Chapter 10, Problem 10.91PAE

The reaction shown below is involved in the refining of iron. (The table that follows provides all of the thermodynamic data you should need for this problem.)

   ${\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+\text{3C}\left(\text{s},\text{graphite}\right)\to \text{4Fe}\left(\text{s}\right)+{\text{3CO}}_{\text{2}}\left(\text{g}\right)$

(a) Find $\Delta \text{H}°$ for the reaction.

(b) $\Delta \text{S}°$ for the reaction above is 557.98 J/K. Find S° for Fe₂O₃(s).

(c) Calculate $\Delta \text{G}°$ for the reaction at the standard temperature of 298 K. (There are two ways that you could do this.)

(d) At what temperatures would this reaction be spontaneous?

Compound	$\Delta {\text{H}}_f°$ (kJ mol-1)	S° (kJ mol-1)	$\Delta {\text{G}}_f°$ (J mol-1 K-1)
Fe2O3(s)	-824.2	?	-742.2
C(s, graphite)	0	5.740	0
Fe(s)	0	27.3	0
CO2(g)	-393.5	213.6	-394.4.83

Interpretation Introduction

Interpretation: ΔS^o for the given reaction is 557.98 J/K. Find ΔS^o for Fe₂O₃(s).

Concept Introduction: $\Delta S^{\circ }{}_{reaction}={\displaystyle \sum m{S}_{products}}–{\displaystyle \sum n{S}_{reactants}}$

where m and n stands for the moles of products and reactants respectively

Answer to Problem 10.91PAE

Solution: ΔS^o for Fe₂O₃(s) = 87.4 JK-1

Explanation of Solution

${\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{ + 3C}\text{(s)}\to \text{4Fe}\left(\text{s}\right){\text{ + 3CO}}_{\text{2}}\left(\text{g}\right)$

ΔS^o for reaction =557.98 JK-1, ΔS^o for C(s) = 5.740 JK-1, ΔS^o for Fe(s) = 27.3 JK-1, ΔS^o for CO₂(g) = 213.6 JK-1

$\Delta S^{\circ }{}_{reaction}={\displaystyle \sum m{S}_{products}}–{\displaystyle \sum n{S}_{reactants}}$

$\begin{array}{l}{\text{ΔS}}^{\circ }\text{reaction = }\left({\text{4 x ΔS}}^{\circ }\text{for Fe}\left(\text{s}\right){\text{ + 3 x ΔS}}^{\circ }{\text{for CO}}_{\text{2}}\left(\text{g}\right)\right)\hfill \end{array}$

$\begin{array}{l}\text{ – }\left({\text{2 x ΔS}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right){\text{ + 3 x ΔS}}^{\circ }\text{for C}\left(\text{s}\right)\right)\hfill \end{array}$

$\begin{array}{l}\text{557}{\text{.98 JK}}^{\text{-1}}\text{= }\left(\text{4 x 27}{\text{.3 JK}}^{\text{-1}}\text{+}\text{3 x 213}{\text{.6 JK}}^{\text{-1}}\right)\hfill \end{array}$

$\begin{array}{l}\text{ – }\left({\text{2 x ΔS}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{ + 3 x 5}{\text{.740 JK}}^{\text{-1}}\right)\hfill \end{array}$

$\begin{array}{l}\text{557}{\text{.98 JK}}^{\text{-1}}{\text{= 750 JK}}^{\text{-1}}\text{- }\left({\text{2 x ΔS}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{ + 17}{\text{.22 JK}}^{\text{-1}}\right)\hfill \end{array}$

$\begin{array}{l}{\text{2 x ΔS}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right){\text{ = 750 JK}}^{\text{-1}}\text{-}\text{557}{\text{.98 JK}}^{\text{-1}}\text{-}\text{17}{\text{.22 JK}}^{\text{-1}}\hfill \\ {\text{2 x ΔS}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{ = 174}{\text{.8 JK}}^{\text{-1}}\hfill \\ {\text{ΔS}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{ = 87}{\text{.4JK}}^{\text{-1}}\hfill \end{array}$

Interpretation Introduction

Interpretation: ΔG^o for the given reaction at the standard temperature of 298K

Concept Introduction: $\Delta G_f^{\circ }{}_{reaction}={\displaystyle \sum m{G}_f^{\circ }{}_{products}}–{\displaystyle \sum n{G}_f^{\circ }{}_{reactants}}$

where m and n stands for the moles of products and reactants respectively

Answer to Problem 10.91PAE

Solution: [G] for the given reaction is 301.2 kJ mol-1

Explanation of Solution

ΔG_f^o for Fe₂O₃ = -742.2 kJ mol-1, ΔG_f° for C(s)= 0, ΔG_f° for Fe(s) = 0, ΔG_f° for CO₂(g) = -394.4 kJ mol-1

$\Delta G_f^{\circ }{}_{reaction}={\displaystyle \sum m{G}_f^{\circ }{}_{products}}–{\displaystyle \sum n{G}_f^{\circ }{}_{reactants}}$

$\begin{array}{l}{\text{ΔG}}^{\circ }\text{reaction = }\left({\text{4 x ΔG}}^{\circ }\text{for Fe}\left(\text{s}\right){\text{ + 3 x ΔG}}^{\circ }{\text{for CO}}_{\text{2}}\left(\text{g}\right)\right)\hfill \end{array}$

$\begin{array}{l}\text{ – }\left({\text{2 x ΔG}}^{\circ }{\text{for Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right){\text{ + 3 x ΔG}}^{\circ }\text{for C}\left(\text{s}\right)\right)\hfill \end{array}{\text{ΔG}}^{\circ }\text{reaction}\text{= }\hfill$

$\begin{array}{l}\left({\text{4 x 0 kJ mol}}^{\text{-1}}\text{+}\text{3 x }-394.4{\text{ kJ mol}}^{\text{-1}}\right)\hfill \end{array}$

$\begin{array}{l}\text{ – }\left(\text{2 x }-742.2{\text{ kJ mol}}^{\text{-1}}{\text{+ 3 x 0 kJ mol}}^{\text{-1}}\right)\hfill \end{array}\begin{array}{l}\text{=}\end{array}\hfill$

$\begin{array}{l}-1183.2{\text{kJ mol}}^{\text{-1}}+1484.4{\text{kJ mol}}^{\text{-1}}\hfill \\ \text{= 301}{\text{.2 kJ mol}}^{\text{-1}}\end{array}\hfill \\ \hfill \\ \hfill$

Interpretation Introduction

Interpretation: the temperature at which this reaction be spontaneous

Concept Introduction: For the reaction to be spontaneous ΔG must be negative. Using the Gibbs free energy equation, ΔG = ΔH − TΔS, we can find the temperature at which this reaction be spontaneous.

$\begin{array}{l}\begin{array}{l}\Delta G=\Delta H–T\Delta S\\ 0=\Delta H–T\Delta S\end{array}\hfill \\ T=\frac{\Delta H}{\Delta S}\hfill \end{array}$

Answer to Problem 10.91PAE

Solution: The temperature at which this reaction will be spontaneous is 5353.5K

Explanation of Solution

$T=\frac{\Delta H}{\Delta S}=\frac{467.9kJmo{l}^{-1}}{87.4J{K}^{-1}}=\frac{467.9\text{x}1000Jmo{l}^{-1}}{87.4J{K}^{-1}}=5353.5K$