Part B For a different reaction, Kc = 120x10^7, kf = 8.74 X 10^3s^-1 , and kr =

Question

Part B For a different reaction, Kc = 120x10^7, kf = 8.74 X 10^3s^-1 , and kr = 7.26x10^-4s^-1 . Adding a catalyst increases the forward rate constant to 4.27x10^6s^-1 . What is the new value of the reverse reaction constant, kr, after adding catalyst? Express your answer with the appropriate units. Part A For a certain reaction, Kc = 6.89x10^-2 and kf = 0.269M^-2. s^-1. Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units.

Explanation / Answer

Part A:-

Kc = Kf/Kr

Thus, (6.89*10-2) = 0.269/Kr

or, Kr = 0.256 M-2 s-1

Part B:-

Sice additional of a ctalyst does not change the value of Kc

Therefore, Kr = Kf/Kc = (4.27*106)/(1.2*107) = 0.356 s-1

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