The below reaction is at equilibrium or not has to be determined. P C l 3 ( g ) + C l 2 ( g ) ⇌ P C l 5 ( g ) , K c = 0.56 Concept Introduction: Consider a gaseous reaction of the form as given below. a A ( g ) + b B ( g ) → c C ( g ) + d D ( g ) The expression of the reaction quotient can be written as given below. Q c = [ C ] c [ D ] d [ A ] a [ B ] b Where, [ C ] and [ D ] are concentrations of product C and D and [ A ] and [ B ] are concentrations of reactant A and B. Direction of a reaction: If Q < K, the reaction has tendency to proceed toward products. If Q = K, the reaction has achieved the equilibrium position. If Q > K, the reaction has tendency to proceed toward reactants. Where, Q is reaction quotient.

Question

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Answer 1

Question

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

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Answer 2

Question

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

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Answer 3

Question

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

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Answer 4

Question

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

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Answer 5

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

Answer 6

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

Answer 7

Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

Concept Introduction:

Consider a gaseous reaction of the form as given below.

aA(g)+bB(g)→cC(g)+dD(g)

The expression of the reaction quotient can be written as given below.

Qc=[C]c[D]d[A]a[B]b

Where, [C] and [D] are concentrations of product C and D and [A] and [B] are concentrations of reactant A and B.

Direction of a reaction:

If Q < K, the reaction has tendency to proceed toward products.

If Q = K, the reaction has achieved the equilibrium position.

If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

Answer 8

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

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

Qc=[PCl5][PCl3][Cl2]

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

By plugging all data in the above equation, the value of Q can be calculated as given below.

Qc=[PCl5][PCl3][Cl2]Qc=3(6)(1)Qc=0.5

Therefore, the reaction quotient for PCl3(g)+Cl2(g)⇌PCl5(g) is 0.5.

Here, Q≠K. So the reaction is not at equilibrium.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

PCl3(g)+Cl2(g)⇌PCl5(g), Kc=0.56

The value of Q for the reaction is 0.5.

Therefore, Qc<K. So the reaction will proceed to form products.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Answer 20

(c)

Expert Solution

Explanation of Solution

The concentrations of each substance can be calculated as given below.

[PCl5]=MolesVolume=1.50 mol0.500 L=3 M[PCl3]=MolesVolume=3.00 mol0.500 L=6 M[Cl2]=MolesVolume=0.5 00mol0.500 L=1 M

An equilibrium table can be set up as given below.

Composition(M)PCl3Cl2PCl5Initial concentration613Change in concentration−x−x+xEquilibrium concentration6−x1−x3+x

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

Kc=(3+x)(6−x)(1−x)

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

Kc=(3+x)(6−x)(1−x)0.56=(3+x)(6−x)(1−x)0.56x2−4.92x+0.36=0x=−(−4.92)±(4.92)2−4×(0.56)×(0.36)2×0.56x=0.071[PCl5]=3+x=3.07 mol.L-1[PCl3]=6 -x=5.93 mol.L-1[Cl2]=1−x=0.93 mol.L-1

Therefore, the equilibrium composition of the reaction is [PCl5]=3.07 mol.L-1, [PCl3]=5.93 mol.L-1, and [Cl2]=0.93 mol.L.

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

Videos

Explanation of Solution

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Explanation of Solution

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Chapter 5 Solutions

Videos

Explanation of Solution

Interpretation: The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Explanation of Solution

Want to see more full solutions like this?

Chapter 5 Solutions

Interpretation:

The direction of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

The equilibrium composition of the reaction PCl3(g)+Cl2(g)⇌PCl5(g) has to be determined.