Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write an equilibrium constant, \(K_{\mathrm{p}},\) for the formation from its gaseous elements of (a) 1 mol \(\mathrm{NOCl}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{ClNO}_{2}(\mathrm{g}) ;\) (c) \(1 \mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g}) ;\) (d) \(1 \mathrm{mol}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

Short Answer

Expert verified
The expressions for \(K_p\) are: \n(a) \(K_p = \frac{P_{NOCl}}{P_{N2}^{0.5} \times P_{O2}^{0.5} \times P_{Cl2}}\) \n(b) \(K_p = \frac{P_{ClNO2}^2}{P_{N2} \times P_{O2}^2 \times P_{Cl2}}\) \n(c) \(K_p = \frac{P_{N2H4}}{P_{N2} \times P_{H2}^2} \n\) (d) \(K_p = P_{NH4Cl}\)

Step by step solution

01

Write the reactions for the formation from gaseous elements

Start by writing the reactions for the formation of each compound from its elements in gaseous state. Every compound forms from its components in their basic states:\n\n(a) \(\mathrm{NOCl(g)}\) forms from half a mole \(\mathrm{N}_{2}\) and \(\mathrm{Cl}_{2}\) and \(\mathrm{O}_{2}\):\n\[\(\begin{equation} \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) + Cl_2(g) \longrightarrow NOCl(g) \end{equation}\)\n(b) \(\mathrm{ClNO}_2(g)\) forms from \(\mathrm{N}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\):\n\[\(\begin{equation} N_2(g) + Cl_2(g) + 2O_2(g) \longrightarrow 2ClNO_2(g) \end{equation}\)\n(c) \(\mathrm{N}_{2}\mathrm{H}_{4}\) forms from \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\):\n\[\(\begin{equation} N_2(g) + 2H_2(g) \longrightarrow N_2H_4(g) \end{equation}\)\n(d) \(\mathrm{NH}_{4}\mathrm{Cl(s)}\) forms from \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\):\n\[\(\begin{equation} NH_3(g) + HCl(g) \longrightarrow NH_4Cl(s) \end{equation}\)
02

Write the expressions for \(K_p\)

For each reaction, \(K_p\) is the ratio of the partial pressure of the product over the partial pressure of the reactants, each raised to their coefficients as power.\n\n(a) For \(\mathrm{NOCl}\):\n\[\(\begin{equation} K_p = \frac{P_{NOCl}}{P_{N2}^{0.5} \times P_{O2}^{0.5} \times P_{Cl2}} \end{equation}\)\n(B) For \(\mathrm{ClNO}_{2}\):\n\[\(\begin{equation} K_p = \frac{P_{ClNO2}^2}{P_{N2} \times P_{O2}^2 \times P_{Cl2}} \end{equation}\)\n(c) For \(\mathrm{N}_{2}\mathrm{H}_{4}):\n\[\(\begin{equation} K_p = \frac{P_{N2H4}}{P_{N2} \times P_{H2}^2} \end{equation}\)\n(d) For \(\mathrm{NH}_{4}\mathrm{Cl}\) (don't include solids in \(K_p\)):\n\[\(\begin{equation} K_p = P_{NH4Cl} \end{equation}\)
03

Fill out the values of pressures in \(K_p\)

If given values, this step would involve filling out the pressures and calculating \(K_p\). However, in this problem, we are not given any pressures and are asked to write the expression only. Thus, this step is not needed in this problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is an essential concept in understanding how reactions occur in chemistry. It describes the state in which both the reactants and products of a chemical reaction are present in concentrations that do not change over time. This balance occurs when the forward reaction rate equals the backward reaction rate.
In the context of the given exercise, we deal with forming compounds from gaseous elements and understanding at what point these reactions reach equilibrium.
At chemical equilibrium, the ratio of the products to reactants remains constant. In the case of gas reactions, this constant is often expressed in terms of partial pressures, denoted by the equilibrium constant "K" or more specifically, "Kp" for gases.
  • This Kp indicates how much product is formed compared to the reactants when equilibrium is achieved.
  • Understanding chemical equilibrium helps predict the concentrations of various species in a chemical system over time.
  • This balances the dynamic forces of the reactions moving forward and the reactions moving backward.
Gaseous Elements
Gaseous elements play a critical role in chemical reactions involving the equilibrium constant Kp. These elements make up the reactants from which products are formed in the reactions observed in the exercise.
They are typically the early stages of most chemical processes, especially in combustion and synthesis reactions.
When calculating the equilibrium constant for reactions involving gaseous elements, it's crucial to consider their individual properties such as:
  • The molar volume at given temperatures and pressures.
  • The general behavior of gases such as diffusion and effusion.
  • Their ability to occupy space based on the ideal gas law.
In the exercise, gaseous [\( \mathrm{N}_2 \), \( \mathrm{O}_2 \), and \( \mathrm{Cl}_2 \)]elements are fundamental in forming compounds such as \( \mathrm{NOCl} \) and \( \mathrm{ClNO}_2 \). Understanding the basics of these elements helps in comprehending how the reactions achieve their equilibrium states.
Partial Pressure
Partial pressure is a crucial concept when discussing gases in a mixture within a chemical reaction. It is defined as the pressure that a gas in a mixture would exert if it alone occupied the entire volume of the original mixture. The idea of partial pressure is vital for calculating the equilibrium constant \( K_{\mathrm{p}} \), which applies specifically to gases.
In the exercise, we derive \( K_{\mathrm{p}} \) expressions using the partial pressures of each gas component:
  • Products and reactants must be raised to the power of their coefficients in the balanced chemical equation.
  • This adjustment reflects how many moles of each gas are involved in the reaction.
  • \( K_{\mathrm{p}} \) offers insights into which direction a reaction is likely to go, helping to predict and control chemical processes.
For example, if you want to find \( K_{\mathrm{p}} \) for forming \( \mathrm{NOCl} \), you need to know the partial pressures of \( \mathrm{N}_2 \), \( \mathrm{O}_2 \), and \( \mathrm{Cl}_2 \), combined according to the formula given in the solution.
Reaction Formation
Reaction formation is the process by which reactants are converted into products. In the context of the exercise, this refers to forming specific compounds from gaseous elements. The reactions involve contemporaneous formation and decomposition processes, each contributing to an overall chemical balance.
The key reactions are written with gaseous reactants on the left-hand side and the desired compound as the product on the right-hand side. The coefficients in these equations give you important information:
  • They tell how many molecules are involved in the balanced reaction formation.
  • The relationship between reactants and products is essential for calculating \( K_{\mathrm{p}} \).
Understanding reaction formation helps predict how products will be made from certain reactant conditions. Each molecule or atom must transform according to its specific reaction characteristics, and equilibrium is the state when no further net change is observed unless conditions are altered.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At \(2000 \mathrm{K}, K_{c}=0.154\) for the reaction \(2 \mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If a \(1.00 \mathrm{L}\) equilibrium mixture at \(2000 \mathrm{K}\) contains \(0.10 \mathrm{mol}\) each of \(\mathrm{CH}_{4}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (a) what is the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) present? (b) Is the conversion of \(\mathrm{CH}_{4}(\mathrm{g})\) to \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) favored at high or low pressures? (c) If the equilibrium mixture at \(2000 \mathrm{K}\) is transferred from a 1.00 L flask to a 2.00 L flask, will the number of moles of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) increase, decrease, or remain unchanged?

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

Determine \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\) \(\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}),\) given the following data at \(298 \mathrm{K}\) $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.0 \times 10^{-9}$$ $$\operatorname{NOCl}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=1.1 \times 10^{2}$$ $$\mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=0.3$$

Given the equilibrium constant values $$\begin{aligned} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{c}}=2.7 \times 10^{-18} \\ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) & \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) K_{\mathrm{c}}=4.6 \times 10^{-3} \\ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.1 \times 10^{-9} \end{aligned}$$ Determine a value of \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free