3-2.3.1 The decomposition of crystalline N₂O5Part Ais an example of a reaction that is thermodynamically favored even though it absorbs heat. At 25 °C we have the following values for the standard state enthalpy and freeenergy changes of the reaction:Calculate AS° at 25 °C.Express your answer with the appropriate units.μAAS⁰ = ValueSubmit Request AnswerUnitsN₂O5 (s)?→ 2NO2(g) + O2(g)ΔΗ° =AG = -3+109.6 kJ/mol-30.5 kJ/mol

The decomposition of crystalline N2O5 N2O5(s)➡️ 2NO2(g) + 1/2 O2(g) At 25°C Standard state enthalpy…

Answered: The decomposition of crystalline N₂O5…

Biochemistry

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ISBN:9781305577206

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Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter21: Photosynthesis

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The decomposition of crystalline N2O5 N2O5(s) → 2NO2(g) + 1/2O2(g) is an example of a reaction that is thermodynamically favored, even though it absorbs heat. At 25 °C we have the following values for the standard state enthalpy and free energy changes of the reaction: ∆H° = +109.6 kJ/mol ∆G° = -30.5 kJ/mol (a) Calculate ∆S ° at 25 °C. (b) Why is the entropy change so favorable for this reaction?
The formation of maltose, a disaccharide, from two glucose molecules, is not energetically favorable. However, if this reaction is coupled with the hydrolysis of ATP, the reaction occurs more favorably. Maltose + H2O = 2 Glucose , ΔG'o = -15.5 KJ/mol or -3.7 kcal/mol a. Determine if the coupled reaction will occur spontaneously at standard state through calculating the Gibbs Free Energy of Reaction. b. Calculate the equilibrium constant for each individual reaction, and for the coupled reaction (using free energy of reaction). Show that the equilibrium constant for the coupled reaction equals the equilibrium constants for the individual reactions multiplied together. c. If the reaction medium contains the following chemical species at their given concentrations (298 K and 1.0 atm, pH = 7.0), will the reaction proceed in the forward or the reverse direction? [Maltose] = [Glucose] = 10.0 mM; [ATP] = 5.0 mM; [ADP] = [Pi] = 20 mM
The hydrolysis of ATP has an enthalpy and entropy of -24.3 kJ/mol and +21.6 J.K-1.mol-1, respectively, at 37 o C. What is the free energy change for the hydrolysis of 5 mols of ATP? Explain what contributes to the negative enthalpy change and positive entropy change in this reaction. What physical characteristics of the reaction would change if an ATP hydrolase enzyme is added to the solution?
If the hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase has a ΔG′∘ of −11.386 kJ/mol at 25 °C, what percentage of substrate remains once the reaction reaches equilibrium assuming no product was initially present? (Round answer to the nearest whole number)
The biochemical standard free energy change for the reaction: A → B is −15.0 kJ/mol. What is the equilibrium constant at 25°C and pH 7? What is the free energy change for the reaction A→B at 37°C when [A]=10.0 mM and [B]=0.100 mM? (Give your answer in kJ/mol)
Calculate the Gibbs free energy change (G) for the following chemical reaction: glutamate + NH3 glutamine + H2OThe reaction occurs at 293 K, the change in heat (H) = 4103 cal, and the change in entropy (S) = 2.4 cal/K.
Consider the following chemical equation whose delta(G) = 9kcal/mol: AC + BD ---> AB + CD what are the reactants and what are the products is this reaction spontaneous? How do you know? Is energy released or consumed by this reaction? If an enzyme, which catalyzes this reaction is added, what will happen to delta (G) If this reaction is coupled to another reaction, whose delta(G) is -12 kcal/mol, what will be the net delta(G) value? will the overall reaction be spontaneous
The phosphoryl group transfer potentials for glucose-1-phosphate and glucose-6-phosphate are 20.9 kJ/mol and 13.8 kJ/mol, respectively. (a) What is the equilibrium constant for the reaction shown below at 25 °C? (b) If a mixture was prepared containing 1 m glucose-6-phosphate and 1 x 10-3 M glucose-1-phosphate, what would be the thermodynamically favored direction for the reaction?
The 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. 21
An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?
If a 0.1 M solution of glucose 1- phosphate at 25 °C is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Calculate Keq and ΔG′° for this reaction.
Based on the definition of kcat, substitute a value that can be measured and yet still represents the value associated with the original concentration of the R. What would the rate or velocity of the reaction be equal to under these circumstances? How can cells increase Vmax? What variable that we could change would directly impact Vmax? Would the value of KM be affected by the ways you determined that Vma,x could be increased? What does this indicate about KM? Thinking about how catalysts work, about the Michaelis-Menten Equation, and the definition of kcat, what specifically does the enzyme change in the reaction mechanism to increase the rate? If an enzyme follows the 2 step mechanism proposed by Michaelis-Menten, what do you know about this enzyme? Be very specific and comprehensive. Please answer very soon will give rating surely

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