(a) Interpretation: The equilibrium constant for the given reaction is to be stated. The value of Δ G ° ' for the given reaction is to be stated. Concept introduction: When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by K eq . The equilibrium constant is independent of the initial amount of the reactant and product.

Question

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Answer 1

Question

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

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Answer 2

Question

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

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Answer 3

Question

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

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Answer 4

Question

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

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Answer 5

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Answer 6

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Answer 7

Chapter 8, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the given reaction is to be stated. The value of ΔG°' for the given reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Answer 8

Expert Solution

Answer to Problem 8P

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is,

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

The value of universal gas constant is 8.314×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression,

Keq=[Forward rate constant][Backward rate constant]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −4 s −1 10 −6 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.314×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Answer 13

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are to be stated. The equilibrium constant for the given enzyme-catalyzed reaction is to be stated. The value of ΔG°' for the given enzyme-catalyzed reaction is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

Answer 15

Expert Solution

Answer to Problem 8P

The rate constants for the given enzyme-catalyzed reaction if an enzyme increases the reaction rate by 100-fold are kF=10−2 s−1 and kR=10−4 s−1.

The equilibrium constant for the given reaction is 100.

The value of ΔG°' for the given reaction is - 11.41 kJ mol−1.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The given reaction for the conversion of substrate (S) to product (P) is:

S⇄10−6 s−110−4 s−1P.

The given rate constant for forward reaction for the conversion of S to P is kF=10−4 s−1.

If an enzyme increases the forward reaction rate by hundred-fold, then the rate of forward reaction becomes kF=10−4 s−1×100=10−2 s−1

The given rate constant for backward reaction for the conversion of P to S is kR=10−6 s−1

If an enzyme increases the backward reaction rate by hundred-fold, then the rate of forward reaction becomes

kF=10−6 s−1×100=10−4 s−1

The value of universal gas constant is 8.315×10−3 kJ mol−1 K−1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273 K

Thus, the given temperature becomes,

25°C=25+273 K=298 K

The value of Keq is calculated by the expression:

Keq=[Concentration of Product][Concentration of Substrate]=kFkR

Substitute the values of rate constants for forward reaction and backward reaction in the above expression.

Keq= 10 −2 s −1 10 −4 s −1=100

Thus, the equilibrium constant for the given reaction is 100.

The value of ΔG°' is calculated by the expression,

ΔG°'=−RTlnKeq

Where,

R is the universal gas constant.
T is the temperature.
Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in the above formula.

ΔG°'=−8.315×10−3 kJ mol−1 K−1×298 K×ln(100)=8.315×10−3 kJ mol−1 K−1×298 K×4.6052=−11.41 kJ mol−1

Thus, the value of Gibbs free energy is - 11.41 kJ mol−1.

Answer 20

Ch. 8 - Prob. 1P Ch. 8 - Prob. 2P Ch. 8 - Prob. 3P Ch. 8 - Prob. 4P Ch. 8 - Prob. 5P Ch. 8 - Prob. 6P Ch. 8 - Prob. 7P Ch. 8 - Prob. 8P Ch. 8 - Prob. 9P Ch. 8 - Prob. 10P

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