- Chapter 1- Some Basic Concepts of Chemistry
- Chapter 2- Structure of The Atom
- Chapter 3- Classification of Elements and Periodicity in Properties
- Chapter 4- Chemical Bonding and Molecular Structure
- Chapter 5- States of Matter
- Chapter 6- Thermodynamics
- Chapter 8- Redox Reactions
- Chapter 9- Hydrogen
- Chapter 10- The sBlock Elements
- Chapter 11- The p Block Elements
- Chapter 12- Organic Chemistry- Some Basic Principles and Techniques
- Chapter 13- Hydrocarbons
- Chapter 14- Environmental Chemistry

Chapter 1- Some Basic Concepts of Chemistry |
Chapter 2- Structure of The Atom |
Chapter 3- Classification of Elements and Periodicity in Properties |
Chapter 4- Chemical Bonding and Molecular Structure |
Chapter 5- States of Matter |
Chapter 6- Thermodynamics |
Chapter 8- Redox Reactions |
Chapter 9- Hydrogen |
Chapter 10- The sBlock Elements |
Chapter 11- The p Block Elements |
Chapter 12- Organic Chemistry- Some Basic Principles and Techniques |
Chapter 13- Hydrocarbons |
Chapter 14- Environmental Chemistry |

A liquid is in equilibrium with its vapours in a sealed container at a fixed temperature. The volume of the container is suddenly increased,

(i) What is the initial effect of the change on the vapour pressure ?

(ii) How do the rates of evaporation and condensation change initially ?

(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?

**Answer
1** :

(i) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.

(ii) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

(iii) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

**What is Kc for the following reaction in astate of equilibrium :****2SO _{2}(g)+ O_{2}(g) **

Given : [SO_{2}] = 0.6 M ; [O_{2}]= 0.82 M ; and [SO_{3}] = 1.90 M

**Answer
2** :

At a certain temperature and total pressureof 10^{5 }Pa, iodine vapour contains 40% by volume ofI atoms

Calculate *K**p*for the equilibrium.

**Answer
3** :

Partialpressure of I atoms,

Partial pressure of I_{2} molecules,

Now,for the given reaction,

Write the expression for the equilibriumconstant, *K**c* for each of the following

reactions:

(ii)

(iii)

(iv)

(v)

**Answer
4** :

Find out the value of *K**c* for each of thefollowing equilibria from the value of *K**p*:

**Answer
5** :

The relation between *K**p* and *K**c* is given as:

*K**p* = *K**c* (RT)^{Δ}*n*

(a) Here,

Δ*n* = 3 – 2 = 1

*R* = 0.0831 barLmol^{–1}K^{–1}

*T* = 500 K

*K**p* = 1.8 × 10^{–2}

Now,

*K**p* =* K**c* (*RT*) ^{Δ}*n*

(b) Here,

Δ*n* = 2 – 1 = 1

*R *= 0.0831 barLmol^{–1}K^{–1}

*T *= 1073 K

*K**p*= 167

Now,

*K**p* =* K**c* (*RT*) ^{Δ}*n*

Both the forward and reverse reactions inthe equilibrium are elementary bimolecular reactions. What is *K**c*, for the reversereaction?

**Answer
6** :
It is given that for the forward reaction is

Then, forthe reverse reaction will be,

Explainwhy pure liquids and solids can be ignored while writing the equilibriumconstant expression?

**Answer
7** :

Fora pure substance (both solids and liquids),

Now,the molecular mass and density (at a particular temperature) of a puresubstance is always fixed and is accounted for in the equilibrium constant.Therefore, the values of pure substances are not mentioned in the equilibriumconstant expression.

Reaction between N_{2} and O_{2} takes place asfollows:

If a mixture of 0.482 mol of N_{2} and 0.933 mol of O_{2} is placed in a 10 Lreaction vessel and allowed to form N_{2}O at a temperature forwhich *K**c* = 2.0 × 10^{–37}, determine thecomposition of equilibrium mixture.

**Answer
8** :

Let the concentration of N_{2}O at equilibrium be *x*.

Thegiven reaction is:

Therefore,at equilibrium, in the 10 L vessel:

The value of equilibrium constant i.e., = 2.0 × 10^{–37} is very small.Therefore, the amount of N_{2} and O_{2} reacted is also verysmall. Thus, *x* can be neglected from the expressions of molarconcentrations of N_{2} and O_{2}.

Then,

Now,

Nitric oxide reacts with Br_{2} and gives nitrosylbromide as per reaction given below:

When 0.087 mol of NO and 0.0437 mol of Br_{2} are mixed in aclosed container at constant temperature, 0.0518 mol of NOBr is obtained atequilibrium. Calculate equilibrium amount of NO and Br_{2}.

**Answer
9** :

Thegiven reaction is:

Now,2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr areformed from 0.0518 mol of NO.

Again,2 mol of NOBr are formed from 1 mol of Br.

Therefore,0.0518 mol of NOBr are formed from molof Br, or

0.0259mol of NO.

Theamount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br_{2}] = 0.0437 mol

Therefore,the amount of NO present at equilibrium is:

[NO]= 0.087 – 0.0518

=0.0352 mol

And,the amount of Br present at equilibrium is:

[Br_{2}] = 0.0437 – 0.0259

At 450 K, *K*_{p}= 2.0 × 10^{10}/bar for the givenreaction at equilibrium.

What is *K**c *at this temperature?

**Answer
10** :

Forthe given reaction,

Δ*n* = 2 – 3 = – 1

*T* = 450 K

*R* = 0.0831 bar L bar K^{–1 }mol^{–1}

= 2.0 × 10^{10} bar ^{–1}

Weknow that,

Name:

Email:

Copyright 2017, All Rights Reserved. A Product Design BY CoreNet Web Technology