonia process becomes
nonspontaneous?
ANSWER:
25 ^\circ \rm C
47 ^\circ \rm C
61 ^\circ \rm C
193 ^\circ \rm C
500 ^\circ \rm C
Correct
Sample Exercise 19.10 Practice Exercise 1 with feedback
Part A ­ Relating \Delta G to a Phase Change at Equilibrium
If the normal boiling point of a liquid is 67 ^\circ \rm C, and the standard molar entropy change for the boiling process is +100 \rm J/K, estimate the
standard molar enthalpy change for the boiling process.
ANSWER:
+6700 \rm J
­6700 \rm J
+34,000 \rm J
­34,000 \rm J
Correct
Sample Exercise 19.11 Practice Exercise 1 with feedback
Part A ­ Calculating the Free­Energy Change under Nonstandard Conditions
Which of the following statements is true?
ANSWER:
If a reaction is spontaneous under standard conditions, it is spontaneous under all conditions.
If Q>1, \Delta G>\Delta G^\circ .
The larger the Q, the larger the \Delta G^\circ .
If Q = 0, the system is at equilibrium.
The free­energy change for a reaction is independent of temperature.
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By observing the free­energy change equation for nonstandard conditions, it can be deduced that as Q~>~1, \Delta G~>~\Delta G^\circ.
\Delta G~=~\Delta G^\circ~+~RT~{\rm ln}Q
Keep in mind that \texttip{Q}{Q} only effects \Delta G (free­energy under non­standard conditions), and \Delta G^\circ is a constant under standard
conditions. \Delta G is dependent on temperature, and the spontaneity of a reaction may change depending on the temperature. Q = K at equilibrium.
Sample Exercise 19.12 Practice Exercise 1 with feedback
Part A ­ Calculating an Equilibrium Constant from \Delta G^\circ
The K_{\rm sp} for a very insoluble salt is 4.2\times10^{­47} at 298 \rm K. What is \Delta G^\circ for the dissolution of the salt in water?
ANSWER:
­265 \rm kJ/mol
­115 \rm kJ/mol
­2.61 \rm kJ/mol
+115 \rm kJ/mol
+265 \rm kJ/mol
Correct
Since the salt is very insoluble, the \Delta G^\circ, should be greater than 0 because it is nonspontaneous.
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