/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q28E Write the expression of the reac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Q28E (page 752)

Write the expression of the reaction quotient for the ionization of HOCN in water.

Short Answer

Expert verified

Therefore,the reaction quotient for the ionization of HOCN in water.

\({Q_c} = \frac{{c\left( {{H_3}{O^ + }} \right) \cdot c\left( {OC{N^ - }} \right)}}{{c(HOCN)}}\).

Step by step solution

01

Reaction regarding ionization of HOCN in water:

First, we will write reaction of dissociation:

\(HOCN(aq) + {H_2}O(l) \to {H_3}{O^ + }(aq) + OC{N^ - }(aq)\)

(Note, there needs to be double arrow, because this is a reverse reaction).

Equation for reaction quotient is the same as for equilibrium constant, but we don't use equilibrium concentrations, but concentrations in any time before equilibrium.

02

The Expression for the reaction quotient for the ionization of HOCN in water:

\({Q_c} = \frac{{c\left( {{H_3}{O^ + }} \right) \cdot c\left( {OC{N^ - }} \right)}}{{c(HOCN)}}\)

Note - water is not included into the equation because it is liquid and its activity is 1 , so it would be like this:

\({Q_c} = \frac{{c\left( {{H_3}{O^ + }} \right) \cdot c\left( {OC{N^ - }} \right)}}{{c(HOCN) \cdot 1}}\).

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 91影视!

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

For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%.

Is \({K_c} > 1,\; < 1\), or \( \approx 1\) for a titration reaction?

What is the pressure of \(BrCl\) in an equilibrium mixture of \(C{l_2}\), \(B{r_2}\), and \(BrCl\) if the pressure of \(C{l_2}\) in the mixture is \(0.115atm\)and the pressure of \(B{r_2}\) in the mixture is \(0.450atm\)?

\(C{l_2}(g) + B{r_2}(g) \rightleftharpoons 2BrCl(g)\)

\({K_P} = 4.7 \times 1{0^{ - 2}}\)

A sample of ammonium chloride was heated in a closed container. NH4 Cl (s)鈬� NH3 (g) + HCl(g)at equilibrium, the pressure of NH3 (g)was found to be 1.75 atm. What is the value of the equilibrium constant, Kp, for the decomposition at this temperature?

Question: At 25 掳C and at 1 atm, the partial pressures in an equilibrium mixture of N2O4 and NO2 are PN2O4= 0.70 atm and PNO2 = 0.30 atm.

(a) Predict how the pressures of NO2 and N2O4 will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?

(b) Calculate the partial pressures of NO2 and N2O4 when they are at equilibrium at 9.0 atm and 25 掳C

Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: \(3{{\rm{C}}_2}{{\rm{H}}_2}(g) \to {{\rm{C}}_6}{{\rm{H}}_6}(g)\). Which value of \({K_c}\) would make this reaction most useful commercially?

\({K_c} \approx 0.01,\;{K_c} \approx 1,\;\;{\rm{or}}\;{K_c} \approx 10.\) Explain your answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

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.