Problem 84 - Chemistry The Science in Context

Question

The equilibrium constant \(K_{\mathrm{p}}\) of the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ is 7.69 at \(830^{\circ} \mathrm{C} .\) If a vessel at this temperature initially contains pure \(\mathrm{SO}_{3}\) and if the partial pressure of \(\mathrm{SO}_{3}\) at equilibrium is 0.100 atm, what is the partial pressure of \(\mathrm{O}_{2}\) in the flask at equilibrium?

Step-by-Step Solution

Answer

Answer: The partial pressure of O2 at equilibrium is approximately 0.0247 atm.

Step 1: Write the expression for Kp

To find the partial pressure of O2 at equilibrium, first, we need to write the expression for the equilibrium constant Kp. Kp can be written as follows: $$ K_p = \frac{[SO_2]^2 \cdot [O_2]}{[SO_3]^2} $$ where the brackets represent the partial pressures of each species at equilibrium.

Step 2: Write the expression for the change in moles of each species

The reaction shows that for every 2 moles of SO3 that react, we produce 2 moles of SO2 and 1 mole of O2. Therefore, we can write the expressions for the changes of the species as follows: Let x be the amount of SO3 that reacts to reach equilibrium. - Initial pressure of SO3: P_0 - Initial pressure of SO2 and O2: 0 - Change in SO3: -2x - Change in SO2: +2x - Change in O2: +x

Step 3: Substitute the change expressions into the Kp expression

We are given that the partial pressure of SO3 at equilibrium is 0.100 atm, so (P_0 - 2x) = 0.100 atm. Now we can substitute the expressions for the changes in the species into the Kp expression: $$ K_p = \frac{([2x]^2) \cdot ([x])}{[0.100]^2} $$

Step 4: Substitute the given value of Kp and solve

We are given that the Kp value at 830°C is 7.69. Plug this value into our equation and solve for x: $$ 7.69 = \frac{(2x)^2 \cdot (x)}{0.0100^2} $$ Rearrange the equation and solve for x: $$ x^3 = \frac{7.69 \cdot 0.0100^2}{4} $$ $$ x \approx 0.0247\,\mathrm{atm} $$

Step 5: Calculate the partial pressure of O2 at equilibrium

Now that we have the value of x, we can determine the partial pressure of O2 at equilibrium. From step 2, we know that the change in O2 is equal to x. Therefore, the partial pressure of O2 at equilibrium is: $$ P_{O_2} = x \approx 0.0247\, \mathrm{atm} $$

Key Concepts

Partial Pressure

Partial pressure is a vital concept in the study of gas-phase reactions. It refers to the pressure exerted by a single gas in a mixture of gases. Imagine a group of different gases living together in a container; each gas behaves as if it has its own space to push and shove. That pushing action is what we call partial pressure. It's important because gases react based on their partial pressures, not just their concentrations.

For example, in the given reaction where sulfur trioxide (SO_{3}) breaks down, the partial pressure of each gas plays a key role in determining the position of equilibrium. If we know the partial pressure of SO_{3}, we can use the equilibrium expression to find out the pressure of other gases at equilibrium, like oxygen (O_{2}).

Equilibrium

At equilibrium, a chemical reaction is in a state where the rate of the forward reaction equals the rate of the reverse reaction, meaning there's no net change in the amounts of products and reactants. Think of it like a tug-of-war where both teams are equally strong; the rope doesn’t move to either side. For the SO_{3} decomposition reaction, equilibrium is expressed by the equilibrium constant (K_{p}), which relates the partial pressures of the products and reactants.

Crucially, K_{p} remains constant at a given temperature, reflecting the precise balance point for that reaction under those conditions. If the temperature stays the same, regardless of how you start the reaction, whether with pure SO_{3} or a mixture, you'll always reach the same equilibrium state defined by K_{p}.

Le Chatelier's Principle

When you mess with a system at equilibrium, it pushes back—that's Le Chatelier's principle. It's like trying to force a patient Zen master out of balance; they'll just calmly step back to the center. This principle suggests that if you change the conditions of a reaction at equilibrium (like the pressure, temperature, or concentration), the reaction shifts to partially counteract that change.

For example, if we increase the pressure of the gas vessel, the system may respond by favoring the direction that produces fewer gas molecules to reduce pressure. Similarly, an increase in temperature for our SO_{3} reaction might shift the equilibrium to absorb the extra heat. Le Chatelier’s principle helps chemists predict how a reaction will respond to different stresses, maintaining equilibrium in the very dynamic world of chemical reactions.

Chemical Reaction

Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. In our example, SO_{3} is the reactant that decomposes into SO_{2} and O_{2}, the products. Every chemical reaction comes with a unique recipe that includes specific amounts of reactants combining in certain conditions.

Moreover, the path a reaction takes can be represented through a chemical equation, translating the language of atoms and molecules into a format that we can understand. Like in our equation, where the coefficient '2' in front of SO_{3} and SO_{2} tells us they’re involved in a 1:1 ratio; for every one part of SO_{3} that reacts, it creates one part of SO_{2}. The oxygen, with no coefficient in front, is like the surprise guest who shows up in half the quantity.