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Chapter 15: Problem 83

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 $$K_{\mathrm{p}}=\frac{\left(x_{\mathrm{NH}_{3}}\right)^{2}}{\left(x_{\mathrm{N}_{2}}\right)\left(x_{\mathrm{H}_{2}}\right)^{2}} \times \frac{1}{\left(P_{\mathrm{tot}}\right)^{2}}$$

Short Answer

Expert verified
The equilibrium constant expression for \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\) in terms of mole fractions and total gas pressure is \(K_P = \frac{{(x_{NH_{3}})^2}}{{x_{N_{2}}(x_{H_{2}})^3}} \times \frac{1}{{(P_{tot})^2}}\). The derivation involves expressing the partial pressures in the initial equilibrium constant expression in terms of mole fractions and total pressure, and simplifying.

Step by step solution

01

Understanding the problem

According to the ideal gas law, for a reaction at equilibrium involving gases, the equilibrium constant \(K_P\) is expressed in terms of partial pressures of the gases. For the given reaction, \(K_P = \frac{{(P_{NH_{3}})^2}}{{P_{N_{2}}(P_{H_{2}})^3}}\), where \(P_X\) is the partial pressure of gas \(X\). The problem requires us to express this equilibrium constant in terms of mole fractions and total pressure.
02

Expressing partial pressure in terms of mole fraction and total pressure

The partial pressure of a gas in a mixture is given by \(P_X = x_X \times P_{tot}\), where \(x_X\) is the mole fraction of gas \(X\) and \(P_{tot}\) is the total pressure of the mixture. Substituting this into the expression for \(K_P\), we get \(K_P = \frac{{(x_{NH_{3}} \times P_{tot})^2}}{{(x_{N_{2}} \times P_{tot})(x_{H_{2}} \times P_{tot})^3}}\).
03

Simplifying the expression

Simplifying the above expression, we get \(K_P = \frac{{(x_{NH_{3}})^2}}{{x_{N_{2}}(x_{H_{2}})^3}} \times \frac{1}{{(P_{tot})^2}}\). This is the required expression for the equilibrium constant in terms of mole fractions and total pressure.

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Most popular questions from this chapter

Equilibrium is established in a 2.50 L flask at \(250^{\circ} \mathrm{C}\) for the reaction $$\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=3.8 \times 10^{-2}$$ How many moles of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present at equilibrium, if (a) 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are initially introduced into the flask? (b) \(0.610 \mathrm{mol} \mathrm{PCl}_{5}\) alone is introduced into the flask?

Would you expect that the amount of \(\mathrm{N}_{2}\) to increase, decrease, or remain the same in a scuba diver's body as he or she descends below the water surface?

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).

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: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$
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