For a certain reaction, K_c = 9.23 Times 10^-7 and k_f = 4.32 Times 10^5 M^-2 s^
ID: 940876 • Letter: F
Question
For a certain reaction, K_c = 9.23 Times 10^-7 and k_f = 4.32 Times 10^5 M^-2 s^-1. Calculate the value of the reverse rate constant, k_r, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M^-2 s^-1 include (multiplication dot) between each measurement. For a different reaction, K_c = 355, k_f = 579s^-1, and k_r = 1.63 s^-1. Adding a catalyst increases the forward rate constant to 1.11 Times 10^5 s^-1. What is the new value of the reverse reaction constant, k_r, after adding catalyst? Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M^-2 s^-1 include A ¢ (multiplication dot) between each measurement Yet another reaction has an equilibrium constant K_c = 4.32 Times 10^5 at 250 degree C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 degree C, what will happen to the equilibrium constant?Explanation / Answer
A) Kc = kf/kr hence 9.23 x 10^-7 = 4.32 x 10^5 /kr
kr = 4.68 x 10^11
B) Kc=Kf/Kr hence 355 = 1.11 x 10^5 / kr ,
kr = 312.7
C) for exothermic reaction Kc will decrease as temperature rises