The equilibrium constant for the dissociation of molecular
iodine...

Question

The equilibrium constant for the dissociation of molecular
iodine, I21g2Δ2 I1g2, at 800 K is Kc = 3.1 * 10-5.
(b) Assuming both forward and reverse reactions are
elementary processes, which reaction has the larger rate
constant, the forward or the reverse reaction?

Accepted Answer

Hi everyone. So as to consider the following reaction below where the K value was eight times 10 to the eight. We were asked to determine whether the four or the reverse reaction has a larger rate constant, assuming that both reactions are elementary reactions. To determine this, we need to look at the equilibrium constant K equilibrium concept. It was the right concept for the forward reaction by by the rate constant for the reverse reaction. Okay, it's gonna be greater than one if the right constant of the forward reaction is greater than the rate constant of the reverse reaction. So this means we have a bigger number on top and we divided by a smaller number on the bottom. And if K is less than one, the rate constant of the forward reaction gonna be less than the rate constant of the reverse reaction. So this means we have a small number on top, divided by a bigger number on the bottom, since K equals eight Times 10 to the eight And eight times 10 to the eight is greater than one. This means that the full reaction has a larger rate constant in a reverse reaction. Thanks for watching my video and I hope it was helpful

The equilibrium constant for the dissociation of molecular iodine, I21g2Δ2 I1g2, at 800 K is Kc = 3.1 * 10-5. (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger rate constant, the forward or the reverse reaction?

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