The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

Question

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Answer 1

Question

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

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Answers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…

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Answer 2

Question

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

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Answer 3

Question

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

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Answer 4

Question

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

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Answer 5

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Answer 6

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Answer 7

Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

Answer 8

(a)

Expert Solution

Answer to Problem 1DE

Solution: The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The relation between the standard free energy change, standard entropy change and standard enthalpy change is given by the formula,

$ΔG0=ΔH0−TΔS0$ ...(1)

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.

The relation between standard free energy change and the equilibrium constant is given by the formula,

$ΔG0=−RTlnK$ ...(2)

Where,

$R$ is the gas constant.
$K$ is the equilibrium constant.

$−RTlnK$

$ΔG0$

$−RTlnK=ΔH0−TΔS0$

Simplify the above equation to calculate the relation between $K, ΔH0$ and $ΔS0$ .

$lnK=ΔH0−TΔS0−RT=ΔS0R−ΔH0RT$

Answer 13

Conclusion

The relation between equilibrium constant, standard enthalpy change and entropy change is given as,

$ΔS∘R−ΔH∘RT=lnK$

Answer 14

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Answer 15

(b)

Interpretation Introduction

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

Answer 16

(b)

Expert Solution

Answer to Problem 1DE

Solution: The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The standard enthalpy and entropy changes by using $K$ and $T$ data is calculated by using the formula,

$lnK=ΔS0R−ΔH0RT$

Where,

$ΔG0$ is the standard free energy change.
$ΔH0$ is the standard enthalpy change.
$T$ is the temperature.
$ΔS0$ is the standard entropy change.
$R$ is the gas constant.
$K$ is the equilibrium constant.

The above equation is modified to draw the graph of $K$ and $T$ data as the equation of line $(y=mx+c)$ as given below.

$lnK=ΔS0R−ΔH0R(1T)$

Where,

$lnK$ is the variable on the $y$ axis of the graph.
$1T$ is the variable on the $x$ axis of the graph.
$−ΔH0R$ is the slope of the graph.
$ΔS0R$ is the intercept on $y$ axis.

Answer 21

Conclusion

The graph of $K$ versus $1T$ is used to calculate the entropy and enthalpy changes for the given process.

Answer 22

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

Answer 23

(c)

Interpretation Introduction

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

Answer 24

(c)

Expert Solution

Answer to Problem 1DE

Solution: The bond breaking in drug candidate and bond formation in binding reaction uses all the energy produced in the binding reaction which makes the enthalpy change of the reaction close to zero and the breaking of bonds increases vibrations in the system which increases randomness; thereby increase in entropy change observed. This is determined by using the calorimetric technique.

Answer 25

(c)

Expert Solution

Answer 26

(c)

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The given reaction of drug candidate with its DNA target involves the breaking of ionic bonds and formation of hydrogen bonds and the dipole-dipole interactions. These two processes occur simultaneously in the binding reaction. Therefore, the exchange of energy also occur in the reaction medium. The energy released in the bond formation is utilized in the bond breaking which makes the overall enthalpy change of the equilibrium reaction zero.

The equilibrium of drug candidate and DNA target with drug-DNA complex involves the breaking of bonds which increases the vibration in the system and increase in randomness of molecules increases the entropy change of the system.

The experiment designed to determine the values of enthalpy and entropy changes is the calorimetric technique. By using this technique the data of equilibrium constant and temperature is observed for the given reaction. The value of enthalpy and entropy change is determined by the slope and intercept of the graph.

Answer 29

Conclusion

The increase in the vibrations due to bond breaking increases entropy change in the system; whereas, the enthalpy change of the bond breaking and bond formation counterbalance each other which leads to the zero enthalpy change of the reaction. The calorimetric method is used to determine the values of enthalpy and entropy change.

Answer 30

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Answer 31

(d)

Interpretation Introduction

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

Answer 32

(d)

Expert Solution

Answer to Problem 1DE

Solution: The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Answer 33

(d)

Expert Solution

Answer 34

(d)

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

The binding of DNA involves formation of bonds which always results in the release of energy. Therefore, the enthalpy change of the DNA binding reaction is high. The entropy change of the reaction is less because the formation of bonds does not involve increased vibrations in the system. The bond breaking increases the vibrations in the system which increases the entropy change that is not involved in DNA binding reaction. Therefore, the entropy change of the DNA binding reaction is small.

The entropy change and enthalpy change for the DNA binding reaction is tested by calorimetric techniques. The temperature of the reaction at regular intervals of time is observed to calculate the enthalpy change of the reaction and the concentration of species is observed at regular intervals to calculate the equilibrium constant of the reaction from which the value of entropy change is calculated.

Answer 37

Conclusion

The process of DNA binding involves the formation of bonds which leads to high enthalpy change and less entropy change of the reaction. The entropy change and enthalpy change of the reaction is tested by using the calorimetric technique.

Answer 38

Ch. 19.1 - The process of iron being oxidized to make iron...Ch. 19.1 - At 1 atm pressure, CO2(s) sublimes at 78oC. Is...Ch. 19.2 - Prob. 19.2.1PE Ch. 19.2 - Prob. 19.2.2PE Ch. 19.3 - Prob. 19.3.1PE Ch. 19.3 - Prob. 19.3.2PE Ch. 19.3 - Prob. 19.4.1PE Ch. 19.3 - Prob. 19.4.2PE Ch. 19.4 - Prob. 19.5.1PE Ch. 19.4 - Using the standard molar entropies in Appendix C,...

The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

Concept explainers

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

Answer to Problem 1DE

Explanation of Solution

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

Answer to Problem 1DE

Explanation of Solution

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

Answer to Problem 1DE

Explanation of Solution

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

Answer to Problem 1DE

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions

The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

Concept explainers

To determine: The equation that relates the equilibrium constant with the standard enthalpy and entropy changes.

Answer to Problem 1DE

Explanation of Solution

To determine: The way to draw the graph of K<m:math><m:mi>K</m:mi></m:math> and T<m:math><m:mi>T</m:mi></m:math> data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.

Answer to Problem 1DE

Explanation of Solution

To determine: The explanation and experiment for enthalpy change for the given blind reason being close to zero and the entropy change being large and positive.

Answer to Problem 1DE

Explanation of Solution

To determine: The explanation and experiment for the value of enthalpy change is large negative and entropy change is small positive for the DNA binding reaction.

Answer to Problem 1DE

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions

To determine: The way to draw the graph of $K$ and $T$ data to calculate the standard entropy and enthalpy changes for the given drug candidate and DNA binding interaction.