The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below. C l 2 ( g ) ⇌ 2 C l ( g ) , K c = 1.2 × 10 − 7

Question

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Answer 1

Question

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

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Answer 2

Question

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

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Answer 3

Question

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

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Answer 4

Question

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

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Answer 5

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

Answer 6

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

Answer 7

Chapter 5, Problem 5I.13E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

Answer 8

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

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

Kc=[Cl]2[Cl2]

Given that, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of Cl2(g) can be calculated as given below.

[Cl2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)Cl2ClInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−7=4x2(0.001)x2=1.2×10−7×10−34x=5.5×10-6[Cl2]=0.001 M[Cl]=2x=2×5.5×10-6=1.1×10-5M

Therefore, the equilibrium concentrations of Cl2 and Cl in a reaction vessel are 0.001M and 1.1×10-5 M respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

Answer 14

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

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

Kc=[F]2[F2]

Given that, 2.0 mmol F2(g) was sealed into a reaction vessel of volume 2.0 L. The concentration of F2(g) can be calculated as given below.

[F2]=molesvolume=2×10−3 mol2.00 L=0.001M

An equilibrium table can be set up as given below.

Composition(M)F2FInitial concentration0.0010Change in concentration−x+2xEquilibrium concentration0.001−x2x

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

Kc=(2x)2(0.001−x)

By assuming x≪0.001, it can be written that 0.001−x≃0.001.

Now, the above equilibrium expression can be solved for x.

Kc=(2x)2(0.001−x)1.2×10−4=4x2(0.001)x2=1.2×10−4×10−34x=5.5×10-8[F2]=0.001 M[F]=2x=2×5.5×10-8=1.1×10-7M

Therefore, the equilibrium concentrations of F2 and F in a reaction vessel are 0.001M and 1.1×10-7 M respectively.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

Answer 20

(c)

Expert Solution

Explanation of Solution

Then dissociation constant of F2 is larger than that of Cl2 at 1000 K. This infers that at 1000 K more F2 will dissociate to its atoms than that of Cl2. Therefore, Cl2 is more stable relative to its atoms at 1000 K.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

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 composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below. C l 2 ( g ) ⇌ 2 C l ( g ) , K c = 1.2 × 10 − 7

Interpretation: The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below. Cl2(g)⇌2Cl(g), Kc=1.2×10−7

Explanation of Solution

Interpretation: The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below. F2(g)⇌2F(g), Kc=1.2×10−4

Explanation of Solution

Interpretation: The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.

Explanation of Solution

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Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

Cl2(g)⇌2Cl(g), Kc=1.2×10−7

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume 2.00 L has to be calculated. The reaction is given below.

F2(g)⇌2F(g), Kc=1.2×10−4

Interpretation:

The one which is thermodynamically more stable relative to its atoms at 1000 K has to be determined.