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Chapter 15: Problem 16

Rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}),\) is caused by the oxidation of iron by oxygen. Write the equilibrium constant expression first in terms of activities, and then in terms of concentration and pressure.

Short Answer

Expert verified
The equilibrium constant expression in terms of activities is \(K_{A} = a(\mathrm{O_{2}})^{3}\) and in terms of concentration and pressure is \(K_{P} = P(\mathrm{O_{2}})^{3}\).

Step by step solution

01

Formulate the chemical equation

The reaction represented can be written as: \(4 \mathrm{Fe}(s) + 3 \mathrm{O}_{2}(g) \leftrightharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\).
02

Write the equilibrium constant expression in terms of activities

The equilibrium constant expression (K) in terms of activities is given by the product of the activities (more precisely, the effective concentrations) of the products divided by the product of the activities of the reactants, each raised to the power of its stoichiometric coefficient. Pure solids and liquids do not appear in the expression. Thus, \(K_{A} = a(\mathrm{O_{2}})^{3}\) since the activity of pure solids is one, and the other reactant, iron, is a pure solid.
03

Write the equilibrium constant expression in terms of concentration and pressure

For a gas, the activity is equal to its partial pressure divided by a standard pressure (usually 1 bar or 1 atm). Thus, in this case, the equilibrium constant expression in terms of concentration and pressure, \(K_{P}\), will also be equal to the partial pressure of oxygen cubed. As such, \(K_{P} = P(\mathrm{O_{2}})^{3}\), which is the same as \(K_{A}\) since solids and pure liquids do not appear in the expression.

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Most popular questions from this chapter

A mixture of \(1.00 \mathrm{mol} \mathrm{NaHCO}_{3}(\mathrm{s})\) and \(1.00 \mathrm{mol}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is introduced into a \(2.50 \mathrm{L}\) flask in which the partial pressure of \(\mathrm{CO}_{2}\) is 2.10 atm and that of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is \(715 \mathrm{mmHg} .\) When equilibrium is established at \(100^{\circ} \mathrm{C},\) will the partial pressures of \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) be greater or less than their initial partial pressures? Explain. $$\begin{array}{r} 2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}}=0.23 \mathrm{at} 100^{\circ} \mathrm{C} \end{array}$$

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?

The following data are given at \(1000 \mathrm{K}: \mathrm{CO}(\mathrm{g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) ; \quad \Delta H^{\circ}=-42 \mathrm{k} \mathrm{J}\) \(K_{\mathrm{c}}=0.66 .\) After an initial equilibrium is established in a \(1.00 \mathrm{L}\) container, the equilibrium amount of \(\mathrm{H}_{2}\) can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a 10.0 L container; (d) in some way other than (a), (b), Or (c).

Starting with \(0.280 \mathrm{mol} \mathrm{SbCl}_{3}\) and \(0.160 \mathrm{mol} \mathrm{Cl}_{2},\) how many moles of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present when equilibrium is established at \(248^{\circ} \mathrm{C}\) in a 2.50 L flask? $$\begin{aligned} \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 2.5 \times 10^{-2} \mathrm{at} \ 248^{\circ} \mathrm{C} \end{aligned}$$

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)
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