4. An enzyme reaction in a biochemical cycle has an equilibrium constant that is 10 times...
Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature (37 °C). In order to be physiologically useful, the reaction needs to occur 2000 times faster than when it is uncatalyzed. Part A By how many kJ/mol must an enzyme lower the activation energy of the reaction to make it useful? Express your answer using two significant figures. PO AQ * o o ? Ea – Eac = kJ/mol
A certain reaction has an enthalpy value of -56.4 kJ moll. Assume this value remains constant in the temperature range being studied. The standard Gibbs free energy for the reaction is -241 kJ mol-'at 25.0 °C. What is the standard Gibbs free energy for the reaction at 50.0 °C?
step by step pls can you only do uestion 2 and 3 for me step by step From the information in the Data section of the textbook, calculate the equilibrium constant at 338 K for the reaction: assuming that the reaction enthalpy is independent of temperature. Answer: 0.0835 The degree of dissociation, a, for the following reaction is 0.655 at 298 K and 1.00 bar total pressure. Find K Answer: 300 Given K 9.18E-8 for the reaction: laqi and the...
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. NH,CI() = NH(aq) + Cl(aq) AGxn = -7.7 kJ/mol Determine the concentration of NH(aq) if the change in Gibbs free energy, AGrxn , for the reaction is –9.53 kJ/mol. [NH] = 0.72 Consider a general reaction enzyme A(aq) = B(aq) The AG® of the reaction is -4.880 kJ mol-....
Question B1. The binding of a substrate (S) to an enzyme (E) can be described by the equilibrium E + S 4 ES At 273 K the equilibrium constant for this reaction has been found to be 0.134 for a particular substrate and enzyme. (a) Calculate the standard Gibbs energy change for the binding reaction at 273 K. [1 mark] (b) Calculate the free energy change for the reaction when the concentrations of the species at 273 K are: [E]...
Explain IN WORDS the biochemical relationship between Gibbs free energy, equilibrium constant and reduction potential?
What is the standard Gibbs free energy of the reaction, H(aq) - H(aq)+F(aq), if its equilibrium constant at 298 Kis 7.2 10 A 177 kJfmol O B 17.9 kJ/mol O C -17.9 kJ/mol O D 584 l/mol
3. + 2.5/10 points Previous Answers McM8 6.P.012. The standard Gibbs-free energy of a system is related to its equilibrium constant through the following equation. AG° = -R·T· In(K) In this equation R is the gas constant, T is the temperature, and the ° next to AG defines the conditions as standard ambient temperature and pressure, i.e. "SATP". (Answer the following questions to three significant figures.) (a) Given an equilibrium constant of 4.53 x 10-6, what is its standard Gibbs-free...
a.) Calculate the equilibrium constant for the following reaction at 298.15 K from cell potential data. Express the answer as lnK. Sn4+ + 2Fe2+ ----> Sn2+ + 2Fe3+ b.) Calculate the standard Gibbs free energy change in kJ/mol at 298.15 K for the following reaction from cell potential data: 3Sn4+ + 2Cr ----> 3Sn2+ + 2Cr3+
For the reaction N2 (g) + 3H2(g) --> 2 NH3 (g), (a) what is the reaction Gibbs free energy at equilibrium in J/mol? The equilibrium constant of the reaction N2 (g) + 3H2(g) --> 2 NH3 (g) at 81 oC is Keq = 478,789. (b) What is the standard reaction Gibbs energy of this reaction in J/mol?