The equilibrium concentrations for all substances have to be calculated.

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Don't used hand raiting and don't used Ai solution

1. Base on this experimental results, how do you know that the product which you are turning in is methyl 3-nitrobenzoate(meta substituted product ) rather than either of the other two products? 2. What observation suggests that at least a small amount of one or both of the other two isomers are in the mother liquor?

Explain Huckel's rule.

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

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.