Animated Solution for Chemistry - Chemical Equilibrium: For the following reactions, equilibrium constants are given : $\text{S}(s)+{\text{O}}_2(g)\rightleftharpoons {\text{SO}}_2(g);K_1=10^{52}$ $2\text{S}(s)+3{\text{O}}_2(g)\rightleftharpoons 2{\text{SO}}_3(g);K_2=10^{129}$ The equilibrium constant for the reaction, $2{\text{SO}}_2(g)+{\text{O}}_2(g)\rightleftharpoons 2{\text{SO}}_3(g)$ is

Imagine you are a chemical architect, and your job is to build a specific target reaction using a set of given reaction "building blocks." In this problem, we are given two foundational reactions along with their equilibrium constants, $K_1$ and $K_2$.

Our ultimate goal is to find the equilibrium constant, $K_3$, for a specific target reaction:

To achieve this, we cannot just blindly mix things together. We must strategically manipulate our given blocks so that when they are combined, they perfectly form our target structure.

Before we start moving pieces around, let's quickly review the golden rules of manipulating equilibrium constants. These rules are the tools in our architectural toolkit:

1. Reversing a Reaction: If you flip a reaction so that reactants become products and vice versa, the new equilibrium constant is the reciprocal of the original. ($K_{new}=\frac{1}{K}$)

2. Multiplying by a Factor: If you multiply the entire stoichiometric equation by a number $n$, the new equilibrium constant is the original raised to the power of $n$. ($K_{new}=K^n$)

3. Adding Reactions: When you add two or more reactions together to form a new overall reaction, you must multiply their individual equilibrium constants. ($K_{new}=K_A\times K_B$)

Now, let's look closely at our target reaction. We need exactly $2{\text{SO}}_2(g)$ on the reactant side.

Where can we find ${\text{SO}}_2$? It's in our first given reaction:

However, there are two problems here: ${\text{SO}}_2$ is on the product side, and there is only one mole of it. To fix this, we must apply our rules. First, we reverse the reaction, and then we multiply it by 2.

Let's see what happens to the equation and its constant:

Because we reversed it and multiplied by 2, our new constant $K_1^{\prime \prime }$ becomes:

Next, we bring in our second given reaction exactly as it is, because it already has the $2{\text{SO}}_3(g)$ we need on the product side:

Let's add our modified first reaction and the second reaction together. Notice the magic that happens: the $2\text{S}(s)$ on the product side of the first reaction perfectly cancels out the $2\text{S}(s)$ on the reactant side of the second. Furthermore, $2{\text{O}}_2(g)$ on the product side cancels with 2 of the 3 moles of ${\text{O}}_2(g)$ on the reactant side, leaving exactly one mole of ${\text{O}}_2(g)$.

The resulting sum is exactly our target reaction!

Since we added the two reactions to get our target, we must multiply their equilibrium constants to find $K_3$:

Now, it's just a matter of plugging in the given values and doing the math. Don't make a silly mistake with the exponents here!

First, square the denominator: $52\times 2=104$.

Finally, subtract the exponents: $129-104=25$.

And there we have it! The equilibrium constant for our target reaction is $10^{25}$, which corresponds perfectly to option (a). This problem beautifully demonstrates how chemical equations can be treated algebraically, provided you follow the strict rules governing their equilibrium constants.