For a certain reaction, Kc = 38.5 and k1 = 6.33x 103 M-2 . s-1 . Calculate the v

Question

For a certain reaction, Kc = 38.5 and k1 = 6.33x 103 M-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. For a different reaction, Kc = 1.45x105, kf = 4.63 x 105s-1, and k, = 3.19s-1 . Adding a catalyst increases the forward rate constant to 8.24x107s-1 . What is the new value of the reverse reaction constant, kr, after adding catalyst? Express your answer with the appropriate units. Yet another reaction has an equilibrium constant Kc = 4.32 x 105 at 25 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) kr=kf/kc =0.1644155844155844 X 10^3, b) kr= kf/kc ==4.63X 10^5/1.45X 10^5 == 3.193103448275862

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