Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd 23rd PDF

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd 23rd...

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2021/3/22

Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd Due: 12:00am on Thursday, May 13, 2021 To understand how points are awarded, read the Grading Policy for this assignment.

Exercise 19.44 - Enhanced - with Feedback MISSED THIS? Read Section 19.6 (Pages 863 - 867) ; Watch KCV 19.6, IWE 19.4 . Calculate the change in Gibbs free energy for each of the following sets of their standard states.)

,

, and

. (Assume that all reactants and products are in

Part A ,

,

Express your answer as an integer. ANSWER: = -46

Correct The change in Gibbs free energy is expressed as follows:

Substituting in the given values results in the following equation:

Be sure to express both

and

in the same units when doing calculations.

Part B ,

,

Express your answer as an integer. ANSWER: = 38

Correct The change in Gibbs free energy is expressed as follows:

Substituting in the given values results in the following equation:

Be sure to express both

and

in the same units when doing calculations.

Part C ,

,

Express your answer as an integer. ANSWER: = 134

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Correct The change in Gibbs free energy is expressed as follows:

Substituting in the given values results in the following equation:

Be sure to express both

and

in the same units when doing calculations.

Part D ,

,

Express your answer as an integer. ANSWER: = -149

Correct The change in Gibbs free energy is expressed as follows:

Substituting in the given values results in the following equation:

Be sure to express both

and

in the same units when doing calculations.

Part E Predict whether or not the reaction in each part will be spontaneous at the temperature indicated. Drag the appropriate items to their respective bins. ANSWER:

Reset

90. 150 391 90. 150 291

90. 150 291

, ,

, ,

90. 150 856

Help

, ,

, ,

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Correct A decrease in Gibbs free energy ( ) corresponds to a spontaneous process, whereas an increase in Gibbs free energy ( ) corresponds to a nonspontaneous process. According to that, then for Part A:

, the process is spontaneous;

for Part B:

, the process is nonspontaneous;

for Part C:

, the process is nonspontaneous; and

for Part D:

, the process is spontaneous.

Exercise 19.46 - Enhanced - with Feedback MISSED THIS? Read Section 19.6 (Pages 863 - 867) ; Watch KCV 19.6, IWE 19.4 . Consider the following reaction: -1269.8

;

-364.6

Assume that all reactants and products are in their standard states.

Part A Calculate the free energy change for the reaction at 27

.

Express your answer using four significant figures. ANSWER: = -1160

Correct The change in Gibbs free energy at constant temperature is calculated using the following equation: Gibbs free energy is also called the chemical potential because it determines the direction of spontaneous change for chemical systems. Chemical systems have an affinity towards a lower Gibbs free energy.

Part B Is the reaction spontaneous? ANSWER: spontaneous nonspontaneous

Correct The reaction to produce calcium oxide is spontaneous because things to know about Gibbs free energy:

. At constant temperature and pressure, there are three

is proportional to the negative entropy of the universe . A decrease in Gibbs free energy corresponds to a spontaneous process. An increase in Gibbs free energy corresponds to a nonspontaneous process.

Exercise 19.52 - Enhanced - with Feedback MISSED THIS? Read Section 19.7 (Pages 867 - 871) ; Watch KCV 19.7 .

Part A For each pair of substances, choose the one that you expect to have the higher standard molar entropy

at 25

.

Match the items in the left column to the appropriate blanks in the sentences on the right. ANSWER:

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Reset

Between

and

,

Help

will have the higher standard molar

entropy because a solution has more entropy than a solid crystal . Between

and

,

will have the higher standard molar

entropy because it has a greater molar mass and complexity . Between

and

,

will have the higher standard molar entropy because

it is in the gas phase .

Correct Between and , has a higher standard molar entropy at 25 because a solution has a higher entropy than a solid crystal. Between and , has a higher standard molar entropy at 25 because it has a greater molar mass and complexity. Between and , has a higher standard molar entropy at 25 because it is in the gas phase.

Part B For each pair of substances, choose the one that you expect to have the higher standard molar entropy

at 25

.

Match the items in the left column to the appropriate blanks in the sentences on the right. ANSWER:

Reset

Between

and

,

Help

will have the higher standard molar entropy because

it has a greater molar mass . Between

and

,

will have the higher standard molar entropy

because it has a greater molar mass and complexity . Between

and

,

will have the higher

standard molar entropy because it has a greater complexity .

Correct Between

and and

, ,

Between greater molecular complexity.

has a higher standard molar entropy at 25 because it has a greater molar mass. Between has a higher standard molar entropy at 25 because it has a greater molar mass and complexity. and , has a higher standard molar entropy at 25 because it has a

Exercise 19.56 - Enhanced - with Feedback https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=9839282

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

MISSED THIS? Read Section 19.7 (Pages 867 - 871) ; Watch IWE 19.5 . Use data from Appendix IIB to calculate

for each of the reactions given.

Part A

Express your answer as an integer. ANSWER: = -288

Correct The standard entropies for each reactant and product are presented in the table below: Reactant or product 240.1 70.0 146 210.8 is determined by subtracting the standard entropies of the reactants multiplied by their stoichiometric coefficients from the standard entropies of the products multiplied by their stoichiometric coefficients:

Part B

Express your answer using one decimal place. ANSWER: = 14.7

Correct The standard entropies for each reactant and product are presented in the table below: Reactant or product 81.2 197.7 23.8 213.8 is determined by subtracting the standard entropies of the reactants multiplied by their stoichiometric coefficients from the standard entropies of the products multiplied by their stoichiometric coefficients:

Part C

Express your answer using one decimal place. ANSWER: = -94.0

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Correct The standard entropiies for each reactant and product are presented in the table below: Reactant or product 248.2 205.2 256.8 is determined by subtracting the standard entropies of the reactants multiplied by their stoichiometric coefficients from the standard entropies of the products multiplied by their stoichiometric coefficients:

Part D

Express your answer using one decimal place. ANSWER: = 119.6

Correct The standard entropies for each reactant and product are presented in the table below: Reactant or product 304.4 130.7 191.6 188.8 is determined by subtracting the standard entropies of the reactants multiplied by their stoichiometric coefficients from the standard entropies of the products multiplied by their stoichiometric coefficients:

Part E Rationalize the sign of

in each part.

ANSWER:

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Reset

no change an increase a decrease

For the reaction in Part A,

Help

is

negative, which is consistent with a decrease

in the number of moles of

gas. For the reaction in Part B,

is small

and positive, which is consistent with no change

in the number of moles of

gas. For the reaction in Part C,

is

negative, which is consistent with a decrease

in the number of moles of

gas. For the reaction in Part D,

is

positive, which is consistent with an increase

in the number of moles of a

complex gas.

Correct The reaction described in Part A has 3 of gas on the left side of the equation and only 1 of gas on the right side of the equation. So is negative, which is consistent with a decrease in the number of moles of gas. In Part B, carbon dioxide molecules have more complexity than carbon monoxide molecules, but chromium oxide is more complex than chromium metal. The result is that the entropy change is slightly positive. You may expect the entropy change to be zero; however, the specific arrangements of reactants and products mean that the entropy is often slightly positive or negative. The reaction described in Part C has 1.5 of gas on the left side of the equation and only 1 of gas on the right side of the equation. So is negative, which is consistent with a decrease in the number of moles of gas. In Part D, although the number of moles of gas does not change, both hydrogen and nitrogen gases are simple gases. Since the reactant has 1 of a more complex gas and the product has 4 of a more complex gas, the entropy change is positive.

Exercise 19.60 - Enhanced - with Feedback MISSED THIS? Read Section 19.8 (Pages 871 - 877) ; Watch IWE 19.6. In photosynthesis, plants form glucose

and oxygen from carbon dioxide and water.

Part A Write a balanced equation for photosynthesis. ANSWER:

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Correct The balanced equation for photosynthesis is Element

Reactants

Products

Carbon

6

6

Oxygen

18

18

Hydrogen

12

12

From the table, you can see that the number of each atom is the same in both the reactants and products.

Part B Calculate

at 15

.

Express your answer in units of kilojoules. ANSWER: =

Part C Calculate

at 15

.

Express your answer in joules per kelvin. ANSWER: =

Part D Calculate

at 15

.

Express your answer in kilojoules. ANSWER: =

Part E Is photosynthesis spontaneous? ANSWER: spontaneous nonspontaneous

Exercise 19.62 - Enhanced - with Feedback MISSED THIS? Read Section 19.8 (Pages 871 - 877) ; Watch IWE 19.6. For each of the following reactions, calculate , , and at 25 . State whether or not the reaction is spontaneous. If the reaction is not spontaneous, would a change in temperature make it spontaneous? If so, should the temperature be raised or lowered from 25

?

Part A

Express your answer to one decimal place. ANSWER:

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd =

Part B Calculate

at 25

.

Express your answer to one decimal place. ANSWER: =

Part C Complete previous part(s) Part D

Express your answer to one decimal place. ANSWER: =

Part E Calculate

at 25

.

Express your answer to one decimal place. ANSWER: =

Part F Complete previous part(s) Part G

Express your answer using three significant figures. ANSWER: =

Part H Calculate

at 25

.

Express your answer to one decimal place. ANSWER: =

Part I Complete previous part(s) Part J

Express your answer to one decimal place. ANSWER: =

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Part K Calculate

at 25

.

Express your answer to one decimal place. ANSWER: =

Part L Complete previous part(s) Part M Complete previous part(s)

Exercise 19.64 - Enhanced - with Feedback MISSED THIS? Read Section 19.8 (Pages 871 - 877) . Use standard free energies of formation to calculate

at 25

for each of the following reactions.

Substance 228.6 237.1 16.4 87.6 137.2 394.4 50.5 209.9 32.0 159.4 1129.1 603.3

Part A

Express your answer to one decimal place and include the appropriate units. ANSWER: =

Part B

Express your answer to one decimal place and include the appropriate units. ANSWER: =

Part C

Express your answer to one decimal place and include the appropriate units. ANSWER: =

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

Part D

Express your answer to one decimal place and include the appropriate units. ANSWER: =

Exercise 19.66 - Enhanced - with Feedback MISSED THIS? Read Section 19.8 (Pages 871 - 877) ; Watch IWE 19.6. Consider the following reaction:

Estimate for this reaction at each of the following temperatures. (Assume that temperature range.)

and

do not change too much within the given

Part A 290 Express your answer using one decimal place. ANSWER: =

Part B 1030 Express your answer using one decimal place. ANSWER: =

Part C 1445 Express your answer using one decimal place. ANSWER: =

Part D Predict whether or not the reaction in each part will be spontaneous. Drag the appropriate items to their respective bins. ANSWER:

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd Reset

Reaction conducted at 290

Help

Reaction conducted at 1030

Reaction conducted at 1445

Exercise 19.70 - Enhanced - with Feedback MISSED THIS? Read Section 19.9 (Pages 878 - 881) ; Watch IWE 19.10 . Consider the evaporation of methanol at 25.0

:

.

Part A Find

at 25.0

.

Express the free energy change in kilojoules to one decimal place. ANSWER: =

The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find under the given nonstandard conditions.

at 25.0

Part B

Express the free energy change in kilojoules to one decimal place. ANSWER: =

Part C

Express the free energy change in kilojoules to one decimal place. ANSWER: =

Part D

Express the free energy change in kilojoules to one decimal place.

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd

ANSWER: =

Part E The vapor pressure of methanol is 143 and standard pressure (760 ).

. Identify the best reason to explain why methanol spontaneously evaporates in open air at 25.0

ANSWER: Heat can be added under these nonstandard conditions to increase the vapor pressure of methanol. Under these nonstandard conditions, the partial pressure of methanol is lower than its vapor pressure. Calculations for a closed container show a partial pressure of methanol lower than its vapor pressure. The calculated standard free energy change corresponds to a spontaneous reaction.

Exercise 19.72 - Enhanced - with Feedback and Hints MISSED THIS? Read Section 19.9 (Pages 878 - 881) ; Watch IWE 19.10 .

Part A Consider the following reaction:

Calculate

for this reaction at 25

for

is

under these conditions:

,

for

is

, and

for

is

.

Express the energy change in kilojoules per mole to one decimal place. You did not open hints for this part. ANSWER: =

Exercise 19.76 - Enhanced - with Feedback MISSED THIS? Read Sections 19.9 (Pages 878 - 884) , 19.10 (Page) ; Watch IWEs 19.10, 19.11 . Consider the following reaction: 81.9 at 25 . Calculate for the reaction at 25

under each of the following conditions.

Part A standard conditions Express your answer in kilojoules. ANSWER: =

Part B at equilibrium Express your answer in kilojoules. ANSWER:

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Chapters 19 Workshop Activity - Gibbs Free Energy March 22nd/23rd =

Part C 2.55 0.317 0.213

; ; .

Express your answer using one significant figure. ANSWER: =

Exercise 19.78 - Enhanced - with Feedback and Hints MISSED THIS? Read Sections 19.8 (Pages 871 - 877) , 19.10 (Pages 881 - 884) ; Watch IWE 19.11 . Estimate the value of...