(a) Explanation (a) Glucose molecule reacts with O 2 to form CO 2 and H 2 O. Hence, the reaction can be written as: C 6 H 12 O 6 ( s ) + 6 O 2 ( g ) ⇌ 6 C O 2 + 6 H 2 O ( l ) Δ G ° ` = − 2870 k J ....... ( i ) ATP molecule reacts with H 2 O molecule and form ADP molecule. The reaction can be written as: A T P 4 − ( a q ) + H 2 O ( l ) ⇌ A D P 3 − ( a q ) + H 2 P O 4 − ( a q ) Δ G ° ` = − 30.5 k J ....... ( i i ) It we reverse this reaction, we get A D P 3 − ( a q ) + H 2 P O 4 − ( a q ) ⇌ A T P 4 − ( a q ) + H 2 O ( l ) Δ G ° ` = 30.5 k J ..... ( i i i ) Now we multiply the equation iii with 32 (since 1 mole of glucose generate 32 mole of ATP), we get: 32 A D P 3 − ( a q ) + 32 H 2 P O 4 − ( a q ) ⇌ 32 A T P 4 − ( a q ) + 32 H 2 O ( l ) Δ G ° ` = ( 30 (b) Interpretation Introduction To determine: The spontaneity of the reaction.

Chemistry, Books a la Carte Edition and Modified Mastering Chemistry with Pearson eText & ValuePack Access Card (7th Edition)

7th Edition

ISBN: 9780134172514

Author: John E. McMurry

Publisher: PEARSON

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Chapter 17, Problem 17.24P

(a)

Interpretation Introduction

To determine:

The overall reaction and calculate ?G°`.

(b)

Interpretation Introduction

To determine:

The spontaneity of the reaction.

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Chapter 17, Problem 17.23P

Chapter 17, Problem 17.25P

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Chapter 17 Solutions

Chemistry, Books a la Carte Edition and Modified Mastering Chemistry with Pearson eText & ValuePack Access Card (7th Edition)

Ch. 17 - Prob. 17.1P Ch. 17 - Prob. 17.2P Ch. 17 - Prob. 17.3A Ch. 17 - Prob. 17.4P Ch. 17 - Prob. 17.5P Ch. 17 - Prob. 17.6A Ch. 17 - Prob. 17.7P Ch. 17 - Prob. 17.8A Ch. 17 - Prob. 17.9P Ch. 17 - APPLY 17.10 Use the values of Hf and S in Appendix...

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The overall reaction and calculate ?G°`.