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Chapter 15: Problem 28

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K_{\mathrm{p}}=23.2\) at \(600 \mathrm{K} .\) Explain which of the fol- lowing situations might equilibrium: (a) \(\quad P_{\mathrm{CO}}=P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{CO}_{2}}=P_{\mathrm{H}_{2}} ; \quad\) (b) \(\quad P_{\mathrm{H}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=\) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{CO}} ; \quad(\mathrm{c}) \quad\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2}}\right)=\left(P_{\mathrm{CO}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}^{2}\right)\) (d) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{H}_{2}} / P_{\mathrm{CO}}\)

Short Answer

Expert verified
Only situation (c), where \((P_{CO2})(P_{H2}) = (P_{CO})(P_{H2O}^2)\) might be at equilibrium.

Step by step solution

01

Identify The Reaction and The Expression For Kp

The chemical reaction is \(CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)\). The expression for Kp would be \(K_p = \frac{[CO_2][H_2]}{[CO][H_2O]}\)
02

Evaluate situation (a)

In this situation, all gaseous substances have the same pressure, meaning \(P_{CO} = P_{H2O} = P_{CO2} = P_{H2}\). Plugging these values into the expression for Kp, it wouldn't be 23.2. Therefore, situation (a) is not at equilibrium.
03

Evaluate situation (b)

In situation (b), \( P_{H2} / P_{H2O} = P_{CO2} / P_{CO} \). This isn't equivalent to the Kp expression and hence the system wouldn't be in an equilibrium state.
04

Evaluate situation (c)

For situation (c), we have \((P_{CO2})(P_{H2}) = (P_{CO})(P_{H2O}^2)\). This situation is the only matching situation to the Kp expression for the given reaction. Therefore, situation (c) is possibly at equilibrium.
05

Evaluate situation (d)

In situation (d), \(P_{CO2} / P_{H2O} = P_{H2} / P_{CO}\). Like situation (b), it does not match the expression for Kp, therefore it isn't at equilibrium.

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Most popular questions from this chapter

Explain how each of the following affects the amount of \(\mathrm{H}_{2}\) present in an equilibrium mixture in the reaction \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) $$ \Delta H^{\circ}=-150 \mathrm{kJ} $$ (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst.

In your own words, define or explain the following terms or symbols: (a) \(K_{\mathrm{p}} ;\) (b) \(Q_{\mathrm{c}} ;\) (c) \(\Delta n_{\text {gas }}\)

Show that in terms of mole fractions of gases and total gas pressure the equilibrium constant expression for $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$$ is $$K_{\mathrm{p}}=\frac{\left(x_{\mathrm{NH}_{3}}\right)^{2}}{\left(x_{\mathrm{N}_{2}}\right)\left(x_{\mathrm{H}_{2}}\right)^{2}} \times \frac{1}{\left(P_{\mathrm{tot}}\right)^{2}}$$

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

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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