CO Combustion: Write Thermochemical Equation & Heat

Hey there, fellow chemistry enthusiasts! Ever wondered how we quantify the heat involved in chemical reactions? Well, today, we're diving deep into the fascinating world of thermochemistry by tackling a super common and important reaction: the combustion of carbon monoxide. You see, understanding how much energy is released or absorbed during a chemical change isn't just academic; it's crucial for everything from designing efficient engines to understanding metabolic processes in our bodies. So, buckle up, because we're going to break down how to construct a thermochemical equation step-by-step, specifically for a scenario where we know exactly how much heat is given off when a certain amount of carbon monoxide burns. This isn't just about memorizing formulas, guys; it's about grasping the core principles that govern energy transformations, which are literally everywhere around us. We're talking about a practical problem here: we're told that when 0.6 moles of carbon monoxide undergoes combustion, a hefty 173.1 kJ of heat is released. Our mission, should we choose to accept it (and we definitely should!), is to use this information to write the full, glorious thermochemical equation for the reaction. This equation won't just tell us what reactants turn into what products; it'll also explicitly state the enthalpy change (that's the fancy term for the heat change at constant pressure) associated with the reaction, usually per mole of reactant or product based on the balanced equation. It’s a bit like getting the full story, not just the plot points. We'll start by making sure we nail the basic chemical equation, then figure out the enthalpy change per mole, and finally, stitch it all together into a perfect thermochemical equation. Trust me, by the end of this, you’ll be a pro at handling similar problems and seeing the real-world value in these seemingly abstract chemical calculations. So, let’s get started on unlocking the energy secrets of carbon monoxide combustion!

Understanding Carbon Monoxide Combustion: The Basics

Alright, let's kick things off by getting cozy with carbon monoxide itself and its combustion process. What exactly is carbon monoxide (CO), you ask? It's a colorless, odorless, and extremely toxic gas. Think of it as the less-than-friendly cousin of carbon dioxide (CO₂). While CO₂ is a key component of our atmosphere and essential for photosynthesis, CO is typically formed during incomplete combustion of carbon-containing fuels. This means if there isn't enough oxygen around for the fuel to burn completely, you end up with CO instead of CO₂. And that, my friends, is a big problem for safety, which is why detectors for CO are so crucial in homes. But when it does burn completely, in the presence of sufficient oxygen, it transforms into carbon dioxide, releasing a significant amount of heat. This combustion reaction is actually a great source of energy, and understanding its thermochemistry is vital in many industrial applications and even in predicting the behavior of fires. The basic chemical transformation we're looking at is carbon monoxide reacting with oxygen to produce carbon dioxide. First, let's write the unbalanced equation: CO(g) + O₂(g) → CO₂(g). Simple enough, right? Now, for the crucial step: balancing the equation. We need to make sure we have the same number of atoms of each element on both sides of the arrow. On the left, we have 1 carbon, 1 oxygen (from CO), and 2 oxygen (from O₂), totaling 3 oxygens. On the right, we have 1 carbon and 2 oxygens. To balance the oxygens, we can put a coefficient of ½ in front of O₂ on the reactant side, making it CO(g) + ½O₂(g) → CO₂(g). This is perfectly valid in thermochemistry, as we often refer to moles rather than individual molecules, and fractions are okay when referring to reactant ratios. However, sometimes for general chemical equations, we prefer integer coefficients. If we want integer coefficients, we'd multiply the entire equation by 2: 2CO(g) + O₂(g) → 2CO₂(g). For calculating enthalpy change per mole of CO, we'll typically use the first balanced equation (CO(g) + ½O₂(g) → CO₂(g)) because it directly relates to one mole of our primary reactant. The problem statement explicitly mentions heat is released. This is a huge clue! When heat is released by a system into its surroundings, we call it an exothermic reaction. In thermochemical terms, an exothermic reaction has a negative enthalpy change (ΔH). Think of it like this: the system is losing energy, so its energy state goes down. So, whatever value we calculate for ΔH, we know it's going to be negative. This understanding of basic balancing and the exothermic nature of combustion is the bedrock upon which we'll build our full thermochemical equation.

Calculating the Enthalpy Change (ΔH) for Molar Combustion

Okay, guys, now comes the exciting part: crunching the numbers to figure out the exact enthalpy change, or ΔH, for our carbon monoxide combustion reaction. Remember, our goal is to write a thermochemical equation that typically shows the ΔH for the reaction as written with its stoichiometric coefficients. Usually, we want this value normalized, often to per mole of a specific reactant or product. The problem gives us a very specific piece of information: when 0.6 moles of carbon monoxide burn, 173.1 kJ of heat is released. This is a direct measurement for a specific amount, but what we need is the molar enthalpy change—that is, the ΔH associated with the combustion of one mole of CO. This is a simple but critical conversion. If 0.6 moles of CO release 173.1 kJ, then to find out how much heat 1 mole of CO releases, we just need to set up a ratio. It's like finding the price per apple if you know the total cost of a bag of apples! So, the heat released per mole of CO can be calculated as follows: (Heat Released) / (Moles of CO). Let's plug in our values: 173.1 kJ / 0.6 mol CO. If you do that calculation, you'll find that 173.1 / 0.6 = 288.5 kJ/mol. This value, 288.5 kJ, is the amount of heat released when one mole of carbon monoxide undergoes complete combustion. Now, for the critical step in thermochemistry: applying the correct sign convention. As we discussed earlier, heat released signifies an exothermic reaction. And for exothermic reactions, the enthalpy change (ΔH) is always, always negative. So, the molar enthalpy change for the combustion of carbon monoxide is ΔH = -288.5 kJ/mol. This negative sign is super important because it tells us that energy is flowing out of the system and into the surroundings. If the problem had said