The equilibrium concentrations for all substances have to be calculated.

Question

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Answer 1

Question

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

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Answer 2

Question

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

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Answer 3

Question

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

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Answer 4

Question

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

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Answer 5

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

Answer 6

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

Answer 7

Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M[CO]=0.1−x=0.0335 M[H2O]=0.1−x=0.0335 M[CO2]=x=0.0665M[H2]=x=0.0665M

Therefore, the equilibrium concentrations of CO, H2O, CO2, and H2 are 0.0335 M, 0.0335 M, 0.0665 M, and 0.0665 M respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

CO(g)+H2O(g)⇌CO2(g)+H2(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

Kc=[CO2][H2][CO][H2O]

Given that, 1.00 mol H2O and 1.00 mol CO are placed in a reaction vessel of volume 10.00 L. Their concentrations can be calculated as given below.

[CO]=molesvolume=1 mol10.00 L=0.1M[H2O]=molesvolume=1 mol10.00 L=0.1M

An equilibrium table can be set up as given below.

Composition(M)COH2OCO2H2Initial concentration0.10.100Change in concentration−x−x+x+xEquilibrium concentration0.1−x0.1−xxx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kc=(x)(x)(0.1−x)(0.1−x)

It is given that, at equilibrium the number of moles of carbon dioxide gas is 0.665 mol. Its concentration is given below.

At equilibrium[CO2]=molesvolume=0.665 mol10.00 L=0.0665MSo, x=0.0665M

By plugging all data in the equilibrium constant expression, the value of Kc can be obtained.

Kc=(x)(x)(0.1−x)(0.1−x)=(0.0665)2(0.1−0.0665)2Kc=3.94

Therefore, the value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3.94.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

The equilibrium concentrations for all substances have to be calculated.

Interpretation: The equilibrium concentrations for all substances have to be calculated.

Explanation of Solution

Interpretation: The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.

Explanation of Solution

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Chapter 5 Solutions

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

Interpretation:

The value of Kc for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) has to be calculated.