The ΔG°\' of the reaction is -7.050 kJ ·mol–1. Calculate the equilibrium constant for the reaction....

The ΔG°\' of the reaction is -7.050 kJ ·mol–1. Calculate the equilibrium constant for the reaction. (Assume a temperature of 25° C.What is ΔG at body temperature (37.0° C) if the concentration of A is 1.8 M and the concentration of B is 0.75 M?

Expert Solution

queen_honey_blossom answered 4 hours ago

c. For this reaction in heart muscle, ΔG°’ = +4.7 kJ/mol but ΔG = -0.6 kJ/mol....

c. For this reaction in heart muscle, ΔG°’ = +4.7 kJ/mol but ΔG = -0.6 kJ/mol. i. Explain, in words, how ΔG can be negative when ΔG°’ is positive. ii. What would be the ratio of 3PG to 2PG if the reaction were at equilibrium at 25°C? iii. What is the actual ratio of 3PG to 2PG in heart muscle (T = 37°C)?

1. Reaction 1 has a ΔG° of –12.3 kJ/mol, and Reaction 2 has a ΔG° of...

1. Reaction 1 has a ΔG° of –12.3 kJ/mol, and Reaction 2 has a ΔG° of 23.4 kJ/mol. Which statement is TRUE of these two reactions? Select one: A. Reaction 1 occurs faster. B. Reaction 2 occurs faster. C. Both reactions occur at the same rate. D. Reaction 2 will not occur. E. It is impossible to know which reaction occurs faster with this information. 2. Fructose-1-phosphate can be hydrolyzed into fructose + inorganic phosphate (Pi) with a ΔG° of...

Consider the reaction: 2 NO2(g) → N2O4(g) Calculate ΔG (in kJ/mol) at 298°K if the equilibrium...

Consider the reaction: 2 NO2(g) → N2O4(g) Calculate ΔG (in kJ/mol) at 298°K if the equilibrium partial pressures of NO2 and N2O4 are 1.337 atm and 0.657 atm, respectively.

The reaction 1,3-bisphosphoglycerate + ADP <=> 3-phosphoglycerate + ATP has ΔG° = -18.8 kJ mol-1. Calculate...

The reaction 1,3-bisphosphoglycerate + ADP <=> 3-phosphoglycerate + ATP has ΔG° = -18.8 kJ mol-1. Calculate ΔGrxn in kJ mol-1 at 37.0 °C when [1,3-bisphosphoglycerate] = 1.90 mmol L-1, [ADP] = 1.20 mmol L-1, [3-phosphoglycerate] = 0.450 mmol L-1, and [ATP] = 4.00 mmol L-1. (R = 8.3145 J mol-1 K-1) The reaction S(aq) + T(aq) <=> U(aq) has K = 17.5. Which of the following is true when [S] = 12.0 mmol L-1, [T] = 19.0 mmol L-1, and...

a)Calculate the equilibrium constant at 10 K for a reaction with ΔHo = 10 kJ and...

a)Calculate the equilibrium constant at 10 K for a reaction with ΔHo = 10 kJ and ΔSo = 100 J/K. b) Calculate the equilibrium constant at 105 K for the thermodynamic data in the previous question.

For a particular reaction at 129.9 °C, ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K).Calculate...

For a particular reaction at 129.9 °C, ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K).Calculate ΔG for this reaction at -3.4 °C.

Given that ΔG∘ = −13.6 kJ/mol, calculate ΔG at 25∘C for the following sets of conditions....

Given that ΔG∘ = −13.6 kJ/mol, calculate ΔG at 25∘C for the following sets of conditions. 1) 30 atm NH3, 30 atm CO2, 4.0 M NH2CONH2 Express the free energy in kilojoules per mole to two significant figures. 2) 8.0×10−2 atm NH3, 8.0×10−2 atm CO2, 1.0 M NH2CONH2 Express the free energy in kilojoules per mole to two significant figures. Is the reaction spontaneous for the conditions in Part A and/or Part B? A) Is the reaction spontaneous for the...

What is the intracellular glucose concentration if the ΔG for the following reaction is -20.1 kJ/mol...

What is the intracellular glucose concentration if the ΔG for the following reaction is -20.1 kJ/mol at 37°C and concentrations for glucose-6-phosphate and phosphate are both 1 mM? glucose-6-phosphate --------> glucose + Pi ΔG° = -13.8 kJ/mol A) 1.9 M B) 87 M C) 1.9 mM D) 27 mM E) 87 mM The Answer is E. I just need help understanding why it is E. please show all your work. I really need to understand the entire process. Please don't...

The reaction glucose + ATP <=> glucose-1-phosphate + ADP has ΔG°' = -9.6 kJ mol-1. This...

The reaction glucose + ATP <=> glucose-1-phosphate + ADP has ΔG°' = -9.6 kJ mol-1. This is a coupled reaction which can be broken into two parts. The first part of the reaction is ATP + H2O <=> ADP + Pi, with ΔG°' = -30.5 kJ mol-1. The second part of the reaction is glucose + Pi <=> glucose-1-phosphate + H2O. Calculate the ΔG°' value for this reaction in kJ mol-1.

For a particular reaction at 129.9 °C, ΔG = 319.61 kJ/mol, and ΔS = 777.00 J/(mol·K)....

For a particular reaction at 129.9 °C, ΔG = 319.61 kJ/mol, and ΔS = 777.00 J/(mol·K). delta g is -47.6 celcius

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