Problem 7.7- Enhanced - with FeedbackDoes entropy increase or decrease in the following processes?Drag the appropriate items to their respective bins.Entropy increasesSubmitProvide FeedbackMg(s) + Cl₂(g) → MgCl2 (s)Request AnswerType here to searchOComplex carbohydrates aremetabolized by the body andconverted into simple sugars.EEntropy decreases1Water freezes in an ice cube tray.&ResetHelp

Entropy (S) is a thermodynamic quantity which is a measure of disorder or randomness present in the…

Does entropy increase or decrease in the following processes? Drag the appropriate items to their respective bins. Entropy increases Subreit Mg(s) + Cl₂(g) → MgCl₂ (s) Complex carbohydrates are metabolized by the body and converted into simple sugars. Entropy decreases Water freezes in an ice cube tray. Reset Help

Does entropy increase or decrease in the following processes? Drag the appropriate items to their respective bins. Entropy increases Subreit Mg(s) + Cl₂(g) → MgCl₂ (s) Complex carbohydrates are metabolized by the body and converted into simple sugars. Entropy decreases Water freezes in an ice cube tray. Reset Help

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter23: Organic Polymers, Natural And Synthetic

Section: Chapter Questions

Problem 46QAP: Glycolysis is the process by which glucose is metabolized to lactic acid according to the equation...

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Consider 2 reactions, for which the standard free energy changes are given: Reaction 1: Glucose-1-phosphate + UTP --> UDP-glucose + pyrophosphate AG° = ~0 kJ/mol Reaction 2: pyrophosphate --> 2 Pi AG = -19.2 kJ/mol Which statement is CORRECT concerning these 2 reactions? A reaction that couples reaction 1 and 2 together is not feasible. O A reaction that couples reaction 1 and 2 together would be very slow. A reaction that couples reaction 1 and 2 together is thermodynamically favorable. O A reaction that couples reaction 1 and 2 together is kinetically fast.
Q2 CH,OH CH,OPO,2- H H OH H OH 0-PO,2- но но OH H. OH OH Glucose-1-phosphate Glucose-6-phosphate Calculate the standard free-energy change (AG°) fo the reaction catalyzed by the enzyme phosphoglucomutase. When at standard temperature (25 °C) and pressure, the equilibrium of each molecule is 1 mM glucose 1-phosphate and 19 mM glucose 6-phosphate. HINT: at 25 °C, RT = 2.479 kJ/mol
Calculate the equilibrium constant at 37oC for the reactionGlucose + 6O2 + 36ADP + 36Phospate → 6CO2+ 42H2O +36ATP ... ... ...Entropy change of Go= − 423 kcal/mole of Glucose
Hi, Please help me answer these two questions. Hope I'd get all the items done. PS. If possible, provide explanations on it. Thank you so much. *** 1. Calculate the value of ΔGo’ for the following reaction. Glucose-1-phosphate → Glucose-6-phosphate Given: Glucose-1-phosphate → Glucose + Pi ΔGo’ = -20.9 kJ/mol Glucose + Pi → Glucose-6-phosphate ΔGo’ = +12.5 kJ/mol 2. The ΔGo’ for Coupled Reactions. Glucose-1-phosphate is converted into fructose-6-phosphate in two successive reactions: Glucose 1-phosphate →glucose 6-phosphate ΔGo’ = -7.3 kJ/mol Glucose 6-phosphate→fructose 6-phosphate ΔGo’ = + 1.7 kJ/mol Calculate the overall ΔGo’ and the equilibrium constant, K’eq for the sum of the two reactions: Glucose 1-phosphate → fructose 6-phosphate
In biological cells, the energy released by the oxidation of foods is stored in adenosine triphosphate (ATP or ATP4). The essence of ATP's action is its ability to lose its terminal phosphate group by hydrolysis and to form adenosine diphosphate (ADP or ADP³-) ATP+ (aq) + H₂0 (1)→ ADP³- (aq) + HPO¯(aq) + H₂O+ (aq) At physiological conditions, pH = 7 and T = 310K (blood temperature), the enthalpy and Gibbs energy of hydrolysis are A, H = -20 and AG = -31, respectively. The hydrolysis of 1.00 mol of ATP, at these conditions, results in the extraction of 31 kJ of energy that can be used to do non-expansion work such as the synthesis of proteins or activation of neurons. mol Calculate the entropy of hydrolysis at pH = 7 and T = 310K
Glucose-1-phosphate = glucose-6-phosphate AG'o = - 7.3 kJ mol-1 Glucose-6-phosphate = fructose-6-phosphate AG'O = + 1.7 kJ mol-1 Calculate the equilibrium constant, K'eg, for the conversion of glucose-1-phosphate to fructose-6-phosphate. You can use a value of RT = 2.48 kJ mol-1.
A protein has a stability ranging from 6 to 15 kcal/mole at 37 OC. Stability is a measure of the equilibrium between the folded and unfolded protein given by the relationship; folded(F) = unfolded (U), k=[U]/[F] For a protein with a stability of 10 kcal/mole, calculate the fraction of unfolded protein that would exist at equilibrium at 37 OC. Use the following equation: ∆GO=RTlnk=-2.3RTlog k, where R=1.98 x 10-3kcal/mole and T is temperature in Kelvin (K) and k is the equilibrium constant.
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P): G6P F6P AG-1.7 kJ Phosphoglucose Isomerase 6 CH2OPO, 2- H. CH2OPO3 ICH,OH H. H. HO. H. 1. H. HO. HO. H. 14 B OH H. H. HO. gluco se-6-phosphate fructo se-6-phosphate Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calc kJ when the concentrations of G6P and F6P are 60.0 ×10M and 14.0 x10 M respectively, assume the temperature is 37 °C. (R = 8.314 J/K-mol) O a. -0.1 kJ Ob.-1.1 kJ Oc.-2.1 kJ O d.-3.1 kJ O e.-4.1 kJ
HO HO OH OH OH Adenosine triphosphate (ATP) is an organic compound that provides energy in many biological reactions. The alcohol group of glucose has ATP alcohol, the glucose structure then has a HO HO OH OH 00 charges. After phosphorylation of the 'OH A) positive charge; negative charge B) positive charge; no charge C) negative charge; negative charge D) negative charge; no charge E) no charge; negative charge [1] +
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.
Which of the following processes have a AS > O? CH3OH(I) → CH3OH(s) N₂(g) + 3 H₂(g) → 2 NH3(g) CH4(g) + H₂O (g) → CO(g) + 3 H₂(g) Na₂CO3(s) + H₂O(g) + CO₂(g) → 2 NaHCO3(s) All of these processes have a DS > 0.
Use the following equilibria2CH4(g) C2H6(g) + H2(g) Kc1 = 9.5 × 10-13CH4(g) + H2O(g) CH3OH(g) + H2(g) Kc2 = 2.8 × 10-21to calculate the value of Kc' for the following reaction:2CH3OH(g) + H2(g) C2H6(g) + 2H2O(g)