If the standard free energy of hydrolysis (AGO) of Phosphoenolpyruvate (PEP) = -61.9 kJ/mole what is the equilibrium ratio of phosphoenolpyruvate to pyruvate under standard conditions when [ATP]/[ADP] = 10? A 1.55 x 10-4 B D E F 2.53 x 10-4 3.06 x 10-5 2.15 x 10-5 7.63 x 10-6 4.25 x 10³ Correct answer
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- For the ammonia synthesis reaction ⇌ Does the entropy effect favor products? Explain your answer. Does the energy effect favor products? Explain your answer. Is the equilibrium concentration of NH3(g) greater at high or low temperature? Explain.Another 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.When a mixture of hydrogen and bromine is maintained at normal atmospheric pressure and heated above 200. °C in a closed container, the hydrogen and bromine react to form hydrogen bromide and a gas-phase equilibrium is established. Write a balanced chemical equation for the equilibrium reaction. Use bond enthalpies from Table 6.2 ( Sec. 6-6b) to estimate the enthalpy change for the reaction. Based on your answers to parts (a) and (b), which is more important in determining the position of this equilibrium, the entropy effect or the energy effect? In which direction will the equilibrium shift as the temperature increases above 200. °C? Explain. Suppose that the pressure were increased to triple its initial value. In which direction would the equilibrium shift? Why is the equilibrium not established at room temperature?
- The 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 answerGiven 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 kJThe hydrolysis of ATP to form ADP requires energy to break the bonds initially, but releases moreenergy as bond are remade making the reaction exothermic overall. The Gibbs free energy for thisreaction is -30.5 kJ/mole, meaning that it is also exergonic (ΔG=-). This energy can be used by otherbiological systems if this reaction is coupled with another less favorable one. In this case, the creation ofglucose-6-phosphate, C6H10O9P, this is a pivotal intermediate in many biochemical pathways. PO43- + C6H12O6 ⇌ H2O + C6H10O9P ΔG° = +13.8 kJ/moleATP + H2O ⇌ ADP + PO43- ΔG° = -30.5 kJ/mole A) What is the overall reaction that takes place? B) Calculate the ΔG° for this reaction C) Will this be a successful coupled reaction?
- Calculate the equilibirum constant (Keq) for formation of glucose-1-phosphate at 37°C according to the chemical reaction Glucose + HPO4 2- ----> Glucose-1-phosphate + H2OConsider the oxidation of NADH by molecular oxygen as carried out via the electron-transport pathway:NADH + H+ + ½ O2 → NAD+ + H2OThe equilibrium constant (Keq) for this reaction is: a 2.05 x 102 b 7.09 x 103 c 12.05 x 109 d 8.54 x 107 e 6.15 x 1013 f 6.02 x 1023 g 2.63 x 1038 h 4.37 x 1019 i 5.0 x 1026HO AG° = HOT HO OH a form of glucose OH a form ß form H - Which of the two forms is most stable: HO HO HO OH B form of glucose H COH At equilibrium at 25 °C, the two forms are present in a ratio of approximately 64 : 36. Calculate the standard free energy (AG°) difference that corresponds to this equilibrium ratio (i.e. more stable form : less stable form). Kcal/mol
- Consider the interconversion shown, which occurs in glycolysis. Fructose 6-phosphate glucose 6-phosphate The equilibrium constant, K'eq, is 1.97 at 25.0 °C. Calculate AG'º for the reaction. AG'° = kJ/mol If the concentration of fructose 6-phosphate is adjusted to 1.5 M and that of glucose 6-phosphate is adjusted to 0.50 M, what is AG? AG = kJ/molThe following equilibrium constants have been determined for hydrosulfuric acid at 25°C: H2S(aq) = H*(aq) + HS (aq) K¸' = 9.5 × 10-8 HS (aq) = H*(aq) + S²¯(aq) K¸" = 1.0 × 10-19 es Calculate the equilibrium constant for the following reaction at the same temperature: H,S(aq) = 2H*(aq) + s²¯(aq) K = x 10Consider a general reaction enzyme A(aq) = B(aq) The AGo of the reaction is-8.290 kJ · mol. Calculate the equilibrium constant for the reaction at 25 °C. %3D What is AG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.7 M and the concentration of B is 0.50 M? kJ mol-! AG = %3D |help terms of use contact us careers privacy policy about s MacBook Pro