[Miffymycat]

Is the time taken to reach equilibrium the same whether starting from reactants or products, under the same conditions?  I think intuitively it is but cannot find a reference or proof.

[Jorriss]

Why do you think it would be the same?

[Miffymycat]

Well I guess my hunch is based on this:  If the forward reaction is faster, the system will proceed initially fast towards equilibrium but then as the slower reverse reaction builds up it will slow down and the last stages before equilibrium will be protracted ... whereas if we start with the reverse reaction it will be initially slow but maintain a more steady approach as the faster reaction builds up! As I say, I have no quantitative reasons just intuition - so how to set about proving this theory - if it possible to do!?

[curiouscat]

Consider  a reaction

A  B

[tex]

r_f=k_f*c_A \\
r_r=k_r*c_B \\

K_{eq}=k_f / k_r \\

Case 1: \\

t=0 \\
c_A=c_0 ; c_B=0;\\
 Find t=t_f \\

Case 2: \\
t=0 \\
c_A=0; c_B=c_0;\\
 Find t=t_r\\

At any t, c_A + c_B = c_0
At equilibrium c_A / c_B = K_eq
[/tex]

Based on this on ought to be able to derive the relation for t. I'm just feeling lazy to do that right now. 

[Miffymycat]

Thanks - thats looks very promising but dont know how to get tf and tr ... here's hoping you get a burst of energy sometime soon!  Thanks again for the post and taking an interest in my query.

[Babcock_Hall]

See equation (46) in chapter 11 of Catalysis in Chemistry and Enzymology, by William P. Jencks.  The rate constant for approach to equilibrium is k_obs = K_f + k_r.