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

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Answer 1 · 2018-05-04T01:21:21

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, #r_1(t) = r_(-1)(t)#, so:

#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 #K#.

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