Question

In the equilibrium described in Example $15-12,$ the percent dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ can be expressed as $$\frac{3.00\times {10}^{-3}\mathrm{mol}{\mathrm{N}}_2{\mathrm{O}}_4}{0.0240\mathrm{mol}{\mathrm{N}}_2{\mathrm{O}}_4\text{ initially }}\times 100\mathrm{\%}=12.5\mathrm{\%}$$ What must be the total pressure of the gaseous mixture if ${\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})$ is to be $10.0\mathrm{\%}$ dissociated at $298\mathrm{K}?$ $${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{NO}}_2(\mathrm{g})K_{\mathrm{p}}=0.113\text{ at }298\mathrm{K}$$

Short answer

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

Step-by-step solution

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 $0.0240-0.125\ast 0.0240=0.0210mol$.

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 $0.10\ast 0.0240=0.0024mol$ of $N_2{O}_4$ will dissociate into $2\ast 0.0024=0.0048mol$ of $N{O}_2$ according to the balanced chemical equation given.

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 $0.0210-0.0024=0.0186mol$. The equilibrium concentration of $N{O}_2$ is simply the moles that were formed through dissociation which is 0.0048 mol.

Use equilibrium expression to calculate pressure

Now we can use the equilibrium constant ($K_p$) expression based on the balanced equation $$N_2{O}_4\rightleftharpoons 2N{O}_2(g)$$ which gives us $$K_p=\frac{[N{O}_2{]}^2}{[N_2{O}_4]}$$. 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.

Key concepts

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:
$$\text{Percent Dissociation}=\left(\frac{\text{Moles dissociated}}{\text{Total initial moles}}\right)\times 100\mathrm{\%}$$
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.