Logout Open In App
Open In App
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

Chapter 13: Q82E (page 760)

Question: Calculate the equilibrium concentrations that result when 0.25 M O2 and 1.0 M HCl react and come to equilibrium.

Short Answer

Expert verified

The equilibrium concentration is

Step by step solution

01

Determine change in concentration:

Given information:


  • The concentration of O2 is 0.25M
  • The concentration of HCL is 1.0M
  • The equilibrium constant is

We have to find the equilibrium concentrations.

Sinceis large, we will assume that the reaction goes to completion and then comes back to equilibrium.

We will assume that the volume of a solution is 1 L, hence the number of moles of O2 is 0.25 mol and the number of moles of HCL is 1.0 mol

Since 4 moles of HCL reacts with 1 mole of O2, to produce 2 moles of Cl2 and 2 moles of H2O 1.0 mole of HCL will react with 0.25 mole of O2, and produce 0.50 moles of Cl2 (0.50 M) and 0.50 moles of H20 (0.50 M)

We have to determine the value of x,

Since is too large, we will neglect the change in concentration of Cl2, and H2O, and get

02

Determine equilibrium concentration of all species:

The change in the concentration obtained

Therefore, the equilibrium concentrations are

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.

Start your free trial

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

Question: Antimony pentachloride decomposes according to this equation:

An equilibrium mixture in a 5.00-L flask at 4480C contains 3.85 g of \({\rm{SbC}}{{\rm{l}}_5}\),9.14 g of \({\rm{SbC}}{{\rm{l}}_3}\)and 2.84 g of \({\rm{C}}{{\rm{l}}_2}\).How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?

Question : A 0.010Msolution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010Msolution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.

(a) Which acid has the larger equilibrium constant for ionization

HA[HA(aq) ⇌ A−(aq) + H+(aq)]or HB[HB(aq) ⇌ H+(aq) + B−(aq)]?

(b) What are the equilibrium constants for the ionization of these acids?

(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A− or B−), and the hydrogen ion (H+). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

How will an increase in temperature affect each of the following equilibrium? How will a decrease in the volume of the reaction vessel affect each?

\(\begin{gathered} (a)2{H_2}O(g) \rightleftharpoons 2{H_2}(g) + {O_2}(g) \hfill \\ \Delta H = 484kJ \hfill \\ (b){N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g) \hfill \\ \Delta H = - 92.2kJ \hfill \\ (c)2Br(g) \rightleftharpoons B{r_2}(g) \hfill \\ \Delta H = - 224kJ \hfill \\ (d){H_2}(g) + {I_2}(s) \rightleftharpoons 2HI(g) \hfill \\ \Delta H = 53kJ \hfill \\ \end{gathered}\)

Convert the values of Kc to values of Kp or the values of Kp to values of Kc .

\((a)\,{N_2}\left( g \right) + 3{H_2}\left( g \right)\rightleftharpoons 2N{H_3}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_C} = 0.50\,\,at\,\,400^\circ C\)

\((b){{\rm{H}}_2}(g) + {{\rm{I}}_2}(g)\rightleftharpoons 2HI(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = 50.2\,at\,{448^\circ }{\rm{C}}\)

\((c)N{a_2}{\rm{S}}{{\rm{O}}_4} \cdot 10{{\rm{H}}_2}O(s)\rightleftharpoons N{a_2}{\rm{S}}{{\rm{O}}_4}(s) + 10{{\rm{H}}_2}O(g){K_P} = 4.08 \times {10^{ - 25}}at\,{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_P} = 0.122\,at\,{50^\circ }{\rm{C}}\)

Question:Calculate the value of the equilibrium constant \({K_P}\) for the reaction \(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\) from these equilibrium pressures: NO, \(0.050atm;C{l_2},0.30atm;NOCl,1.2atm\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation
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