Two possible ways of producing iron from iron ore are (a) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\frac{3}{2} \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{CO}_{2}(g)\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\) Which of these reactions proceeds spontaneously at the lower temperature?

Short Answer

Expert verified
To determine which reaction proceeds spontaneously at a lower temperature, follow these steps: 1. Calculate the enthalpy change (\(\Delta H\)) for both reactions using the enthalpy of formation values found in a standard enthalpy table. 2. Calculate the entropy change (\(\Delta S\)) for both reactions using the entropy values found in a standard entropy table. 3. Calculate the Gibbs free energy change (\(\Delta G\)) for both reactions using the formula \(\Delta G = \Delta H - T\Delta S\) and a certain temperature in Kelvin. 4. Compare the \(\Delta G\) values between reactions. The reaction with the lower \(\Delta G\) value at the desired temperature will proceed spontaneously at a lower temperature.

Step by step solution

01

Calculate the enthalpy change (\(\Delta H\)) for both reactions

To calculate the enthalpy change, we need to know the enthalpy of formation for each compound involved in the reactions. Using a standard enthalpy table, we can find these values and use them to calculate \(\Delta H\) for each reaction. (a) \(\Delta H_{a} = [2 \times \Delta H_f(\mathrm{Fe}_{(s)}) + \frac{3}{2} \times \Delta H_f(\mathrm{CO}_{2_{(g)}})] - [\Delta H_f(\mathrm{Fe}_2\mathrm{O}_{3_{(s)}}) + \frac{3}{2} \times \Delta H_f(\mathrm{C}_{(s)})]\) (b) \(\Delta H_{b} = [2 \times \Delta H_f(\mathrm{Fe}_{(s)}) + 3 \times \Delta H_f(\mathrm{H}_{2}\mathrm{O}_{(g)})] - [\Delta H_f(\mathrm{Fe}_2\mathrm{O}_{3_{(s)}}) + 3 \times \Delta H_f(\mathrm{H}_{2_{(g)})]\)
02

Calculate the entropy change (\(\Delta S\)) for both reactions

Similarly, for calculating the entropy change, we need the entropy values for each compound involved in the reactions. Using a standard entropy table, we can find these values and use them to calculate \(\Delta S\) for each reaction. (a) \(\Delta S_{a} = [2 \times S_f(\mathrm{Fe}_{(s)}) + \frac{3}{2} \times S_f(\mathrm{CO}_{2_{(g)}})] - [S_f(\mathrm{Fe}_2\mathrm{O}_{3_{(s)}) + \frac{3}{2} \times S_f(\mathrm{C}_{(s)})]\) (b) \(\Delta S_{b} = [2 \times S_f(\mathrm{Fe}_{(s)}) + 3 \times S_f(\mathrm{H}_{2}\mathrm{O}_{(g)})] - [S_f(\mathrm{Fe}_2\mathrm{O}_{3_{(s)}) + 3 \times S_f(\mathrm{H}_{2_{(g)})]\)
03

Calculate the Gibbs free energy change (\(\Delta G\)) for both reactions

Using the calculated enthalpy and entropy changes, we can calculate the Gibbs free energy change for both reactions at a certain temperature. Be sure to convert the temperature into Kelvin. (a) \(\Delta G_{a} = \Delta H_{a} - T\Delta S_{a}\) (b) \(\Delta G_{b} = \Delta H_{b} - T\Delta S_{b}\)
04

Compare the Gibbs free energy change (\(\Delta G\)) between reactions

Finally, we can compare the Gibbs free energy change between reactions. The reaction with the lower \(\Delta G\) value at the desired temperature will proceed spontaneously at a lower temperature. If \(\Delta G_{a} < \Delta G_{b}\), then reaction (a) proceeds spontaneously at a lower temperature. If \(\Delta G_{a} > \Delta G_{b}\), then reaction (b) proceeds spontaneously at a lower temperature. After determining which reaction has a lower \(\Delta G\) value at the desired temperature, we can conclude which reaction proceeds spontaneously at a lower temperature.

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)

Phosgene, \(\mathrm{COCl}_{2}\), can be formed by the reaction of chloroform, \(\mathrm{CHCl}_{3}(l)\), with oxygen: $$ \begin{gathered} 2 \mathrm{CHCl}_{3}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{COCl}_{2}(g)+2 \mathrm{HCl}(g) \\ \Delta H^{\circ}=-353,2 \mathrm{~kJ} ; \Delta G^{\circ}=-452.4 \mathrm{~kJ} \text { at } 25^{\circ} \mathrm{C} \end{gathered} $$ (a) Calculate \(\Delta S^{\circ}\) for the reaction. Is the sign reasonable? (b) Calculate \(S^{\circ}\) for phosgene. (c) Calculate \(\Delta H_{f}^{\circ}\) for phosgene.

In each of the following pairs, choose the substance with a lower entropy. (a) \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(10^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}(l)\) at \(30^{\circ} \mathrm{C}\) (b) C (graphite), \(\mathrm{C}\) (diamond) (c) \(\mathrm{Cl}_{2}(l), \mathrm{Cl}_{2}(g)\), both at room temperature

For the reaction $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ \(K=2.2 \times 10^{11}\) at \(473 \mathrm{~K}\) and \(4.6 \times 10^{8}\) at \(533 \mathrm{~K}\), Calculate \(\Delta G^{\circ}\) at both temperatures.

Fill in the blanks: (a) \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) become equal at _____ \(K\). (b) \(\Delta G^{\circ}\) and \(\Delta G\) are equal when \(Q=\) _____. (c) \(S^{\circ}\) for steam is _____ than \(S^{\circ}\) for water.

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