   # Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as ATP ( a q ) + H 2 O ( l ) → ADP ( a q ) + H 2 PO 4 − ( a q ) where ADP represents adenosine diphosphate. For this reaction, ∆ G ° =−30.5 kJ/mol. a. Calculate K at 25°C. b. If all the free energy from the metabolism of glucose C 6 H 12 O 6 ( s ) + 6 O 2 ( g ) → 6 CO 2 ( g ) + 6 H 2 O ( l ) goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 75E
Textbook Problem
46 views

## Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as ATP ( a q ) + H 2 O ( l ) → ADP ( a q ) + H 2 PO 4 − ( a q ) where ADP represents adenosine diphosphate. For this reaction, ∆G° =−30.5 kJ/mol.a. Calculate K at 25°C.b. If all the free energy from the metabolism of glucose C 6 H 12 O 6 ( s ) + 6 O 2 ( g ) → 6 CO 2 ( g ) + 6 H 2 O ( l ) goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?

(a)

Interpretation Introduction

Interpretation: The reaction of hydrolysis of adenosine and its ΔG° value is given. The value of K is to be calculated at given temperature. Another reaction of breakdown of glucose is given. The number of ATP molecules produced from every molecule of glucose is to be calculated.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

ΔG=0Q=K

ΔG°=RTln(K)

### Explanation of Solution

Given

The value ΔG° for the given reaction is 30.5kJ/mol .

Given temperature is 25°C .

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 30.5kJ into joule is,

30.5kJ=(30.5×103)J=30.5×103J

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 25°C into Kelvin is,

T(K)=T(°C)+273T(K)=(25+273)K=298K

(b)

Interpretation Introduction

Interpretation: The reaction of hydrolysis of adenosine and its ΔG° value is given. The value of K is to be calculated at given temperature. Another reaction of breakdown of glucose is given. The number of ATP molecules produced from every molecule of glucose is to be calculated.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

ΔG=0Q=K

ΔG°=RTln(K)

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