/*! 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} Problem 124 One mole of a compound AB reacts... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One mole of a compound AB reacts with one mole of compound CD according to the equation: \(\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{CB}\) When equilibrium had been established it was found that \(3 / 4\) mole each of reactant \(\mathrm{AB}\) and \(\mathrm{CD}\) has been converted to AD and CB. There is no change in volume. The equilibrium constant of the reaction is? (a) 9 (b) \(\frac{1}{9}\) (c) \(9 / 16\) (d) \(16 / 9\)

Short Answer

Expert verified
The equilibrium constant is 9.

Step by step solution

01

Understanding Initial Conditions

Initially, you have 1 mole each of AB and CD. Therefore, the initial concentration of AB and CD is each 1 mole per liter, assuming 1-liter volume.
02

Calculating Changes at Equilibrium

At equilibrium, it is given that 3/4 mole each of AB and CD has reacted. Thus, the concentration of AB and CD each decreases by 3/4 mole, leaving 1 - 3/4 = 1/4 mole of each at equilibrium.
03

Concentration of Products at Equilibrium

Since 3/4 mole of each reactant reacts, it forms 3/4 mole of each product, AD and CB. Therefore, the equilibrium concentration of AD and CB is 3/4 mole per liter.
04

Writing the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB} \) is given by: \[ K_c = \frac{[\mathrm{AD}][\mathrm{CB}]}{[\mathrm{AB}][\mathrm{CD}]} \] Substitute the concentrations at equilibrium into this expression.
05

Substituting Equilibrium Concentrations

Plug the known concentrations into the equilibrium expression: \[ K_c = \frac{(3/4)(3/4)}{(1/4)(1/4)} \] Simplify and calculate.
06

Calculating the Equilibrium Constant

Calculate the values: \[ K_c = \frac{9/16}{1/16} = 9 \]. Therefore, the equilibrium constant \( K_c \) is 9.

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 91Ó°ÊÓ!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant yet are not necessarily equal. The reaction does not stop; instead, it proceeds indefinitely, where the forward and backward processes continue at the same rate. This creates a balance allowing specific proportions of substances, characteristic to the equilibrium state, to be maintained.

In the exercise, once the chemical system involving the reaction between AB and CD reaches equilibrium, the conversion of reactants to products proceeds at the same speed as their reconversion into reactants. Understanding equilibrium is vital for controlling reaction conditions and predicting the concentration of substances needed to achieve desired results.
Mole Calculations
Mole calculations are fundamental in chemistry, providing a method to express the amount of a substance. A mole represents Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles of the substance. In this context, moles allow chemists to translate between the microscopic scale of atoms and molecules to the macroscopic amounts they measure in the lab.
  • The initial problem provides 1 mole of AB and 1 mole of CD, which simplifies calculations, allowing a straightforward determination of changes in quantities.
  • As 3/4 mole of AB and CD are transformed, these calculations involve subtracting and adding mole quantities to determine both reactant and product concentrations at equilibrium.
Always remember: maintaining proper mole calculations ensures accuracy when predicting and measuring chemical reactions.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Stoichiometry uses balanced chemical equations to ensure the correct proportions are maintained. This relationship guides how much of each reactant is required, and how much product will form.

In our provided reaction, \( \mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB} \), stoichiometry is essential for understanding how 1 mole of AB reacts with 1 mole of CD to form 1 mole each of AD and CB. This relationship is shown by the balanced equation, which helps predict the amounts of product formed from given amounts of reactants.
  • The calculated equilibrium concentrations stem directly from these stoichiometric relationships.
  • By comprehending these connections, you can successfully navigate mole changes in both laboratory settings and theoretical exercises.
Equilibrium Concentrations
At equilibrium, the concentrations of reactants and products have settled into a steady state. These concentrations depend on reactions' specifics and initial conditions. Knowing the equilibrium constant, \( K_c \), and substituting equilibrium concentrations helps predict the direction and extent of reactions.

For AB and CD, the equilibrium concentrations were determined by noting that 3/4 of each had reacted, leaving 1/4 mole of reactants, while forming 3/4 mole of products, AD and CB.

The equilibrium constant expression, \( K_c = \frac{[\mathrm{AD}][\mathrm{CB}]}{[\mathrm{AB}][\mathrm{CD}]} \), involves these concentrations. You substitute and calculate values to find \( K_c \), simplifying to give a numeric value, in this case 9.
  • Understanding equilibrium concentrations helps anticipate how changes can shift the equilibrium position or impact \( K_c \).
  • This knowledge aids in chemical manufacturing, environmental science, and various strategic applications utilizing chemical processes.

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

In a reversible reaction, the catalyst: (a) Decreases activation energy of forward reaction (b) Increases activation energy of forward reaction (c) Decreases activation energy of both forward and backward reactions (d) Increases activation energy of backward reaction

A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves: (a) The equilibrium has been established in the solution (b) Radioactive sugar can displace non-radioactive sugar from its solution.

In which of the following cases does the reaction go farthest to completion: (a) \(\mathrm{K}=1\) (b) \(\mathrm{K}=10\) (c) \(\mathrm{K}=10^{-2}\) (d) \(\mathrm{K}=10^{2}\)

What is the correct sequence of active masses in increasing order in gaseous mixture, containing one gram per litre of each of the following: 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{N}_{2}\) 3\. \(\mathrm{H}_{2}\) 4\. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: (a) \(3,1,4,2\) (b) \(3,4,2,1\) (c) \(2,1,4,3\) (d) \(4,2,1,3\)

In a reaction \(\mathrm{A}_{2}(\mathrm{~g})+4 \mathrm{~B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}<0\) The formation of \(\mathrm{AB}_{4}(\mathrm{~g})\) will be favoured by: (a) Low temperature and high pressure (b) High temperature and high pressure (c) Low temperature and low pressure (d) High temperature and low pressure

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.