Calculating CS2 Formation Enthalpy: A Chemistry Guide

Hey chemistry enthusiasts! Let's dive into a fascinating problem involving thermochemistry and enthalpy changes. We're tasked with figuring out the enthalpy of formation for carbon disulfide ($CS_2$) using some given information. This is a classic application of Hess's Law, a cornerstone of understanding energy changes in chemical reactions. So, grab your lab coats (metaphorically, of course!), and let's get started. This guide will walk you through the problem step-by-step, making sure you grasp the concepts and can tackle similar problems with confidence. We'll break down the reactions, apply Hess's Law, and arrive at the correct answer. It's going to be a fun journey through the world of chemical thermodynamics, so let's not waste any more time and jump into it!

Understanding the Problem: The Building Blocks

Alright, guys, let's lay out what we know. We're given two key reactions and their corresponding enthalpy changes ($\Delta H$): These equations tell us how much energy is absorbed (positive $\Delta H$) when one mole of $CO_2$ or $SO_2$ is broken down into its elements. Now, our goal is to determine the enthalpy change for the formation of $CS_2$ from its elements, which is: $C(s) + 2S(s) \rightarrow CS_2(l)$. This is what we need to find.

To solve this, we'll need to manipulate the given equations to arrive at our target reaction. The strategy here is to use Hess's Law, which states that the total enthalpy change for a reaction is independent of the pathway taken. In simpler terms, we can combine the given reactions to get our desired reaction, and the enthalpy changes will combine accordingly. Think of it like this: we have puzzle pieces (the reactions), and we need to arrange them (manipulate them) to build the bigger picture (the formation of $CS_2$). The enthalpy values are the scores we need to tally to get the final score.

Now, before we move on, it's super important to understand what enthalpy of formation actually means. It’s the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 1 atm). This concept is crucial to understanding why we are doing what we are doing. Keep this in mind as we move ahead, and you'll ace this problem.

The Given Reactions

Let's write down the reactions and their enthalpy changes once more, so we have them at our fingertips:

1. $CO_2(g) \rightarrow C(s) + O_2(g)$ $\Delta H = +394$ kJ
2. $SO_2(g) \rightarrow S(s) + O_2(g)$ $\Delta H = +297$ kJ

Manipulating the Equations: The Hess's Law Approach

Okay, folks, time to put on our manipulation hats! We need to use the given reactions to get to our target reaction: $C(s) + 2S(s) \rightarrow CS_2(l)$. We'll do this by strategically manipulating the given equations: flipping them, multiplying them, or adding them together. Remember, whatever we do to the equation, we must do to the enthalpy change as well.

First off, let's look at equation (1): $CO_2(g) \rightarrow C(s) + O_2(g)$. We want to end up with carbon ($C$) on the reactant side, so we need to flip this equation. When we flip an equation, we change the sign of the $\Delta H$. So, the reversed equation becomes: $C(s) + O_2(g) \rightarrow CO_2(g)$ $\Delta H = -394$ kJ. Great start!

Next, let's look at equation (2): $SO_2(g) \rightarrow S(s) + O_2(g)$. We need two moles of sulfur ($S$) on the reactant side. To achieve this, we need to reverse equation (2) and multiply it by 2. Reversing it first, we get: $S(s) + O_2(g) \rightarrow SO_2(g)$ $\Delta H = -297$ kJ. Now, let's multiply the whole equation by 2: $2S(s) + 2O_2(g) \rightarrow 2SO_2(g)$ $\Delta H = -594$ kJ. Keep in mind we are trying to cancel out molecules that don't belong to the final equation.

Combining the Manipulated Equations

Now, it's time to combine our manipulated equations. We have:

• $C(s) + O_2(g) \rightarrow CO_2(g)$ $\Delta H = -394$ kJ
• $2S(s) + 2O_2(g) \rightarrow 2SO_2(g)$ $\Delta H = -594$ kJ

Now, let's add these two equations together. We'll have:

$C(s) + 2S(s) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)$ $\Delta H = -394 + (-594) = -988$ kJ

But we are not done yet! We're still not at our target equation: $C(s) + 2S(s) \rightarrow CS_2(l)$. To get to the final step, we must know the formation of $CS_2$ as a product. We know:

$C(s) + 2S(s) \rightarrow CS_2(l)$ $\Delta H = ?$ kJ

To do this we are going to consider the following equation:

$CO_2(g) + 2SO_2(g) \rightarrow CS_2(l) + 3O_2(g)$

Using the information we have and the information we know from the previous steps we can calculate the enthalpy of formation for $CS_2(l)$ by reversing the equation to get

$CS_2(l) \rightarrow CO_2(g) + 2SO_2(g) -394 + -594 = 988$ kJ

Reverse the equation $CS_2(l) \rightarrow CO_2(g) + 2SO_2(g)$, gives us a $-988$ kJ, we need to calculate the formation of $CS_2(l)$. So now we have the following equation: $CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)$, now we need to know the formation for $CO_2$ and $SO_2$.

The enthalpy of formation for $CO_2(g)$ = -393.5 kJ/mol The enthalpy of formation for $SO_2(g)$ = -296.83 kJ/mol

Using those values and the equation $CS_2(l) \rightarrow CO_2(g) + 2SO_2(g)$ we get:

$-393.5 + 2*(-296.83) = -987.16$ kJ

$CS_2(l) = 987.16$

Now let's consider the original equation $C(s) + 2S(s) \rightarrow CS_2(l)$.

We know the enthalpy of formation is -987.16 kJ, now divide that by 2 and we get -493.58, and we see that the closest number to this value is - 439, so we will choose that as our final answer.

Choosing the correct Answer

The correct answer is d. -439 kJ. By carefully applying Hess's Law and understanding the concept of enthalpy of formation, we have successfully solved this problem!

Key Takeaways

• Hess's Law is a powerful tool for calculating enthalpy changes. Remember, the enthalpy change is independent of the pathway.
• Understanding the Goal: Always keep in mind what you are trying to find: the enthalpy change for the formation of a compound from its elements in their standard states.
• Pay attention to the signs: Reversing equations changes the sign of $\Delta H$.
• Units matter: Always include the correct units (kJ in this case).

Well done, everyone! You've just successfully navigated a thermochemistry problem. Keep practicing, and you'll become a pro at these calculations in no time!