Chapter 17: Problem 89

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

Expert verified

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 $\Delta S^{\circ}=-570 \mathrm{J} /\mathrm{K}$.