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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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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle used to describe the behavior of gases. It connects various properties of gases, such as pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and moles of gas (\(n\)). The formula \(PV = nRT\) is the Ideal Gas Law, where \(R\) is the gas constant.
The key takeaway is that the law helps us understand how changes in one property of a gas can affect the others. For instance, if the volume of a gas container decreases, the pressure inside may increase if temperature remains constant. In chemical equilibrium involving gases, we often use this law to determine or predict gas behavior under different conditions.
In our equilibrium problem, we use the concept that a gas's partial pressure at equilibrium can be expressed in terms of its mole fraction and the total pressure via the Ideal Gas Law. This simplification is key to converting expressions of equilibrium constants from partial pressures to mole fractions, which is more useful in specific equilibrium calculations.
Mole Fractions
Mole fractions provide a way to express the concentration of a component in a mixture. A mole fraction (\(x_X\)) is the ratio of the moles of one component to the total moles of all components in the mixture. Mathematically, the mole fraction of species \(X\) is given by \(x_X = \frac{n_X}{n_{total}}\).
Using mole fractions is particularly useful in gas mixtures and can help to simplify calculations. In an equilibrium system, knowing the mole fractions allows us to determine the partial pressures of the gases using the relation \(P_X = x_X \times P_{tot}\), where \(P_{tot}\) is the total pressure.
Understanding mole fractions is crucial as it ties into how we express the equilibrium constant for reactions involving gases. By expressing equilibrium constants, such as \(K_P\), in terms of mole fractions, we can more easily analyze and understand the composition of the gas mixture at equilibrium.
Chemical Equilibrium
Chemical equilibrium takes place when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of the reactants and products over time. This condition is crucial for dynamic processes, particularly in gases where pressures and volumes are influenced by equilibrium reactions.
The equilibrium constant, \(K_P\), quantifies the ratio of the concentrations or pressures of products and reactants at equilibrium. For gaseous systems, it is often expressed using partial pressures. However, as seen in our example, we can convert \(K_P\) into a form that uses mole fractions. This allows more versatile applications and interpretations of equilibrium conditions.
The concept of chemical equilibrium is essential not only for predicting the outcome of reactions but also for understanding how changes in conditions like pressure, temperature, or volume will affect a reaction. Mastery of these concepts and their interrelations—even in their mathematical forms like equilibrium constant expressions—enables one to manipulate and predict chemical processes in various scientific and industrial contexts.

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Most popular questions from this chapter

The decomposition of \(\mathrm{HI}(\mathrm{g})\) is represented by the equation $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$$ \(\mathrm{HI}(\mathrm{g})\) is introduced into five identical \(400 \mathrm{cm}^{3}\) glass bulbs, and the five bulbs are maintained at \(623 \mathrm{K}\) Each bulb is opened after a period of time and analyzed for \(I_{2}\) by titration with \(0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) $$\begin{array}{l} \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \end{array}$$ Data for this experiment are provided in the table below. What is the value of \(K_{\mathrm{c}}\) at \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

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?

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) \(2.8 \times 10^{5}\)

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.

The reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \quad \Delta H^{\circ}=\) \(+181 \mathrm{kJ},\) occurs in high-temperature combustion processes carried out in air. Oxides of nitrogen produced from the nitrogen and oxygen in air are intimately involved in the production of photochemical smog. What effect does increasing the temperature have on (a) the equilibrium production of \(\mathrm{NO}(\mathrm{g})\) (b) the rate of this reaction?

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