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.

$a A + b B k_{1} ⇌ c C + d D$ $aA + bB stackrel(k_1" ")(rightleftharpoons) cC + dD$
$_{- 1}^{k_{- 1}}$ $" "" "" "" "^(k_(-1))$

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

$r_{1} (t) = k_{1} {[A]}^{a} {[B]}^{b}$ $r_1(t) = k_1[A]^a[B]^b$
$r_{- 1} (t) = k_{- 1} {[C]}^{c} {[D]}^{d}$ $r_(-1)(t) = k_(-1)[C]^c[D]^d$

At equilibrium, $r_{1} (t) = r_{- 1} (t)$ $r_1(t) = r_(-1)(t)$ , so:

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

From this, we obtain:

$K \equiv \frac{k_{1}}{k_{- 1}} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$ $K -= k_1/(k_(-1)) = ([C]^c[D]^d)/([A]^a[B]^b)$

We know that rate constants are temperature-dependent, and thus, so is $K$ $K$ .

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