Adenosine triphosphate (ATP) is a compound that provides energy for biomechanical reactions in the body when it undergoes hydrolysis. For the hydrolysis of ATP at 37 °C (normal body temperature), AH; = -20.0 kJ-mol and AS; = +34.0 J-K.mol. Assuming that these quantities are independent of temperature, calculate the temperature above which the equilibrium constant for the hydrolysis of ATP becomes greater than one. T =

Adenosine triphosphate (ATP) is a compound that provides energy for biomechanical reactions in the body when it undergoes hydrolysis. For the hydrolysis of ATP at 37 °C (normal body temperature), AH; = -20.0 kJ-mol and AS; = +34.0 J-K.mol. Assuming that these quantities are independent of temperature, calculate the temperature above which the equilibrium constant for the hydrolysis of ATP becomes greater than one. T =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

For a certain reaction, ΔHº = −1.8 x 102 kJ mol−1 and the equilibrium constant at 800. K is 9.1 × 102 . Estimate the temperature at which the equilibrium constant is 1.0 × 106 .
he decomposition of a generic diatomic element in its standard state is represented by the equation 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.20 kJ·mol−1 at 2000. K and −53.38 kJ·mol−1 at 3000. K. Determine the value of the thermodynamic equilibrium constant, ?, at each temperature. At 2000. K, Δ?f=4.20 kJ·mol−1. What is ? at that temperature? ? at 2000. K= At 3000. K, Δ?f=−53.38 kJ·mol−1. What is K at that temperature? ? at 3000. K=
Consider the oxidation of a generic metal M into a metal oxide as shown in the reaction. 2 M(s) +20₂(g) → M₂O4(s) For this reaction, AH = -4.961 kJ/mol and AS ixn = 12.12 J/(mol-K) at 298 K. What is the standard change in Gibbs free energy for the reaction in the forward direction? AGixn= What is the equilibrium constant of this reaction at 298 K? Kp = What is the equilibrium pressure of O₂(g) over M(s) at 298 K? MacBook Pro tyne URI + kJ/mol O
Consider the following reversible reaction at equilibrium: C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l). Given that this reaction is exothermic, if heat is added to the equilibrium system, how is the stress relieved?
The decomposition of a generic diatomic element in its standard state is represented by the equation X₂(g) → X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.71 kJ - mol-¹ at 2000. K and −55.51 kJ · mol-¹ at 3000. K. Determine the value of the thermodynamic equilibrium constant, K, at each temperature. At 2000. K, AG₁ = 4.71 kJ · mol-¹. What is K at that temperature? K at 2000. K= At 3000. K, AGf = −55.51 kJ - mol-¹. What is K at that temperature? K at 3000. K =
Determine the standard Gibbs free energy of reaction, in kJ/mole, for the reaction of aqueous sodium carbonate with hydrochloric acid. G values for the reactants and products are given below. NaCO(aq) + 2 HCl(aq) 2 NaCl(aq) + HO() + CO(g) NaCO(aq) -1051.6HCl(aq) -131.2NaCl(aq) -393.1HO() -237.1CO(g) -394.4
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction N, (g) + 3 H, (g) = 2 NH,(g) the standard change in Gibbs free energy is AG° = –72.6 kJ/mol. What is AG for this reaction at 298 K when the partial pressures are PN, = 0.500 atm, PH, = 0.500 atm, and PNH, = 0.600 atm? AG = kJ/mol
Calculate the standard change in Gibbs free energy, AGixn, for the given reaction at 25.0 °C. Consult the table of thermodynamic properties for standard Gibbs free energy of formation values. KCI(s) K+ (aq) + Cl(aq) AGxn= Determine the concentration of K+ (aq) if the change in Gibbs free energy, AGran, for the reaction is -8.21 kJ/mol. [K+] = kJ/mol M
For a certain chemical reaction, the equilibrium constant K = ×7.7106 at 5.00 °C . Calculate the standard Gibbs free energy of reaction.Round your answer to 2 significant digits.
For a certain chemical reaction, the standard Gibbs free energy of reaction is 129. kJ . Calculate the temperature at which the equilibrium constant K = ×6.010−25 .Round your answer to the nearest degree.
For a certain chemical reaction, the equilibrium constant K = 2.8 × 104 at 5.00 °C. Calculate the standard Gibbs free energy of reaction. Round your answer to 2 significant digits. AG° = kJ x10 X Ś
For a certain chemical reaction, the standard Gibbs free energy of reaction is - 149. kJ. Calculate the temperature at which the equilibrium constant K = 1.3 × 1026. Round your answer to the nearest degree. T=