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The oxidation of glucose to CO2 and water is a major source of energy in aerobic organisms. It is a reaction favored mainly by a large negative enthalpy change.
a. At 37o C, what is the value for
b. In the overall reaction of aerobic
c. What is the efficiency of the process in terms of the percentage of the available free energy change captured in ATP?
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Chapter 3 Solutions
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- Glucose 1-phosphate is converted into fructose 6-phosphate in two successive reactions. Reaction 1: Glucose 1-phosphate → glucose 6-phosphate AG'° = -7.3 kJ/mol Reaction 2: Glucose 6-phosphate → fructose 6-phosphate AGʻ° = +1.7 kJ/mol Using the values given, calculate the standard free-energy change, AGʻ°, ,for the sum of the two reactions: sum > Sum: Glucose 1-phosphate → fructose 6-phosphate AG'°, kJ/mol sum Calculate the equilibrium constant, K', eq > for the sum of the two reactions. K'e eq IIarrow_forward[References) The AG" for the reaction ATP + H₂O ADP + P + H+ is -30.5 kJ mol'¹, Other organophosphate species also undergo hydrolysis of the phosphate moiety via a similar reaction. Determine the value of AG for the following reactions and indicate if the reaction will proceed spontaneously in the direction written if the reactants and products are initially in a 1:1 molar ratio. a. ATP + Glucose ADP+ Glucose-1-phosphate AG⁹¹ = kJ mol-1 The reaction is (The hydrolysis reaction for glucose-1-phosphate is Glucose-1-phosphate + H₂0 Glucose + P + H+; AG-20.9 kJ mol ¹.) b. ATP + Glutamate Carbamoyl phosphate + ADP AG The reaction is kJ mol-1 (The hydrolysis reaction for carbamoyl phosphate is Carbamoyl phosphate + H₂0 Glutamate + P, + H+ ; AG¹ = -51.4 kJ mol ¹.) Submit Answer Try Another Version 3 item attempts remainingarrow_forwardA reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by 03(g) + NO(g) → O2(g) + NO2(g) The rate law for this reaction is rate of reaction = k[03][NO] Given that k = 2.71 × 106 M−¹.s¯¹ at a certain temperature, calculate the initial reaction rate when [03] and [NO] remain essentially constant at the values [03]0 = 5.56 × 10-6 M and [NO]o = 6.37 x 10-5 M, owing to continuous production from separate sources. initial reaction rate: Calculate the number of moles of NO2 (g) produced per hour per liter of air. NO2 produced: M.s-1 mol·h¹.L-1arrow_forward
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- Cytochrome c oxidase catalyses the transfer of electrons from ferrocytochrome c to molecular oxygen. The balanced equation is shown below. 4 Cytochrome c(Fe 2") + O2 + 4 H* → 4 Cytochrome c(Fe * ) +2H:0 The relevant half reactions are: Cytochrome c(Fe * ) + e Cytochrome c(Fe ) E, = +0.254V ½ O2 + 2 H* + 2 e H20 E.' = +0.816V (F = Faraday's constant = 96.485 kJ /V. mol, R= 8.314 J/mol.K) (i) What is the standard free energy change for this reaction? (ii) What is the equilibrium constant (Keg) for this reaction?arrow_forwardConsider the following reaction: Glucose-1-phosphate → Glucose-6-phosphate ΔG° = −7.1 kJ/mol What is the equilibrium constant for this reaction at 25oC?arrow_forwardThe conversion of glucose-1-phosphate to glucose-6-phosphate by the enzyme phosphoglucomutase has a △G°' of -7.6 kJ/mol. Calculate the equilibrium constant for this reaction at 298 K and a pH of 7. (R = 8.315 J/K-mol) A. 0.003 B. 0.047 C. 1.00 D. 21arrow_forward
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