(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.

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

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

Students have asked these similar questions

Show work. don't give Ai generated solution....give correct solution

Biochemistry What is the process of "transamination" in either the muscles or the liver, that involves keto acid or glutamic acid? Please explain how the steps work. Thank you!

Biochemistry Please help. Thank you What is the importance of glutamic acid in the metabolism of nitrogen from amino acids? (we know therole; it’s used to remove the nitrogen from amino acids so that the remaining carbon skeleton can bebroken down by the “usual” pathways, but what is the important, unique role that only glutamicacid/glutamate can do?)

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.

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 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.

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 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.

Want to see more full solutions like this?

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

Answer 5

Chapter 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...

Concept explainers

Videos

Explanation of Solution

Explanation of Solution

Want to see more full solutions like this?

Chapter 3 Solutions