(a) Interpretation: The value of ΔG 0 for the folding of lysozyme at 25 0 C should be calculated. Concept introduction: The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows: ΔG o = ΔH o -TΔS o Here, ΔG o is standard Gibbs free energy change ΔH o is standard enthalpy change ΔS o is standard entropy change If ΔG is negative, the forward reaction is spontaneous. If ΔG is zero, the system is at equilibrium. If ΔG is positive, the reaction is spontaneous in the reverse direction.

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Answer 1

Question

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

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Students have asked these similar questions

The beta-lactamase hydrolyzes the lactam-ring in penicillin. Describe the mechanism of hydrolysis, insuring to include the involvement of S, D, & K in the reaction sequence. Please help

To map the active site of beta-lactamase, the enzyme was hydrolyzed with trypsin to yield a hexapeptide (P1) with the following amino acids. Glu, Lys, Leu, Phe, Met, and Ser. Treatment of P1 with phenyl isothiocyanate yielded a PTH derivative of phenylalanine and a peptide (P2). Treatment of P1 with cyanogenbromide gave an acidic tetrapeptide (P3) and a dipeptide (P4).Treatment of P2 with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis, yields N-2,4-dinitrophenyl-Glu. P1, P2, and P3 contain the active site serine. Why doesn't D in this hexapeptide not participate in the hydrolysis of the beta-lactam ring even though S, K, and D are involved in the catalyst?

To map the active site of -lactamase, the enzyme was hydrolyzed with trypsin to yield a hexapeptide (P1) with the following amino acids. Glu, Lys, Leu, Phe, Met, and Ser. Treatment of P1 with phenyl isothiocyanate yielded a PTH derivative of phenylalanine and a peptide (P2). Treatment of P1 with cyanogenbromide gave an acidic tetrapeptide (P3) and a dipeptide (P4).Treatment of P2 with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis, yields N-2,4-dinitrophenyl-Glu. P1, P2, and P3 contain the active site serine. Using the experimental results described above derive the primary sequence of the active site hexapeptide. Please help!

Answer 2

Question

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

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Answer 3

Question

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

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Answer 4

Question

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

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Answer 5

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

Answer 6

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Answer 7

Chapter 3, Problem 1P

Interpretation Introduction

(a)

Interpretation:

The value of ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ should be calculated.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Answer 8

Expert Solution

Explanation of Solution

Given Information:

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation.

$un folded protein ⇌ Folded protein$

The values of ${ΔH}^{0} {and ΔS}^{0}$ for the folding of the protein lysozyme are as follows:

$\begin{array}{l} {ΔΗ}^{o} = -280 kJ/mol \\ {ΔS}^{o} = -790 J/mol .K \end{array}$

Explanation:

The ${ΔG}^{0}$ for the folding of lysozyme at $25^{0} C$ can be calculated using following formulae.

${ΔG}^{0} {= ΔH}^{0} {- TΔS}^{0}$ ……(1)

From the given,

$\begin{matrix} {ΔΗ}^{0} = -280 kJ/mol \\ {ΔS}^{0} = -790 J/mol .K \\ = -0 .79 kJ/mol .K \end{matrix}$

Convert temperature from degree Celsius to Kelvin,

$\begin{matrix} T (K) {= 25}^{o} C + 273 .15 \\ = 298 .15 K \end{matrix}$

Substitute the values in the equation (1)

$\begin{matrix} {ΔG}^{0} = - 280 kJ/mol - (298 .15 K) (- 0 .79 kJ/mol .K) \\ = - 44 .58 kJ/mol \end{matrix}$

Therefore, ${ΔG}^{0}$ for the folding of lysozyme at $25^{o} C$ is $- 44 .58 kJ/mol$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

Answer 13

Interpretation Introduction

(b)

Interpretation:

The temperature at which the unfolding of lysozyme is expected to become favorable should be determined.

Concept introduction:

The standard Gibbs free energy change is related to standard enthalpy change and standard entropy change as follows:

${ΔG}^{o} {= ΔH}^{o} {-TΔS}^{o}$

Here,

${ΔG}^{o}$ is standard Gibbs free energy change

${ΔH}^{o}$ is standard enthalpy change

${ΔS}^{o}$ is standard entropy change

If $ΔG$ is negative, the forward reaction is spontaneous.

If $ΔG$ is zero, the system is at equilibrium.

If $ΔG$ is positive, the reaction is spontaneous in the reverse direction.

Answer 14

Expert Solution

Explanation of Solution

Explanation:

If ${ΔG}^{o} = +ve$ then the unfolding of lysozyme is favorable.

Temperature can be calculated at ${ΔG}^{o} = 0$

Thus,

$\begin{matrix} {ΔH}^{0} {= TΔS}^{0} \\ T = \frac{{ΔH}^{0}}{{ΔS}^{0}} \end{matrix}$

Putting the values,

$\begin{matrix} T = \frac{- 280}{- 0 .79} \\ = 354 .4 K \end{matrix}$

Therefore,at $354 .4 K$ temperature the unfolding of lysozyme is favorable.

Answer 15

Expert Solution

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 3 - Prob. 1P Ch. 3 - Given the following reactions and their...Ch. 3 - The decomposition of crystalline N2O5...Ch. 3 - The oxidation of glucose to CO2 and water is a...Ch. 3 - Prob. 5P Ch. 3 - In another key reaction in glycolysis,...Ch. 3 - Assume that some protein molecule, in its folded...Ch. 3 - When a hydrophobic substance like a hydrocarbon is...Ch. 3 - It is observed that as temperature is increased,...Ch. 3 - Suppose a reaction has Ho and So values...

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