The generic reaction 2AB has the following rate laws: forward reaction:reverse r

Question

The generic reaction 2AB has the following rate laws: forward reaction:reverse reaction:rate=kf[A]2rate=kr[B] where kf is the rate constant for the forward reaction and kr is the rate constant for the reverse reaction. At equilibrium, the two rates are equal and so kf[A]2=kr[B]. The equilibrium constant for a reaction is related by the law of mass action to the rate constants for the forward and reverse reactions: Kc=[B][A]2=kfkr

At a certain temperature the initial concentration of NO was 0.400 M and that of Br2 was 0.265 M . At equilibrium the concentration of NOBr was found to be 0.250 M. Kc = 19.8

At this temperature the rate constant for the reverse reaction is 300. M1s1 . What is kf for the reaction?

Explanation / Answer

Equilibrium constant Kc = 19.8

formula : Kc = Kf / Kr

Kf = forward rate constant

Kr = reverse rate constant

Kc = 19.8

Kr = 300.

Kc = Kf / Kr

19.8 = Kf / 300

Kf = 5940

forward rate constant Kf = 5940 M-1s-1

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