Understanding Pressure Effects in Chemical Equilibrium

In the reaction heat + C + CO2 ->equil-> 2CO, if pressure is increased while temperature remains constant, what happens to the concentration of CO?

• A. CO will increase
• B. CO will remain unchanged
• C. CO will decrease
• D. The reaction will stop entirely

Answer: (C)

In the given reaction, when considering the equilibrium: \[ \text{heat} + C + \text{CO}_2 \rightleftharpoons 2\text{CO} \] the equilibrium involves one mole of CO2 on the reactant side and two moles of CO on the product side. This indicates that for every mole of CO2 that reacts, two moles of CO are produced. According to Le Chatelier's principle, if the pressure of a system at equilibrium is increased, the system will respond by shifting the equilibrium in the direction that produces fewer moles of gas to counteract the change. In this case, since there are two moles of gas on the product side (2 CO) and one mole of gas on the reactant side (1 CO2), an increase in pressure will favor the formation of CO2, the side with fewer gas molecules. As a result, this shift will lead to a decrease in the concentration of CO, as the system moves towards the left to establish a new equilibrium under the increased pressure condition.

When studying for the American Chemical Society (ACS) Chemistry Exam, grasping the effects of pressure on chemical equilibrium is crucial. Whether you're feeling a little nervous or confidently prepared, understanding how pressure influences reactions can give you a solid edge. Don’t you just love it when chemistry not only makes sense but becomes practical for exams?

Take a classic example to illustrate the principle. Imagine a reaction like this:

[

\text{heat} + C + \text{CO}_2 \rightleftharpoons 2\text{CO}

]

In this scenario, we see that there’s one mole of carbon dioxide ((CO_2)) and two moles of carbon monoxide ((CO)). Now, if we crank up the pressure while keeping the temperature the same, what do you think happens to the concentration of (CO)? A lot of students might instinctively choose option A, believing that increasing the pressure would favor the products.

But wait—when we apply Le Chatelier's principle, things take a different turn. This principle tells us that a system at equilibrium will strive to counteract any changes imposed on it, such as pressure increases. So, if we jack up the pressure, the reaction will shift toward the side with fewer moles of gas to help alleviate that pressure.

Let's break it down. On the product side, we have two moles of CO, but on the reactant side, there’s just one mole of (CO_2). So, what happens under increased pressure? The equilibrium shifts to favor (CO_2), which is our reactant side containing fewer gas moles. This shift means that, unfortunately, the concentration of (CO) will decrease as the system adjusts to establish a new equilibrium under the increased pressure condition.

Isn't it fascinating? And this principle isn't just a dry fact—it's a powerful concept that applies not only in the exam room but in real-life chemistry labs too. It reminds us that chemistry is all around us and constantly in motion, even when it feels like it's set in stone.

Understanding these nuances can be the difference between simply passing the ACS exam and excelling at it. Picture yourself answering confidently, knowing how these fundamental principles work together in a real-world setting. By grasping these concepts, you're not just preparing for a test; you're equipping yourself with insights that will serve you well in your chemistry journey—both in academia and beyond!

So, the next time you take a moment to study for your exam, let these principles about pressure, concentration, and equilibrium settle in. They’re not just chemical equations; they’re the building blocks of understanding how the universe operates on a molecular level.