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

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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 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.

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