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Chapter 17: Problem 81

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

Short Answer

Expert verified
The \(\Delta S^{\circ}\) for the target reaction \(\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g)\) is \(-784 \mathrm{~J}/\mathrm{K}\).

Step by step solution

01

Identify the given equations and their \(\Delta S^{\circ}\)

For this problem, we are given the following two reactions: 1. \(\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g)\) with \(\Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K}\) 2. \(\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l)\) with \(\Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K}\) The target reaction is: $$\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g)$$
02

Manipulate the given reactions to match the target reaction

To arrive at the target reaction, we need to manipulate the given reactions in a way that can be added together. Observe that reversing reaction 1 gives us the gaseous form of \(\mathrm{Fe}(\mathrm{CO})_{5}\), which is present in the target reaction. Let's reverse reaction 1 and change the sign of its \(\Delta S^{\circ}\): $$\mathrm{Fe}(\mathrm{CO})_{5}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l)$$ with \(\Delta S^{\circ}=-107 \mathrm{~J} / \mathrm{K}\) Now, we can add reaction 2 and the reversed reaction 1 to obtain the target reaction: $$\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l)$$ with \(\Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K}\) $$+\mathrm{Fe}(\mathrm{CO})_{5}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l)$$ with \(\Delta S^{\circ}=-107 \mathrm{~J} / \mathrm{K}\) $$\implies \mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g)$$
03

Find the \(\Delta S^{\circ}\) for the target reaction

Now that we have the target reaction, we can calculate its \(\Delta S^{\circ}\) by adding the \(\Delta S^{\circ}\) values of the manipulated given reactions: $$\Delta S^{\circ}_{\text{target}} = \Delta S^{\circ}_{2} + \Delta S^{\circ}_{\text{reversed 1}}$$ $$\Delta S^{\circ}_{\text{target}} = -677 \mathrm{~J} / \mathrm{K} + (-107 \mathrm{~J} / \mathrm{K})$$ $$\Delta S^{\circ}_{\text{target}} = -784 \mathrm{~J} / \mathrm{K}$$ Therefore, the \(\Delta S^{\circ}\) for the target reaction \(\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g)\) is \(-784 \mathrm{~J}/\mathrm{K}\).

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Most popular questions from this chapter

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ}\) ? a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Consider the following reaction: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}\), \(P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Consider the following reaction: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{298}=0.090 $$ For \(\mathrm{Cl}_{2} \mathrm{O}(g)\), $$ \begin{aligned} \Delta G_{\mathrm{f}}^{\circ} &=97.9 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ} &=80.3 \mathrm{~kJ} / \mathrm{mol} \\ S^{\circ} &=266.1 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \end{aligned} $$ a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=-R T \ln (K)\) b. Use bond energy values (Table 8.4) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and \(\mathrm{b}\) to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\). e. Estimate the value of \(K\) at \(500 . \mathrm{K}\). f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when \(P_{\mathrm{H}_{2} \mathrm{O}}=18\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=\) \(2.0\) torr, and \(P_{\mathrm{HOCl}}=0.10\) torr.

Hydrogen cyanide is produced industrially by the following exothermic reaction: Is the high temperature needed for thermodynamic or kinetic reasons?

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$ \begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2} $$
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