In the equilibrium described in Example \(15-12,\) the percent dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be expressed as $$\frac{3.00 \times 10^{-3} \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}}{0.0240 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4} \text { initially }} \times 100 \%=12.5 \%$$ What must be the total pressure of the gaseous mixture if \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) is to be \(10.0 \%\) dissociated at \(298 \mathrm{K} ?\) $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.113 \text { at } 298 \mathrm{K} $$

Short Answer

Expert verified
The total pressure of the gaseous mixture, to achieve 10% dissociation at 298 K, is approximately 3.98 atm.

Step by step solution

01

Determine initial concentration of \(N_{2}O_{4}\)

First, let's determine the initial concentration of \(N_{2}O_{4}\). Since its original dissociation is 12.5%, it means 0.125 mole of \(N_{2}O_{4}\) would dissociate. Subtract this from the total initial moles of \(N_{2}O_{4}\) which gives us an initial \(N_{2}O_{4}\) concentration of \(0.0240 - 0.125*0.0240 = 0.0210 \, mol\).
02

Calculate the dissociation in moles

Now, if the desired dissociation is 10%, we calculate this as a percentage of the total initial moles of \(N_{2}O_{4}\). That means \(0.10 * 0.0240 = 0.0024 \, mol\) of \(N_{2}O_{4}\) will dissociate into \(2*0.0024 = 0.0048 \, mol\) of \(NO_{2}\) according to the balanced chemical equation given.
03

Calculate new concentrations at equilibrium

Subtract the moles of \(N_{2}O_{4}\) that dissociate from its initial concentration to calculate its equilibrium concentration which is \(0.0210 - 0.0024 = 0.0186 \, mol\). The equilibrium concentration of \(NO_{2}\) is simply the moles that were formed through dissociation which is 0.0048 mol.
04

Use equilibrium expression to calculate pressure

Now we can use the equilibrium constant (\(K_p\)) expression based on the balanced equation \[N_{2}O_{4} \rightleftharpoons 2NO_{2}(g)\] which gives us \[K_p = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]}\]. Suitably replacing the concentrations and the provided \(K_p\) value, the equation becomes: \(0.113 = \frac{[0.0048]^2}{0.0186}\). Solving this for P, the total pressure of the gas mixture, gives us the final answer.

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Key Concepts

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

Percent Dissociation
Percent dissociation represents the ratio of dissociated molecules to the initial number of molecules, usually expressed as a percentage. It's an indicator of the extent to which a compound segregates into its elements or simpler compounds when in a dynamic equilibrium in a chemical reaction.

For instance, if we start with a certain number of moles of a compound and some fraction of it dissociates, the percent dissociation can be calculated using the following formula:
\[\text{Percent Dissociation} = \left(\frac{\text{Moles dissociated}}{\text{Total initial moles}}\right) \times 100\%\]
It's crucial for students to understand that the percent dissociation might vary if the conditions such as temperature or pressure change, as reflected in an equilibrium scenario. In a textbook exercise, students might encounter tasks asking them to relate pressure variables with percent dissociation, which fundamentally tests their grasp of both these concepts and the interplay between them.
Equilibrium Constant (Kp)
The equilibrium constant, expressed as Kp when dealing with gaseous mixtures, is a measure of the concentration of the products to the concentration of reactants at the state of equilibrium, with each raised to the power of their stoichiometric coefficients and taking into account the partial pressures. The general expression for a reaction is:
\[K_p = \frac{{(P_{\text{product}})^n}}{{(P_{\text{reactant}})^m}}\]where \(P\) stands for partial pressure, and \(m\) and \(n\) are the stoichiometric coefficients.

For chemical reactions involving gases, Kp is temperature-dependent and gives insight into the position of the equilibrium. A higher Kp value tends to favor products, indicating a greater extent of reaction proceeding to the right. Conversely, a lower Kp means the reactants are favored. Mastery of this concept allows students to predict the behavior of a gaseous system in equilibrium when external conditions, such as temperature, are varied.
Concentration at Equilibrium
Concentration at equilibrium refers to the stable concentrations of reactants and products in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. This state does not imply that the reactants and products have the same concentration, but that their concentrations cease to change over time.

It's essential to understand how to calculate these equilibrium concentrations, especially in the context of the equilibrium constant expression. The process typically involves writing the balanced chemical equation, setting up an 'ICE' table (Initial, Change, Equilibrium) to track the changes in concentration due to the reaction, and finally, plugging these equilibrium concentrations into the Kp expression to solve for unknowns.

For example:
  • Determine initial concentrations of all species in the reaction.
  • Calculate the change in concentrations as the system shifts towards equilibrium.
  • Derive the concentrations at equilibrium using the initial amounts and the calculated changes.
By understanding these steps, students can successfully navigate problems that require manipulating equilibrium expressions and can make quantitative predictions about the system's behavior under various conditions.

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Most popular questions from this chapter

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

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?

In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

The Deacon process for producing chlorine gas from hydrogen chloride is used in situations where \(\mathrm{HCl}\) is available as a by-product from other chemical processes. $$\begin{aligned} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-114 \mathrm{kJ} \end{aligned}$$ A mixture of \(\mathrm{HCl}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{Cl}_{2}\) is brought to equilibrium at \(400^{\circ} \mathrm{C}\). What is the effect on the equilibrium amount of \(\mathrm{Cl}_{2}(\mathrm{g})\) if (a) additional \(\mathrm{O}_{2}(\mathrm{g})\) is added to the mixture at constant volume? (b) \(\mathrm{HCl}(\mathrm{g})\) is removed from the mixture at constant volume? (c) the mixture is transferred to a vessel of twice the volume? (d) a catalyst is added to the reaction mixture? (e) the temperature is raised to \(500^{\circ} \mathrm{C} ?\)

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) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

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