For the hydrolysis of ATP, the equilibrium constant K should be founded at 298 K . Concept introduction: Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy. Free energy change ΔG : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant K , ΔG = ΔG o + RT ln ( K ) ( o r ) ΔG o = − RT ln ( K ) Where, T is the temperature ΔG is the free energy ΔG 0 is standard free energy values.

Question

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Answer 1

Question

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

The temperature changes from $20oC to 37oC$ each of $K$ changes.

The calculations at the new temperature $T= (273 + 37 =310.0 K)$

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Case-1

$ln K = ΔGo-RT= (−30.5 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K = 11.8339K= e11.8339= 1.37847×105= 1.38×105.$

Case-2

$ln K = ΔGo-RT= (13.8 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =−5.3543576K= e−5.3543576= 4.7275×10−3= 4.73×10-3.$

Case-3

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =6.4795K= e6.4795= 6.51645×102= 6.52×102.$

Hence, the each cases equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

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Answer 2

Question

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

The temperature changes from $20oC to 37oC$ each of $K$ changes.

The calculations at the new temperature $T= (273 + 37 =310.0 K)$

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Case-1

$ln K = ΔGo-RT= (−30.5 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K = 11.8339K= e11.8339= 1.37847×105= 1.38×105.$

Case-2

$ln K = ΔGo-RT= (13.8 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =−5.3543576K= e−5.3543576= 4.7275×10−3= 4.73×10-3.$

Case-3

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =6.4795K= e6.4795= 6.51645×102= 6.52×102.$

Hence, the each cases equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

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Answer 3

Question

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

The temperature changes from $20oC to 37oC$ each of $K$ changes.

The calculations at the new temperature $T= (273 + 37 =310.0 K)$

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Case-1

$ln K = ΔGo-RT= (−30.5 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K = 11.8339K= e11.8339= 1.37847×105= 1.38×105.$

Case-2

$ln K = ΔGo-RT= (13.8 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =−5.3543576K= e−5.3543576= 4.7275×10−3= 4.73×10-3.$

Case-3

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =6.4795K= e6.4795= 6.51645×102= 6.52×102.$

Hence, the each cases equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

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Answer 4

Question

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

The temperature changes from $20oC to 37oC$ each of $K$ changes.

The calculations at the new temperature $T= (273 + 37 =310.0 K)$

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Case-1

$ln K = ΔGo-RT= (−30.5 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K = 11.8339K= e11.8339= 1.37847×105= 1.38×105.$

Case-2

$ln K = ΔGo-RT= (13.8 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =−5.3543576K= e−5.3543576= 4.7275×10−3= 4.73×10-3.$

Case-3

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =6.4795K= e6.4795= 6.51645×102= 6.52×102.$

Hence, the each cases equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

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Answer 5

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

The temperature changes from $20oC to 37oC$ each of $K$ changes.

The calculations at the new temperature $T= (273 + 37 =310.0 K)$

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Case-1

$ln K = ΔGo-RT= (−30.5 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K = 11.8339K= e11.8339= 1.37847×105= 1.38×105.$

Case-2

$ln K = ΔGo-RT= (13.8 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =−5.3543576K= e−5.3543576= 4.7275×10−3= 4.73×10-3.$

Case-3

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =6.4795K= e6.4795= 6.51645×102= 6.52×102.$

Hence, the each cases equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Answer 6

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Answer 7

Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298 K$ .

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

Answer 8

(a)

Expert Solution

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo=-30.5 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (−30.5 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = 12.3104K= e12.3104K= 2.2199×105 = 2.22×105.$

The hydrolysis of ATP is equilbrium constant value is $K=2.22×105_.$

Answer 13

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Answer 14

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

Answer 15

(b)

Expert Solution

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)−−−−−−−[1]ΔGo= 13.8 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $K$

We know that equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation,

$ln K = −ΔGoRT= (13.8 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K = −5.569969K= e−5.569969K= 3.8105985×10−3 K= 3.81×10-3.$

Hence, the ATPs, dehydration equilbrium is $K=3.81×10-3_.$

Answer 20

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

Answer 21

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298 K$ .

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

Answer 22

(c)

Expert Solution

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $K=8.46×102_.$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The chemical equation for dehydration of ATP is,

$(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

Free energy equation is,

$ΔG=0= ΔGo+ RT ln (K)ΔGo=− RT ln (K)ΔGo=−16.7 kJ/molΔGo= -16.7 kJR= 8.314 J/mol×KT= 273 K+25 K =298 K.$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Rearrange the above equation to calculate $K$

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)(298 K))(103J1 kJ)ln K =6.74047K= e6.74047= 8.45958×102= 8.46×102.$

Hence, the coupled reaction equilbrium value is $K=8.46×102_.$

Answer 27

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$

The temperature changes from $20oC to 37oC$ each of $K$ changes.

The calculations at the new temperature $T= (273 + 37 =310.0 K)$

The equilibrium equation,

$ΔGrxno =−RT ln (K)$

Case-1

$ln K = ΔGo-RT= (−30.5 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K = 11.8339K= e11.8339= 1.37847×105= 1.38×105.$

Case-2

$ln K = ΔGo-RT= (13.8 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =−5.3543576K= e−5.3543576= 4.7275×10−3= 4.73×10-3.$

Case-3

$ln K = ΔGo-RT= (−16.7 kJ/mol−(8.314 J/mol⋅K)( 310 K))(103J1 kJ)ln K =6.4795K= e6.4795= 6.51645×102= 6.52×102.$

Hence, the each cases equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Answer 28

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from $20oC to 37oC$ in each cases a,b and c

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

$ΔG =∑nΔGf° (products)-∑mΔGf° (reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔGo+ RT ln (K) (or)ΔGo=−RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ is standard free energy values.

Answer 29

(d)

Expert Solution

Answer to Problem 20.99P

Given each reactions equilibrium changes has $K= 1.38×105, 4.73×10-3 and 6.52×102.$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The chemical equation of ATP is,

$(1). ATP4−(aq) + H2O(l) ⇌ ADP3−(aq) + HPO42−(aq) + H+(aq) ΔGo=−30.5 kJ(2). Glucose+ HPO42−(aq) ⇌ [Glucose phosphate]− + H2O(l) ΔGo= 13.8 kJ(3). Glucose + ATP4−(aq) ⇌ [Glucose phosphate]− + ADP3−(aq) ΔGo= −16.7 kJ$