Question

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

Short answer

The standard entropy change for the formation of \(\mathrm{Fe}(\mathrm{CO})_{5}(g)\) from \(\mathrm{Fe}(s)\) and \(5 \mathrm{CO}(g)\) is \(\Delta S^{\circ}=-570 \mathrm{J} /\mathrm{K}\).

Step-by-step solution

: We are given two reactions: 1. \(\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow\mathrm{Fe}(\mathrm{CO})_{5}(g)\), where \(\Delta S^{\circ}=?\) 2. \(\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow\mathrm{Fe}(\mathrm{CO})_{5}(g)\), where \(\Delta S^{\circ}=107 \mathrm{J} /\mathrm{K}\) 3. \(\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow\mathrm{Fe}(\mathrm{CO})_{5}(l)\), where \(\Delta S^{\circ}=-677 \mathrm{J} /\mathrm{K}\) #Step 2: Combine the reactions#

: We can obtain the desired reaction by adding reaction 2 and reaction 3: \((\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow\mathrm{Fe}(\mathrm{CO})_{5}(g)) + (\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow\mathrm{Fe}(\mathrm{CO})_{5}(l))\) When we add these reactions, the \(\mathrm{Fe}(\mathrm{CO})_{5}(l)\) terms on the right side of reaction 3 and left side of reaction 2 cancel, resulting in: \(\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow\mathrm{Fe}(\mathrm{CO})_{5}(g)\) #Step 3: Combine the corresponding entropy changes#

: Now we need to combine the entropy changes for these reactions to get the entropy change of the desired reaction. Since entropies are state functions, we can simply add the entropy changes of reactions 2 and 3: \(\Delta S^{\circ}_{total} = \Delta S^{\circ}_{2} + \Delta S^{\circ}_{3} = 107 \mathrm{J} /\mathrm{K} - 677 \mathrm{J} /\mathrm{K}\) #Step 4: Calculate the desired entropy change#

: Finally, we can calculate the entropy change for the desired reaction: \(\Delta S^{\circ}_{total} = 107 \mathrm{J} /\mathrm{K} - 677 \mathrm{J} /\mathrm{K} = -570 \mathrm{J} /\mathrm{K}\) Thus, the standard entropy change for the formation of \(\mathrm{Fe}(\mathrm{CO})_{5}(g)\) from \(\mathrm{Fe}(s)\) and \(5 \mathrm{CO}(g)\) is \(\Delta S^{\circ}=-570 \mathrm{J} /\mathrm{K}\).

Key concepts

Understanding Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In chemical systems, thermodynamics helps us understand how energy and entropy interact within chemical reactions. Entropy, a key thermodynamic quantity, measures the level of disorder or randomness in a system.
Thermodynamics encompasses several laws that describe how energy works within systems. Important to our context, the Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time. It only remains constant if all processes are reversible. For real-world chemical reactions which are often irreversible, entropy tends to increase.
Thermodynamics also allows us to predict whether a given chemical process or reaction is energetically feasible. In the context of the exercise, we apply thermodynamics to calculate the standard entropy change, which reflects the energy dispersal among the reactants and products during a chemical reaction. This is crucial for understanding the feasibility and spontaneity of reactions.

The Role of State Functions

In thermodynamics, state functions are properties that depend only on the current state of a system, not on the path used to get there. Entropy is one of these state functions. This means the state function’s value is determined by the system's state, independent of how the system reached that state.
State functions play a vital role in thermodynamics because they allow us to calculate changes in properties regardless of the path taken by a reaction. In our scenario, entropy being a state function means we can add the entropy changes of consecutive reactions to find the total change, which is exactly what was done in the solution calculating the standard entropy change for the conversion Fe(s) + 5 CO(g) to Fe(CO)_5(g).
Understanding the concept of state functions simplifies the analysis of thermodynamic processes, making it possible to break down complex processes into simpler steps.

Entropy Calculations Explained

Entropy calculations involve determining the disorder change or "randomness" change within a chemical reaction. When calculating changes in entropy, note that entropy values are generally given in units such as Joules per Kelvin (J/K). The standard entropy change, denoted as \( \Delta S^{\circ} \), represents the change in entropy from reactants to products under standard conditions (1 atmosphere, 298 K usually).
In the given exercise, we used entropy data from specific reactions to determine the overall entropy change:

• First, note the entropy change from Fe(CO)_5(l) to Fe(CO)_5(g), which was given directly as \(107 \text{ J/K} \).
• Next, add the given entropy change for the reaction Fe(s) + 5 CO(g) to Fe(CO)_5(l), which is \(-677 \text{ J/K} \).

Add these changes, keeping the sign in mind, to find the overall \( \Delta S^{\circ} \) for the target reaction:
\( \Delta S^{\circ} = 107 \text{ J/K} - 677 \text{ J/K} = -570 \text{ J/K} \).
This result indicates a decrease in disorder from reactants to products in the gaseous form of iron pentacarbonyl.