An uncatalyzed reaction has a rate of 4.2 x 10^-7 sec-1 at 25 degrees Celsius. When an enzyme is added the rate is 3.2 x 10^4 sec-1. Calculate the difference in activation energy between the catalyzed and uncatalyzed reactions. (Please show work so I can see what you did).
A. 22 kJ/mol
B. 25 kJ/mol
C. 43 kJ/mol
D. 62 kJ/mol
An uncatalyzed reaction has a rate of 4.2 x 10^-7 sec-1 at 25 degrees Celsius. When...
An uncatalyzed reaction has a rate of 4.4 x 10-7 s–1 at room temperature (25 °C). When an enzyme is added the rate is 3.1 x 104 s–1. If the height of the activation barrier for the uncatalyzed rate is 28.6 kcal/mol, what is the height of the activation barrier for the catalyzed rate? Report your answer in terms of kcal/mol to the nearest tenths. Also, assume the pre-exponential terms for the uncatalyzed and catalyzed reactions are the same.
calculate the rate constant at 25 degrees celsius for a reaction that has a rate constant of 5.5x10^-5 s^-1 at 100 degrees celsius and an activation energy of 40.0 KJ/mol
The activation energy of a certain uncatalyzed biochemical reaction is 46.7 kJ/mol. In the presence of a catalyst at 39ºC, the rate constant for the reaction increases by a factor of 2030 as compared with the uncatalyzed reaction. Assuming the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction. Activation energy = kJ/mol
The uncatalyzed decomposition of CH3CHO has an activation energy of 188 kJ mol-1. I2 catalyzes the reaction, and at 600 K the catalyzed reaction is ten million times faster than the uncatalyzed reaction. Therefore, assuming that the pre-exponential factor, A, is the same for both the uncatalyzed and catalyzed reactions, the activation energy of the catalyzed reaction is, in kJ mol- 80.4 B. 71.0 C. 108 D. 117 E. 172
A particular first-order reaction has a rate constant of 1.35 x 10^2s-1 at 25.0 degrees celsius. What is the magnitude of k at 65.0 degrees celsius if Ea = 55.5 kJ/mol?
Be sure to answer all parts. The rate constant of a reaction is 4.2 x 10's at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation. *10 ,1
Ignore #9
actually ignore the whe whole question because is
missing information! Thank You
9. Which represents the catalyzed reaction activation energy? a. AGN(double dagger) b. AGE(double dagger) C.S(double dagger) d. ES(double dagger) 10. For the reactions in the diagram, the rate of the catalyzed reaction is 955 times that of the uncatalyzed reaction. What is the stabilization energy difference between these reactions at standard temperature? a. 17.0 kJ/mol b. 18.6 kJ/mol C. 19.0 kJ/mol d. 21.3 kJ/mol
Suppose that a catalyst lowers the activation barrier of a reaction from 121 kJ mol−1 to 55 kJ mol−1 By what factor would you expect the reaction rate to increase at 25 ∘C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)
4. Suppose that a reaction R → P proceeded according to the energy diagram shown. a. [4 pts] A catalyst lowers the activation barrier of a reaction from 125 kJ/mol to 55 kJ/mol. Write out a reaction mechanism for the catalyzed reaction. b. [6 pts] Sketch what the catalyzed reaction pathway would look like on an energy diagram. All relative heights should make sense for catalyzed pathways for full credit. c. [6 pts] By what factor would you expect the...
Calculate the rate constant at 225 degrees C for a reaction that has a rate constant of 8.1 x 10^-4 s^-1 at 95 degrees C and an activation energy of 95.0 kj/mol