# How do the rates of the forward and reverse reactions compare at a state of dynamic chemical equilibrium?

##### 1 Answer

May 4, 2018

They are equal, and NONzero.

At **dynamic chemical equilibrium**, the rates of the forward and reverse reactions are ** equal to each other**, i.e.

#aA + bB stackrel(k_1" ")(rightleftharpoons) cC + dD#

#" "" "" "" "^(k_(-1))#

For this, assuming the equilibrium consists of elementary reactions, the **forward rate law** and **reverse rate law** are:

#r_1(t) = k_1[A]^a[B]^b#

#r_(-1)(t) = k_(-1)[C]^c[D]^d#

At equilibrium,

#k_1[A]^a[B]^b = k_(-1)[C]^c[D]^d#

From this, we obtain:

#K -= k_1/(k_(-1)) = ([C]^c[D]^d)/([A]^a[B]^b)#

We know that rate constants are **temperature-dependent**, and thus, so is

It is also important to note that the rates of the forward and reverse reactions **MUST be nonzero** to have a dynamic chemical equilibrium.