Chemistry Atoms First2e
2nd Edition
ISBN: 9781947172647
Author: OpenStax
Publisher: OpenStax College
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Chapter 12, Problem 47E
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
In this process, ATP becomes ADP summarized by the following equation:
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
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In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the
following equation: Glu+ ATP→G6) + ADP AG° = - 17KJ
In this process, ATP becomes ADP summarized by the following equation: ATP→ ADP A G° =- 30 kJ
What is the standard free energy change for the following reaction: Glu →G6P AG° =?
A 17 k)
B) -13 k)
c) 13 kJ
-17 kJ
Cells use the hydrolysis of adenosine triphosphate (ATP) as a source of energy. The conversion of ATP to ADP has a standard free-energy change of -30.5 kJ>mol. If all the free energy from the metabolism of glucose, C6H12O61s2 + 6 O21g2 ¡ 6 CO21g2 + 6 H2O1l2 goes into the conversion of ADP to ATP, how many moles of ATP can be produced for each mole of glucose?
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:Glu+ATP⟶G6P+ADP ΔG°=−17 kJ
In this process, ATP becomes ADP summarized by the following equation:ATP⟶ADP ΔG°=−30 kJ
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:Glu⟶G6P ΔG°=?
Chapter 12 Solutions
Chemistry Atoms First2e
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- Glycolysis is the process by which glucose is metabolized to lactic acid according to the equation C6H12O6(aq)2C3H6O3(aq) G=198 kJ at pH 7.0 and 25°C Glycolysis is the source of energy in human red blood cells. In these cells, the concentration of glucose is 5.0103 M, while that of lactic acid is 2.9103 M. Calculate AG for glycolysis in human blood cells under these conditions. Use the equation G=G+RT In Q, where Q is the concentration quotient, analogous to K.arrow_forwardAnother step in the metabolism of glucose, which occurs after the formation of glucose6-phosphate, is the conversion of fructose6-phosphate to fructose1,6-bisphosphate(bis meanstwo): Fructose6-phosphate(aq) + H2PO4(aq) fructose l,6-bisphosphate(aq) + H2O() + H+(aq) (a) This reaction has a Gibbs free energy change of +16.7 kJ/mol of fructose6-phosphate. Is it endergonic or exergonic? (b) Write the equation for the formation of 1 mol ADP fromATR for which rG = 30.5 kJ/mol. (c) Couple these two reactions to get an exergonic process;write its overall chemical equation, and calculate theGibbs free energy change.arrow_forwarda) As stated in question 5a, glucose (C6H12O6(s)) is a source of cellular energy. Calculate the standard free energy for the metabolism of glucose: Given ΔG°C6H12O6(s) = -911kJ/mol; ΔG°O2(g) = 0kJ/mol; ΔG°CO2 = -394kJ/mol; ΔG°H2O(l) = -237kJ/molC6H12O6(s) + 6O2(g) ➝ 6CO2(g) + 6H2O(l) b) Cells couple the hydrolysis of adenosine triphosphate (ATP) into adenosine diphosphate (ADP) andinorganic phosphate (PO43–) to drive chemical reactions (i.e. as a source of chemical energy). The reaction is: ATP(aq) + H2O(l) ➝ ADP(aq) + PO43–(aq) Calculate K for this reaction if ΔG° = –30.5 kJ/mole. c) If all of the energy from glucose metabolism went into ATP synthesis from ADP and inorganicphosphate, how many molecules of ATP could be generated from each molecule of glucose?arrow_forward
- In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:Glu + ATP⟶G6P + ADP ΔG°298 = −17 kJIn this process, ATP becomes ADP summarized by the following equation:ATP⟶ADP ΔG°298 = −30 kJDetermine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:Glu⟶G6P ΔG°298 = ?arrow_forwardIn glycolysis, glucose is broken down to yield energy. If the equilibrium constant of the isomerization of glucose-6-phosphate to fructose-6-phosphate is 0.5, would this reaction proceed in the forward or reverse direction? What is the free energy of the reaction? Using standard conditions (1 M) as an initial state, what are the equilibrium concentrations?arrow_forwardGiven that the reaction of 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 3 H20 (g) ΔΗ - - 906 kJ What would AH for NO (g) + ? H20 (g) → NH3 (g) + O2 (g) be? 2. 906 kJ 226.5 kJ -226.5kJ 453 kJarrow_forward
- ATP → ADP + Pi (ΔG°’ = -30.5 kJ/mol) In cells, the [ATP] is 0.5 mM, the [ADP] is 4.5 mM, and the [Pi] is 0.65 mM. Calculate the free energy change for ATP hydrolysis under cellular conditions. Enter the numerical value in units of kJ/mol and round to the nearest tenths place.arrow_forwardThe Keq for the isomerization of glucose-6-phosphate to fructose-6-phosphate is 0.504. What is the ΔG for the reaction if the concentration of glucose-6-phosphate is 0.01 M and the concentration of fructose-6-phosphate is 0.05 M? This reaction is in the cell (this is your temperature = 30 degree celcius ). the answer should be 0.588 kcal/mol - please show all steps I am not sure how to get that answerarrow_forwardGiven that S(g) + O2(g) → SO2(g) A,G° = -300.1 kJ/mol-rxn 2 S(g) + 3 O2(g) → 2 SO3(g) A,G° = -742.1 kJ/mol-rxn calculate AG° of the following reaction: SO2(g) + 1/2 O2(g) → SO3(g) O -1042.2 kJ/mol-rxn O -71.0 kJ/mol-rxn O +2.47 kJ/mol-rxn +1042.2 kJ/mol-rxn O +71.0 kJ/mol-rxnarrow_forward
- Please answer this question. For the aqueous reaction: dihydroxyacetone phosphate ↽−−⇀ glyceraldehyde−3−phosphatedihydroxyacetone phosphate the standard change in Gibbs free energy is Δ?°′=7.53 kJ/molΔG°′=7.53 kJ/mol. Calculate ΔG for this reaction at 298K when [dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00200arrow_forwarda 75 kg adult requires a calorie intake of 2000 Cal of food every 24 hours. The food is metabolized, and the free energy is used to synthesis ATP for useful work. {each Calorie listed on food labels corresponds to 1.0 Cal in biochemical terms}. Assuming that the efficiency of converting food energy into ATP is 50%. Calculate the weight of ATP used by this human in 24 hours and the percentage of the person’s body weight that this represents. {the formula mass of ATP is 505 g/mol and each mole of ATP is equivalence to 7.30 Cal}arrow_forwardThe standard free energy variation, at 25 ºC, for equilibrium: Glucose-6-phosphate (G-6-P) Glucose-1-phosphate (G-1-P) is ΔGº '= + 7280 J / mol. Calculate a) The equilibrium constant of the reaction. b) The real change in free energy when one mole of G-6-P is transformed into G-1-P, both concentrations remaining constant and equal to 10mM and 2 mM respectively.arrow_forward
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