Animated Solution for Chemistry - Chemical Kinetics: For the reaction $2{\text{H}}_2(g)+2\text{NO}(g)\to {\text{N}}_2(g)+2{\text{H}}_2\text{O}(g)$ the observed rate expression is, $\text{rate}=K_f[\text{NO}{]}^2[{\text{H}}_2]$. The rate expression for the reverse reaction is

When we first look at the chemical equation $2{\text{H}}_2(g)+2\text{NO}(g)\rightleftharpoons {\text{N}}_2(g)+2{\text{H}}_2\text{O}(g)$, our instinct might be to assume it's a simple, single-step process. If that were true, the forward rate law would perfectly mirror the stoichiometry: $r_f=K_f[{\text{H}}_2{]}^2[\text{NO}{]}^2$.

However, the problem throws a curveball. We are given the observed forward rate expression:

Notice the difference? The power of $[{\text{H}}_2]$ is $1$, not $2$. This discrepancy is a massive clue: this is a complex, multi-step reaction. Because of this, we cannot simply guess the reverse rate law by looking at the products. We need a more robust, foolproof method.

To find the reverse rate law, we must invoke the ultimate equalizer in chemistry: Chemical Equilibrium.

At equilibrium, the chaotic dance of molecules reaches a perfect balance. The rate at which reactants form products exactly equals the rate at which products revert to reactants. Mathematically, this is expressed as:

Furthermore, the equilibrium constant $K_c$ bridges the gap between thermodynamics and kinetics. By the Law of Mass Action, $K_c$ is the ratio of product concentrations to reactant concentrations, raised to their stoichiometric coefficients:

Simultaneously, $K_c$ is also the ratio of the forward and reverse rate constants:

Now, we have two distinct expressions for the same constant, $K_c$. Let's equate them to unlock the relationship between the rate constants and the concentrations:

Our goal is to find the reverse rate, $r_b$. To do this, we first need to isolate the components of our known forward rate, $r_f$. Let's cross-multiply to group the terms:

Look closely at the left side of the equation: $K_f[{\text{H}}_2{]}^2[\text{NO}{]}^2$. It looks incredibly similar to our given forward rate law, $r_f=K_f[\text{NO}{]}^2[{\text{H}}_2]$, but it has an extra $[{\text{H}}_2]$ term.

To make the left side perfectly match $r_f$, we simply divide both sides of the equation by $[{\text{H}}_2]$:

Suddenly, the left side is exactly $r_f$. And since we established earlier that $r_f=r_b$ at equilibrium, the entire right side of the equation must represent the reverse rate law!

This elegant algebraic manipulation reveals the hidden mechanics of the reverse reaction. The inverse dependence on $[{\text{H}}_2]$ suggests a fascinating mechanism where hydrogen might actually inhibit one of the reverse elementary steps. By trusting the fundamental laws of equilibrium, we bypassed the complexity of the reaction mechanism and arrived flawlessly at the correct answer.