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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dissolution of Ammonia
When ammonia gas (NH₃(g)) dissolves in water, it transforms into its aqueous form, NH₃(aq). This process establishes a dynamic equilibrium between the two states:
  • Ammonia, being a gas, tends to dissolve in water because water molecules surround the gas molecules, moving them into the solution.
  • This dissolution is reversible, as dissolved ammonia can escape back into the gas phase.
Understanding this equilibrium is crucial because it explains how gases interact with liquids, which is important in various chemical and environmental contexts.
The balance between the dissolving and escaping ammonia molecules explains why not all the gas will be dissolved at a particular moment.
Equilibrium reflects a steady state where the concentration of ammonia in both phases doesn't change unless external conditions alter the balance. In problems like these, we often use the equilibrium constant to quantify this relationship.
Partial Pressure Calculation
Partial pressure refers to the pressure exerted by a specific gas in a mixture of gases. When dealing with ammonia dissolving in water, the partial pressure of gaseous ammonia in the air above the solution becomes a critical factor.
To calculate the partial pressure in such problems, we follow these general steps:
  • Use the known concentration of the dissolved form of the gas (ammonia in this case).
  • Apply the given equilibrium constant that relates concentration and pressure.
  • Solve for the unknown pressure using the equilibrium constant formula.
For instance, if we have a specific concentration of aqueous ammonia, we can insert it into the equation \( K = \frac{[NH_{3}(\text{aq})]}{P_{NH_{3}(\text{g})}} \) and find the partial pressure by rearranging it.
The partial pressure calculation provides insight into how much of a gas is present in a closed environment, helping predict its behavior in natural and artificial settings.
Aqueous Equilibrium Expression
The aqueous equilibrium expression relates the concentrations and partial pressures of components in a reaction at equilibrium. For the dissolution of ammonia in water, our equilibrium equation is: \( K = \frac{[NH_{3}(\text{aq})]}{P_{NH_{3}(\text{g})}} \).
Here's what each component stands for:
  • \([NH_{3}(\text{aq})]\) is the concentration of ammonia dissolved in water.
  • \(P_{NH_{3}(\text{g})}\) is the partial pressure of ammonia gas over the solution.
  • \(K\) is the equilibrium constant, signifying the ratio of these values at equilibrium.
This expression is a cornerstone of understanding how gases dissolve in liquids and maintain equilibrium.
Using the constant, scientists and students can predict changes in concentration or pressure when parameters are altered, making it a vital tool in chemistry.
Remember that altering temperature or pressure might change \(K\), so it is essential to consider the conditions under which it is applied.
Understanding these principles enables better predictions and manipulations of reactions not only in labs but in real-world scenarios as well, such as the environmental impact seen in acid rains.

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Most popular questions from this chapter

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

What is the apparent molar mass of the gaseous mixture that results when \(\mathrm{COCl}_{2}(\mathrm{g})\) is allowed to dissociate at \(395^{\circ} \mathrm{C}\) and a total pressure of 3.00 atm? $$\begin{aligned} \operatorname{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+& \mathrm{Cl}_{2}(\mathrm{g}) \\ & K_{\mathrm{p}}=4.44 \times 10^{-2} \mathrm{at} 395^{\circ} \mathrm{C} \end{aligned}$$ Think of the apparent molar mass as the molar mass of a hypothetical single gas that is equivalent to the gaseous mixture.

Based on these descriptions, write a balanced equation and the corresponding \(K_{\mathrm{p}}\) expression for each reversible reaction. (a) Oxygen gas oxidizes gaseous ammonia to gaseous nitrogen and water vapor. (b) Hydrogen gas reduces gaseous nitrogen dioxide to gaseous ammonia and water vapor. (c) Nitrogen gas reacts with the solid sodium carbonate and carbon to produce solid sodium cyanide and carbon monoxide gas.

One of the key reactions in the gasification of coal is the methanation reaction, in which methane is produced from synthesis gas-a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons & \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H &=-230 \mathrm{kJ} ; K_{\mathrm{c}}=190 \mathrm{at} 1000 \mathrm{K} \end{aligned}$$ (a) Is the equilibrium conversion of synthesis gas to methane favored at higher or lower temperatures? Higher or lower pressures? (b) Assume you have 4.00 mol of synthesis gas with a 3:1 mol ratio of \(\mathrm{H}_{2}(\mathrm{g})\) to \(\mathrm{CO}(\mathrm{g})\) in a 15.0 L flask. What will be the mole fraction of \(\mathrm{CH}_{4}(\mathrm{g})\) at equilibrium at \(1000 \mathrm{K} ?\)

Equilibrium is established in the reversible reaction \(2 \mathrm{A}+\mathrm{B} \rightleftharpoons 2 \mathrm{C} .\) The equilibrium concentrations are \([\mathrm{A}]=0.55 \mathrm{M},[\mathrm{B}]=0.33 \mathrm{M},[\mathrm{C}]=0.43 \mathrm{M}\) What is the value of \(K_{c}\) for this reaction?

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