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.