Animated Solution for Chemistry - Chemical Equilibrium: Consider the following reversible chemical reactions, $A_2(g)+B_2(g)\rightleftharpoons 2AB(g)$ ...(i) with equilibrium constant $K_1$ $6AB(g)\rightleftharpoons 3A_2(g)+3B_2(g)$ ...(ii) with equilibrium constant $K_2$ The relation between $K_1$ and $K_2$ is

The Art of Reaction Manipulation

Chemical equilibrium is a delicate dance of molecules, and the equilibrium constant ($K$) is the mathematical choreographer that describes this balance. In this problem, we are presented with two related chemical reactions and asked to find the mathematical bridge connecting their equilibrium constants, $K_1$ and $K_2$.

Let's break down the thought process step-by-step and uncover the elegant rules that govern these transformations.

We are given two reversible reactions:

Reaction 1: This reaction has an equilibrium constant denoted as $K_1$.

Reaction 2: This reaction has an equilibrium constant denoted as $K_2$.

Our objective is to express $K_2$ in terms of $K_1$. To do this, we must first write out the explicit mathematical expressions for both constants using the Law of Mass Action.

The Law of Mass Action states that for a general reaction $aA+bB\rightleftharpoons cC+dD$, the equilibrium constant $K_c$ is given by the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients:

Applying this fundamental law to our first reaction, we get:

Now, let's construct the expression for the second reaction. Here, the products are $A_2$ and $B_2$, both with a coefficient of 3, and the reactant is $AB$ with a coefficient of 6. Therefore:

Now comes the crucial step: finding the relationship between these two expressions. Let's take a closer look at the expression for $K_2$. Notice that every exponent in the numerator and denominator is a multiple of 3. We can factor out this common power of 3:

Does the term inside the parentheses look familiar? It is exactly the reciprocal (the upside-down version) of our expression for $K_1$!

Since $K_1=\frac{[AB{]}^2}{[A_2][B_2]}$, it follows that:

By substituting $\frac{1}{K_1}$ into our factored equation for $K_2$, we arrive at the final relationship:

Using the standard rules of exponents, we can rewrite this as:

This perfectly matches option (c).

While deriving the relationship algebraically is a great exercise, there are two powerful shortcuts you should always keep in your toolkit for competitive exams like JEE and NEET:

1. Reversing a Reaction: If you reverse a chemical equation, the new equilibrium constant is the reciprocal of the original constant ($K_{new}=\frac{1}{K_{old}}$). 2. Multiplying by a Factor: If you multiply a chemical equation by a stoichiometric factor '$n$', the new equilibrium constant is the original constant raised to the power of '$n$' ($K_{new}=(K_{old}{)}^n$).

If we look at our original problem through the lens of these rules, we can see that Reaction 2 is simply Reaction 1 reversed and then multiplied by 3.

Reversing gives us $K^{-1}$, and multiplying by 3 raises it to the power of 3, instantly giving us the result $K^{-3}$. Mastering these rules will save you precious time and prevent silly algebraic mistakes!