Math Problem Statement

The conversion of glucose (C6H12O6) into lactic acid (2CH3CH(OH)COOH) drives the phosphorylation of 2 moles of ADP to ATP and has a ΔG° = –135 kJ/mol.

The conversion of ATP to ADP has a ΔG° = –30.5 kJ/mol. $$ATP4−(*aq*)+H2​O(*l*)ADP3−(*aq*)+HPO2−4​(*aq*)+H+(*aq*)   2nd attempt See Hint See Periodic Table What is ΔG° for the conversion of glucose into lactic acid? $$C6​H12​O6​(*aq*)2CH3​CH(OH)COOH(*aq*)    kJ     

Solution

To determine the ΔG\Delta G^\circ for the conversion of glucose (C6H12O6C_6H_{12}O_6) into lactic acid (2CH3CH(OH)COOH2 \, CH_3CH(OH)COOH), we start by analyzing the given information and the reactions.


Given:

  1. ΔG\Delta G^\circ for the phosphorylation of 2 moles of ADP to ATP during the glucose conversion is 135kJ/mol-135 \, \mathrm{kJ/mol}: C6H12O6(aq)+2HPO42(aq)+2ADP3(aq)+2H+(aq)2CH3CH(OH)COOH(aq)+2ATP4(aq)+2H2O(l)C_6H_{12}O_6(\text{aq}) + 2 \, HPO_4^{2-}(\text{aq}) + 2 \, ADP^{3-}(\text{aq}) + 2 \, H^+(\text{aq}) \rightarrow 2 \, CH_3CH(OH)COOH(\text{aq}) + 2 \, ATP^{4-}(\text{aq}) + 2 \, H_2O(\text{l})

  2. ΔG\Delta G^\circ for ATP hydrolysis is 30.5kJ/mol-30.5 \, \mathrm{kJ/mol}: ATP4(aq)+H2O(l)ADP3(aq)+HPO42(aq)+H+(aq)ATP^{4-}(\text{aq}) + H_2O(\text{l}) \rightarrow ADP^{3-}(\text{aq}) + HPO_4^{2-}(\text{aq}) + H^+(\text{aq})


Step 1: Energy contribution of ATP synthesis

In the glucose-to-lactic acid reaction, 2 moles of ATP are synthesized. Since the hydrolysis of 1 mole of ATP releases 30.5kJ/mol-30.5 \, \mathrm{kJ/mol}, synthesizing 1 mole of ATP requires +30.5kJ/mol+30.5 \, \mathrm{kJ/mol}. For 2 moles: ΔG for ATP synthesis=2×(+30.5)=+61.0kJ/mol.\Delta G^\circ \text{ for ATP synthesis} = 2 \times (+30.5) = +61.0 \, \mathrm{kJ/mol}.


Step 2: Determine ΔG\Delta G^\circ for glucose-to-lactic acid conversion

The net energy released by the conversion of glucose into lactic acid, after accounting for ATP synthesis, is given as 135kJ/mol-135 \, \mathrm{kJ/mol}. Since this value includes the energy cost of ATP synthesis: ΔG(glucose to lactic acid only)=ΔG(overall)+ΔG(ATP synthesis).\Delta G^\circ (\text{glucose to lactic acid only}) = \Delta G^\circ (\text{overall}) + \Delta G^\circ (\text{ATP synthesis}).

Substitute the values: ΔG(glucose to lactic acid only)=135+61.0=196kJ/mol.\Delta G^\circ (\text{glucose to lactic acid only}) = -135 + 61.0 = -196 \, \mathrm{kJ/mol}.


Final Answer:

The ΔG\Delta G^\circ for the conversion of glucose into lactic acid (C6H12O62CH3CH(OH)COOHC_6H_{12}O_6 \rightarrow 2 \, CH_3CH(OH)COOH) is: 196kJ/mol.\boxed{-196 \, \mathrm{kJ/mol}}.


Let me know if you'd like further clarifications or have questions! 😊

Related Questions:

  1. How does the energy cost of ATP synthesis compare to ATP hydrolysis?
  2. Why is ΔG\Delta G^\circ negative for exergonic reactions like glucose breakdown?
  3. What role does phosphorylation play in cellular metabolism?
  4. How can the efficiency of ATP production from glucose be quantified?
  5. How would ΔG\Delta G^\circ change if fewer moles of ATP were synthesized?

Tip: When solving thermodynamic problems, carefully isolate each energy component for clarity!

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Energy Calculations
Chemical Equilibria

Formulas

ΔG° (overall) = ΔG° (glucose to lactic acid only) + ΔG° (ATP synthesis)
ΔG° (ATP synthesis) = 2 × (ΔG° for ATP hydrolysis)

Theorems

Law of Conservation of Energy
Gibbs Free Energy Equation

Suitable Grade Level

Grades 11-12, College Level