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!

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

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.

In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?

Write the equilibrium constant expression for the following reaction, $$\begin{array}{r} \mathrm{Fe}(\mathrm{OH})_{3}+3 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ K=9.1 \times 10^{3} \end{array}$$ and compute the equilibrium concentration for \(\left[\mathrm{Fe}^{3+}\right]\) at \(\left.\mathrm{pH}=7 \text { (i.e., }\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\right)\)

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?

Refer to Example \(15-4 . \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) at \(747.6 \mathrm{mmHg}\) pressure and a \(1.85 \mathrm{g}\) sample of \(\mathrm{I}_{2}(\mathrm{s})\) are introduced into a \(725 \mathrm{mL}\) flask at \(60^{\circ} \mathrm{C} .\) What will be the total pressure in the flask at equilibrium? $$\begin{aligned} \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{S}(\mathrm{s}) & \\ K_{\mathrm{p}}=& 1.34 \times 10^{-5} \mathrm{at} 60^{\circ} \mathrm{C} \end{aligned}$$

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