Animated Solution for Chemistry - Chemical Equilibrium: The equilibrium constant $K_c$ for the reaction $S{O}_3(g)\rightleftharpoons S{O}_2(g)+\frac{1}{2}{O}_2(g)$ is $4.9\times 10^{-2}$. The value of $K_c$ for the reaction $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$ will be

In chemical equilibrium, the equilibrium constant $K_c$ is a powerful tool that tells us the ratio of products to reactants at equilibrium. But what makes it truly fascinating is how it responds when we mathematically manipulate the chemical equation.

We are given an initial reaction:

The equilibrium constant for this reaction is given as $K_{c1}=4.9\times 10^{-2}$.

Our goal is to find the equilibrium constant, let's call it $K_{c2}$, for a new target reaction:

Notice how the target reaction is related to the initial one. The products and reactants have swapped places, and the stoichiometric coefficients have changed. This means we need to perform a series of mathematical transformations on our initial equation to reach the target equation.

Let's break down the transformation into two logical steps.

Step 1: Reversing the Reaction First, we need $S{O}_2$ and $O_2$ on the reactant side. To achieve this, we reverse the initial reaction:

According to the laws of chemical equilibrium, when you reverse a reaction, the new equilibrium constant becomes the reciprocal of the original constant. So, our intermediate constant is:

Step 2: Multiplying by a Factor Next, we look at the coefficients. Our reversed reaction has $1$ mole of $S{O}_2$, but our target reaction requires $2$ moles. Therefore, we must multiply the entire reversed reaction by $2$:

When a chemical equation is multiplied by a factor $n$, the equilibrium constant is raised to the power of $n$. Since we multiplied by $2$, we must square our intermediate constant:

This is our master equation! It perfectly links the unknown constant to the known one.

Now comes the execution phase. Let's substitute the given value of $K_{c1}$ into our master equation:

To make the calculation smoother, let's handle the power of ten first. Bringing $10^{-2}$ to the numerator gives us $10^2$, which is $100$:

Now, we square both the numerator and the denominator. The numerator is simply $100^2=10000$. For the denominator, $4.9^2$ is approximately $24.01$:

Dividing $10000$ by a number slightly larger than $24$ will give us a result slightly less than $\frac{10000}{24}$. Since $\frac{1000}{24}\approx 41.6$, our result should be around $416$.

Rounding to the nearest integer, we get $416$. This perfectly matches option (a). The beauty of this problem lies in trusting the properties of the equilibrium constant and executing the algebra with precision!