Question

At a given temperature, the elementary reaction A --->B in the forward direction is first order in A with a rate constant of 1.60*10^2 s^-1. The reverse reaction is first order in B and the rate constant is 9.30*10^-2 s^-1
What is the value of the equilibrium constant for the reaction A --->B at this temperature?


What is the value of equilibrium constant for the reaction B-->A at this temperature?

(and all reactions are in equilibrium of course i just did not know how to make the second arrow =)

Answer 1

For ,  s^-1

For ,  s^-1

 

So for,

The Equilibrium Constant  is defined as the forward rate divided by the reverse rate:

 

Do the same for , except in this case would be equal to 

Answer 2

The first order rate constant for elementary reaction A- Bis k 1.60x10 s The reverse first order rate constant for the above elementary reaction is k,-9.3x10-2 s-1 The equillibrium constant for the reaction A - B is given below as follows: K- 1.60x 102 s 9.3x10 s K 1720.43 The equillibrium constant for the reaction B -> A is given below as follows: h, 9.3x102 s1 1.60x10 s1 K-5.8125