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What must be the temperature if the following reaction has \(\Delta G^{\circ}=-45.5 \mathrm{kJ}, \Delta H^{\circ}=-24.8 \mathrm{kJ},\) and \(\Delta S^{\circ}=15.2 \mathrm{JK}^{-1} ?\) $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})$$

Short Answer

Expert verified
The temperature at which this reaction would occur is approximately 1365 K.

Step by step solution

01

Assign given variables

Assign the given variables to the correct symbols. Here, we have \(\Delta G^{\circ} = -45500 \, J\), \(\Delta H^{\circ} = -24800 \, J\), and \(\Delta S^{\circ} = 15.2 \, J/K\). Note that we've converted the energy units from kJ to J for consistency.
02

Apply Gibbs-Helmholtz equation

The Gibbs-Helmholtz equation is \( \Delta G = \Delta H - T \Delta S \). We can rearrange this to solve for temperature, getting \( T = (\Delta H - \Delta G) / \Delta S \).
03

Substitute values and calculate temperature

Substitute the given values into the rearranged Gibbs-Helmholtz equation to find the temperature. So, \( T = ((-24800 \, J) - (-45500 \, J)) / 15.2 \, J/K \), which simplifies down to about 1365 K.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs-Helmholtz Equation
The Gibbs-Helmholtz Equation is a fundamental formula that relates the change in Gibbs Free Energy of a system (\(\triangle G\)) with its enthalpy change (\(\Delta H\)) and entropy change (\(\Delta S\)) at constant temperature (\(T\)). This equation is expressed as \(\Delta G = \Delta H - T\Delta S\).
The concept is crucial in determining whether a process will occur spontaneously under constant temperature and pressure. A negative \(\Delta G\) indicates a spontaneous process, while a positive \(\Delta G\) suggests that the process is non-spontaneous and will require input of energy to proceed.
In the context of the textbook exercise, the Gibbs-Helmholtz Equation helps us find the temperature at which the chemical reaction is in equilibrium—meaning that \(\Delta G\) is zero. By rearranging the equation, we can solve for \(T\) as follows: \(T = (\Delta H - \Delta G) / \Delta S\). It's important to keep units consistent, hence in the step-by-step solution, all energy units were converted from kJ to J to ensure accurate temperature calculation. This equation is vital in thermochemistry and offers students a valuable tool for understanding energy changes during reactions.
Thermodynamic Temperature Calculation
Thermodynamic temperature calculation is a process used to quantify the absolute temperature of a system. The absolute temperature, measured in kelvin (K), is a key component in numerous thermodynamic equations, including the Gibbs-Helmholtz Equation discussed earlier.
To understand the temperature at which a reaction is at equilibrium, we utilize the rearranged Gibbs-Helmholtz Equation, as shown in the solution steps. Temperature directly influences the spontaneity of chemical processes; higher temperatures can increase the entropy (\(\Delta S\)), potentially making an otherwise non-spontaneous reaction spontaneous due to the \(T\Delta S\) term becoming larger than the enthalpy change (\(\Delta H\)).
In the given exercise, the calculation shows that when the variables from the reaction are plugged into the equation, the thermodynamic temperature of equilibrium is approximately 1365 K. This type of calculation is practical when predicting reaction behavior under different temperature conditions and is critical for scientists and engineers designing processes in various fields.
Chemical Reaction Equilibrium
Chemical reaction equilibrium refers to the state in which the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant over time. The position of the equilibrium can be influenced by a variety of factors, including temperature, which we've explored through thermodynamic temperature calculation.
In equilibrium considerations, Gibbs Free Energy is a valuable indicator. A \(\Delta G = 0\) corresponds to a system at equilibrium. This state of equilibrium can be understood through Le Chatelier's principle, which tells us that a system at equilibrium will adjust concentrations, pressure, or temperature to counteract any changes imposed upon it.
Therefore, knowing the temperature at which a chemical reaction reaches equilibrium, as determined by the Gibbs-Helmholtz Equation, gives us the ability to predict how the system will behave under different conditions and to control the reaction to favor the formation of either reactants or products. Such insights are essential in industrial chemistry where yield optimization is necessary for cost-effective production.

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Most popular questions from this chapter

\(\mathrm{H}_{2}(\mathrm{g})\) can be prepared by passing steam over hot iron: \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) (a) Write an expression for the thermodynamic equilibrium constant for this reaction. (b) Explain why the partial pressure of \(\mathrm{H}_{2}(\mathrm{g})\) is independent of the amounts of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present. (c) Can we conclude that the production of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) could be accomplished regardless of the proportions of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present? Explain.

Sodium carbonate, an important chemical used in the production of glass, is made from sodium hydrogen carbonate by the reaction \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Data for the temperature variation of \(K_{\mathrm{p}}\) for this reaction are \(K_{\mathrm{p}}=1.66 \times 10^{-5}\) at \(30^{\circ} \mathrm{C} ; 3.90 \times 10^{-4} \mathrm{at}\) \(50^{\circ} \mathrm{C} ; 6.27 \times 10^{-3}\) at \(70^{\circ} \mathrm{C} ;\) and \(2.31 \times 10^{-1}\) at \(100^{\circ} \mathrm{C}\) (a) Plot a graph similar to Figure \(19-12,\) and determine \(\Delta H^{\circ}\) for the reaction. (b) Calculate the temperature at which the total gas pressure above a mixture of \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is \(2.00 \mathrm{atm}\).

Titanium is obtained by the reduction of \(\mathrm{TiCl}_{4}(1)\) which in turn is produced from the mineral rutile \(\left(\mathrm{TiO}_{2}\right)\) (a) With data from Appendix D, determine \(\Delta G^{\circ}\) at 298 K for this reaction. $$\mathrm{TiO}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{TiCl}_{4}(1)+\mathrm{O}_{2}(\mathrm{g})$$ (b) Show that the conversion of \(\mathrm{TiO}_{2}(\mathrm{s})\) to \(\mathrm{TiCl}_{4}(1)\) with reactants and products in their standard states, is spontaneous at \(298 \mathrm{K}\) if the reaction in (a) is coupled with the reaction $$2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})$$

The following data are given for the two solid forms of \(\mathrm{HgI}_{2}\) at \(298 \mathrm{K}\). $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ} & \Delta G_{f,}^{\circ} & S^{\circ} \\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{HgI}_{2} \text { (red) } & -105.4 & -101.7 & 180 \\ \mathrm{Hg} \mathrm{I}_{2} \text { (yellow) } & -102.9 & (?) & (?) \\ \hline \end{array}$$ Estimate values for the two missing entries. To do this, assume that for the transition \(\mathrm{HgI}_{2}(\mathrm{red}) \longrightarrow\) \(\mathrm{HgI}_{2}(\text { yellow }),\) the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) have the same values that they do at the equilibrium temperature of \(127^{\circ} \mathrm{C}\).

Following are some standard Gibbs energies of formation, \(\Delta G_{f}^{2},\) per mole of metal oxide at \(1000 \mathrm{K}: \mathrm{NiO},\) \(-115 \mathrm{kJ} ; \mathrm{MnO},-280 \mathrm{kJ} ; \mathrm{TiO}_{2},-630 \mathrm{kJ} .\) The standard Gibbs energy of formation of \(\mathrm{CO}\) at \(1000 \mathrm{K}\) is \(-250 \mathrm{kJ}\) per mol CO. Use the method of coupled reactions (page 851 ) to determine which of these metal oxides can be reduced to the metal by a spontaneous reaction with carbon at \(1000 \mathrm{K}\) and with all reactants and products in their standard states.

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