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For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº = -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1
(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)
(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)
- a)-46.9 KJ mol-1
- b)+46.9 KJ mol-1
- c)-35 KJ mol
- d)+35 KJ mol
Correct answer is option 'A'. Can you explain this answer?
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°C and pH 7, the standard Gibbs free energy change (∆G°) is -30.5 kJ/mol. This means that the hydrolysis of 1 mole of ATP releases 30.5 kJ of energy.
The hydrolysis of ATP is an exergonic reaction, meaning it releases energy. This energy is used to drive various cellular processes, such as muscle contraction, active transport, and synthesis of macromolecules.
The hydrolysis of ATP involves the breaking of a high-energy phosphate bond, resulting in the formation of ADP (adenosine diphosphate) and inorganic phosphate (Pi). This reaction is catalyzed by the enzyme ATPase.
The released energy from ATP hydrolysis is often used to power cellular processes through the transfer of the phosphate group to other molecules. For example, when ATP donates its phosphate group to glucose, it forms ADP and glucose-6-phosphate, providing energy for the initial steps of glycolysis.
Overall, the hydrolysis of ATP plays a crucial role in cellular energy metabolism, providing the necessary energy for the functioning of various cellular processes.
The hydrolysis of ATP is an exergonic reaction, meaning it releases energy. This energy is used to drive various cellular processes, such as muscle contraction, active transport, and synthesis of macromolecules.
The hydrolysis of ATP involves the breaking of a high-energy phosphate bond, resulting in the formation of ADP (adenosine diphosphate) and inorganic phosphate (Pi). This reaction is catalyzed by the enzyme ATPase.
The released energy from ATP hydrolysis is often used to power cellular processes through the transfer of the phosphate group to other molecules. For example, when ATP donates its phosphate group to glucose, it forms ADP and glucose-6-phosphate, providing energy for the initial steps of glycolysis.
Overall, the hydrolysis of ATP plays a crucial role in cellular energy metabolism, providing the necessary energy for the functioning of various cellular processes.
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For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer?
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For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer?.
For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer?.
Solutions for For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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ample number of questions to practice For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº= -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)a)-46.9 KJ mol-1b)+46.9 KJ mol-1c)-35 KJ mold)+35 KJ molCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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