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Chapter 15: Problem 22

Write the equilibrium constant expression for the dissolution of ammonia in water: $$\mathrm{NH}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{aq}) \quad K=57.5$$ Use this equilibrium constant expression to estimate the partial pressure of \(\mathrm{NH}_{3}(\mathrm{g})\) over a solution containing \(5 \times 10^{-9} \mathrm{M} \mathrm{NH}_{3}(\text { aq }) .\) These are conditions similar to that found for acid rains with a high ammonium ion concentration.

Short Answer

Expert verified
The partial pressure of the gaseous ammonia (\( \mathrm{NH}_{3}(\mathrm{g}) \)) over the solution is \( 8.7 \times 10^{-11} \) atm.

Step by step solution

01

Equilibrium Constant Expression

The first thing to define is the equilibrium constant expression for the dissolution of ammonia in water. The equilibrium for a gas dissolving in water can be given by: \[ \text{Gas} (\text{g}) \rightleftharpoons \text{Gas} (\text{aq}) \]and the equilibrium constant expression (in terms of partial pressures) for this can be written as: \[ K = \frac{[NH_{3}(\text{aq})]}{P_{NH_{3}(\text{g})}} \] where \( [NH_{3}(\text{aq})] \) is the concentration of aqueous ammonia and \(P_{NH_{3}(\text{g})}\) is the partial pressure of gaseous ammonia. The given value of the equilibrium constant,K, is 57.5.
02

Plugging Numbers into Equation

We are given that the aqueous concentration of ammonia is \( 5 \times 10^{-9} \) M. We can substitute this, and the provided equilibrium constant into our equilibrium expression, and solve for the partial pressure of ammonia. So, we can write:\[57.5 = \frac{5 \times 10^{-9}}{P_{NH_{3}(\text{g})}}\] Consequently, \(P_{NH_{3}(\text{g})}\) can be found as: \( P_{NH_{3}(\text{g})} = \frac{5 \times 10^{-9}}{57.5} \)
03

Calculating the Partial Pressure

We now calculate the partial pressure of the gas by calculating the value obtained: \( P_{NH_{3}(\text{g})} = \frac{5 \times 10^{-9}}{57.5} = 8.7 \times 10^{-11} \, \text{atm} \). Thus, the partial pressure of the gaseous ammonia over the solution is estimated to be \( 8.7 \times 10^{-11} \, \text{atm} \).

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Most popular questions from this chapter

Write the equilibrium constant expression for the following reaction, $$\begin{array}{r} \mathrm{Fe}(\mathrm{OH})_{3}+3 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ K=9.1 \times 10^{3} \end{array}$$ and compute the equilibrium concentration for \(\left[\mathrm{Fe}^{3+}\right]\) at \(\left.\mathrm{pH}=7 \text { (i.e., }\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\right)\)

In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?

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

For the reaction \(2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) \(K_{\mathrm{c}}=1.8 \times 10^{-6}\) at \(184^{\circ} \mathrm{C} . \mathrm{At} 184^{\circ} \mathrm{C},\) the value of \(K_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g})\) is (a) \(0.9 \times 10^{6}\) (b) \(7.5 \times 10^{2}\) (c) \(5.6 \times 10^{5}\) (d) \(2.8 \times 10^{5}\)

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?
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