Using Appendix 4 and the following data, determine S^∘ for Fe(CO)5(g) Fe(s)+5 CO(g) ⟶Fe(CO)5

Name: Problem 83, Chapter 17: Chemistry
Uploaded: 2020-12-08T20:18:05-08:00
Duration: 2 min 23 s
Channel: Lottie Adams
Description: Using Appendix 4 and the following data, determine S^∘ for Fe(CO)5(g) Fe(s)+5 CO(g) ⟶Fe(CO)5(g) ΔS^∘=? Fe(CO)5(l) ⟶Fe(CO)5(g) ΔS^∘=107 J / K Fe(s)+5 CO(g) ⟶Fe(CO)5(l) ΔS^∘=-677 J / K

step 1 So what we have here is we have two reactions that we need to be considering in this thermodynam

This problem has been solved

100% free to try

Question

Using Appendix 4 and the following data, determine $S^{\circ}$ for $\mathrm{Fe}(\mathrm{CO})_{5}(g)$ $$\begin{aligned}\mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\ \mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\ \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K} \end{aligned}$$

   Using Appendix 4 and the following data, determine $S^{\circ}$ for $\mathrm{Fe}(\mathrm{CO})_{5}(g)$
$$\begin{aligned}\mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\
\mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\
\mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K}
\end{aligned}$$

Chemistry

Steven S. Zumdahl,… 8th Edition

Chapter 17, Problem 83

Instant Answer

Solved by Expert Lottie Adams

12/08/2020

Step 1

We can see that the first reaction is the desired reaction, and the second and third reactions can be combined to form the first reaction. To do this, we can add the second and third reactions together: Fe(CO)5(l) → Fe(CO)5(g) : ΔS° = 107 J/K Fe(s) + 5 CO(g) → Show more…

Show all steps

Thanks for your feedback!

Using Appendix 4 and the following data, determine $S^{\circ}$ for $\mathrm{Fe}(\mathrm{CO})_{5}(g)$ $$\begin{aligned}\mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\ \mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\ \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K} \end{aligned}$$

Play audio

Feedback

Lottie Adams and 74 other subject Chemistry 102 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Phase Transition Entropy

Phase transition entropy refers to the change in entropy when a substance transitions from one state of matter to another, such as from liquid to gas. This concept is important because phase changes often result in significant increases in disorder, affecting the overall entropy change in reactions.

Hess's Law

Hess's Law states that the total entropy change for a reaction is the sum of the entropy changes of the individual steps that make up the overall reaction. It allows the calculation of unknown entropy changes by combining several reactions with known entropy changes.

Standard Entropy Change

This concept involves the measure of disorder or randomness in a system under standard conditions. It is used to express the change in entropy when a reaction occurs at standard temperature and pressure, and it is crucial for predicting the spontaneity of reactions.

Please give Ace some feedback