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Chapter 19: Problem 53

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

Short Answer

Expert verified
For part (a), upon solving, we will find that the partial pressure of oxygen at equilibrium is extremely low. For part (b), to obtain significant quantities of oxygen, Lavoisier likely increased the temperature or decreased the pressure to shift the reaction equilibrium toward the products.

Step by step solution

01

Formulate the Equation for the Equilibrium Constant \(K_p\)

Express the equilibrium constant in terms of partial pressures. For the given reaction \(\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\), the equilibrium constant \(K_p\) is given by \(K_p = P_{\mathrm{O}_{2}(\mathrm{g})}^{1/2}\), where \(P_{\mathrm{O}_{2}(\mathrm{g})}\) is the partial pressure of oxygen.
02

Calculate \(K_p\) Using Gibbs Free Energy Change Equation

From thermodynamics, we have \(\Delta G^\circ = -RT \ln(K_p)\) where \(\Delta G^\circ\) is the standard Gibbs free energy change, \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(K_p\) is the equilibrium constant. Given \(\Delta G^\circ=+58.54 \mathrm{kJ/mol} = +58.54 \times 10^3 \mathrm{J/mol}\), \(R = 8.314 \mathrm{J/mol.K}\), and \(T = 298 \mathrm{K}\) (corresponding to \(25^{\circ} \mathrm{C}\)), we can solve for \(K_p\).
03

Calculate the Partial Pressure of \(\mathrm{O}_{2}\)

From Step 2, we will get a value for \(K_p\). Since \(K_p = P_{\mathrm{O}_{2}(\mathrm{g})}^{1/2}\), we can find \(P_{\mathrm{O}_{2}(\mathrm{g})}\) by squaring \(K_p\). This will confirm if the partial pressure of oxygen at equilibrium is very low.
04

Discuss the Conditions for Obtaining Significant Quantities of Oxygen

Part (b) requires understanding of the effect of pressure and temperature on the reaction. In order to obtain significant amounts of oxygen from the decomposition of \(\mathrm{HgO}\), Lavoisier would have needed to manipulate these conditions. If \(K_p\) is small, this means the equilibrium lies to the left, favoring the reactants. Hence, to push the reaction to the right, either increase the temperature or decrease the pressure. Specific conditions would depend on practical constraints.

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Most popular questions from this chapter

From the data given in the following table, determine \(\Delta S^{\circ} \quad\) for the reaction \(\quad \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) All data are at \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

Arrange the entropy changes of the following processes, all at \(25^{\circ} \mathrm{C},\) in the expected order of increasing \(\Delta S,\) and explain your reasoning: (a) \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{atm}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{atm})\) (b) \(\mathrm{CO}_{2}(\mathrm{s}, 1 \mathrm{atm}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}, 10 \mathrm{mm} \mathrm{Hg})\) (c) \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{atm}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 10 \mathrm{mmHg})\)

What values of \(\Delta H, \Delta S,\) and \(\Delta G\) would you expect for the formation of an ideal solution of liquid components? (Is each value positive, negative, or zero?)

Explain why you would expect a reaction of the type $\mathrm{AB}(\mathrm{g}) \longrightarrow \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})$ always to be spontaneous at high rather than at low temperatures.

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 \(\Delta S^{\circ}=15.2 \mathrm{JK}^{-1} ?\) $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})$$
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