[Big-Daddy]

The question is:

There is a reversible reaction of form aA + bB    cC + dD (a, b, c and d are stoichiometric coefficients). The rate of the forward reaction is k₁*[A]^nA*[ B]^nB (nA and nB are orders of the reaction). The rate of the backward reaction is k₂*[C]^nC*[D]^nD.

K_eq=([C]^c*[D]^d)/([A]^a*[ B]^b)

If the "rate of reaction" is defined as the rate of loss of A, then what is the differential rate equation?

I have worked out an expression for K_eq in terms of k₁, k₂ and the various concentrations, but I'm not sure how this helps me reach the differential rate equation for loss of A. That expression is:

k₁/k₂=K_eq*(([C]^nC-c*[D]^nD-d)/([A]^nA-a*[ B]^nB-b))

if you need it. Thanks for the help.

Edit: This website will provide valuable background if you have not seen it yet (look at the section on "Equilibrium reactions or opposed reactions": http://en.wikipedia.org/wiki/Rate_equation).