Equilibrium constant vs Reaction rate constant

Question

For a reaction, e.g., $$a X + b Y → c Z$$ Its reaction rate constant is $${\displaystyle r=k_f(T)[\mathrm {X} ]^{m}[\mathrm {Y} ]^{n}}$$ where the exponents m and n are called partial orders of reaction and are not generally equal to the stoichiometric coefficients a and b.

However, for the equilibrium constant, the exponents must be stoichiometric, e.g., $$K_c=\frac{[Z]^c}{[X]^a [Y]^b}$$

However, for reversible chemical reaction, $K_c=k_f/k_b$ which means that stoichiometric coefficients are required for equilibrium. Does this mean that when the exponents of the rate reaction equation are similar to that of stoichiometric coefficients, the reaction is in equilibrium?

Answer 1

If there are more complex reactions, reaction rate equations do not follow the numerator and denominator of the equilibrium constant equation.

• They may not match stoichiometric coefficients.
• The may have non-integer exponents.
• They may contain additive and/or ratio terms.

Such reactions are not elementary and have a reaction schema.

Therefore, such forward/backward reaction rates, involving initial reactants and final products, do not belong to the opposite directions of the same elementary reaction.

The ratio of their rate constants then need not to match the equation for the equilibrium constant.

As example, see surprisingly complex forward reaction rate equation(small PDF) for the reaction

$$\ce{H2(g) + Br2(g) <=> 2 HBr(g)}$$

$$\frac{\mathrm{d}[\ce{HBr}]}{\mathrm{d}t}=\frac{k_1[\ce{H2}]{[\ce{Br2}]}^{\frac 12}}{1 + k_2\frac{[\ce{HBr}]}{[\ce{Br2}]}}$$ (simplified)

Answer 2

There is no contradiction if the partial order of reaction differ from the stochiometric parameters of the reaction. This will depend on the reaction mechanism.

If you consider a reaction that is kinetically reversible, for example $$ \ce{A + B <=> C + D} $$ both the forward and backward steps of the reaction are elementary.Since the reactions are elementary in both the reactions, the partial orders of the reactions coincide with the stoichiometric coefficients.

$$ \begin{align} \ce{A + B -> C + C} & \quad v_f = k_f |A||B|\\ \ce{C + D -> A + B} & \quad v_b = k_b |C||D| \end{align} $$

And the expressión of $K_c$ is consistent with the one you expected.

However, if the reaction mechanism does not correspond to a secuence of reversible steps you can not use the stoichiometric coefficients to predict the form of the rate equations. An example of this case would be ammonium equilibrium

$$\ce{3 H_2(g) + N_2(g) <=> 2NH_3(g)}$$