Animated Solution for Chemistry - Chemical Equilibrium: For the reaction, $2{\text{SO}}_2(g)+{\text{O}}_2(g)\rightleftharpoons 2{\text{SO}}_3(g)$ In an equilibrium mixture, the partial pressures are $p_{{\text{SO}}_3}=43\text{ kPa}$; $p_{{\text{O}}_2}=530\text{ Pa}$ and $p_{{\text{SO}}_2}=45\text{ kPa}$. The equilibrium constant $K_p=\cdots \times 10^{-2}$. (Nearest integer)

Imagine a closed container where sulfur dioxide and oxygen are reacting to form sulfur trioxide. They have reached a state of dynamic equilibrium. This means the rate at which sulfur trioxide is being formed is exactly equal to the rate at which it is decomposing back into sulfur dioxide and oxygen.

In this state of perfect balance, the partial pressures of the gases remain constant. We are given these partial pressures: $p_{{\text{SO}}_3}=43\text{ kPa}$, $p_{{\text{SO}}_2}=45\text{ kPa}$, and $p_{{\text{O}}_2}=530\text{ Pa}$. Our goal is to find the equilibrium constant, $K_p$.

Before we jump into any formulas, look closely at the units. This is where mistakes happen! The pressures for sulfur dioxide and sulfur trioxide are in kilopascals (kPa), but oxygen is in pascals (Pa).

If we substitute these values directly into our equation, our result will be completely wrong. We must convert the pressure of oxygen to kilopascals to keep everything consistent. Since $1\text{ kPa}=1000\text{ Pa}$, we divide $530$ by $1000$ to get $0.53\text{ kPa}$. Now, all our pressures are in the same unit, and we are ready to proceed.

Now, let's write down the expression for the equilibrium constant in terms of partial pressures, $K_p$. According to the Law of Mass Action, $K_p$ is the ratio of the product of partial pressures of the products to the reactants, where each pressure is raised to the power of its stoichiometric coefficient from the balanced chemical equation.

For the reaction $2{\text{SO}}_2(g)+{\text{O}}_2(g)\rightleftharpoons 2{\text{SO}}_3(g)$, the expression is:

Notice how the pressures of ${\text{SO}}_3$ and ${\text{SO}}_2$ are squared because of the coefficient $2$ in the balanced equation.

Let's substitute the values we have into our $K_p$ equation. We put $43$ for sulfur trioxide, $45$ for sulfur dioxide, and $0.53$ for oxygen.

Time for some arithmetic. The square of $43$ is $1849$, and the square of $45$ is $2025$.

Next, we multiply $2025$ by $0.53$ in the denominator, which gives us $1073.25$.

Now, we divide $1849$ by $1073.25$. This yields approximately $1.7228$.

The question asks for the answer in the format of an integer multiplied by $10^{-2}$. So, we rewrite $1.7228$ as $172.28\times 10^{-2}$.

Rounding to the nearest integer, we get $172$.

We have our answer, but let's think a bit deeper. What if we suddenly halved the volume of the container? According to Boyle's Law, all partial pressures would double.

If we calculate the reaction quotient $Q_p$ at this new instant, the squared terms in the numerator and denominator would each gain a factor of $4$, which cancel out. However, the single pressure term for oxygen in the denominator would gain a factor of $2$. This means $Q_p$ would become half of $K_p$.

Since $Q_p<K_p$, the equilibrium would shift forward to produce more sulfur trioxide, perfectly aligning with Le Chatelier's principle which states that an increase in pressure favors the side with fewer moles of gas!