Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes

Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes

Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes. Chemical Equilibrium, a condition in the course of a reversible chemical reaction in which no net change occurs in the amounts of reactants and products. A reversible chemical reaction is one in which the products, as soon as they are formed, react to produce the original reactants. Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes.

Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes

Class 10th Chemistry Chapter 1st Chemical Equilibrium
Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes.

Class 10th Chemistry Chapter 1st Chemical Equilibrium

Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes. At equilibrium, the two opposite reactions proceed at equal rates or rates, and therefore there is no net change in the amounts of substances involved. At this point, the reaction can be considered to be complete; that is, for some specified reaction condition, the maximum conversion of reactants to products has been reached.Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes.

Class 10th Chemistry Chapter 1st Chemical Equilibrium
Class 10th Chemistry Chapter 1st Chemical Equilibrium Notes

See more: Biology Notes Class 10 Chapter 18 Pharmacology 2021

Class 10th Biology Notes: Biology Notes GASEOUS EXCHANGE Chapter 10

Chapter No.09

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Characteristics of reversible reactions: reversible reactions. “Backward directions are called reversible reactions.” OR products and the products react to form the reactants back are called Reversible reactions:

Ii.Reaction of hydrogen with oxygen:

Ii.Reaction of sodium with chlorine:

Ii.Reaction of zinc with hydrochloric acid:

i. Reaction of carbon with oxygen:

Characteristics of irreversible reactions:

Examples:

A few examples of irreversible reactions are given below: products only and products are not changed to reactants are called irreversible reactions.ti

Ans. Irreversible reactions:

Q.1. what are irreversible and reversible reactions?

iv. A state of dynamic equilibrium is established in all reversible reactions. iii. Their reversibility can exist in closed containers only.

v. These reactions are, represented by double arrow sign

ii. These reactions never ao to completion i.e. the reactants are not fully These reactions proceed in two directions i.e. forward and backward 17ev,Fsible reactions have the following characteristics: converted into products

Iv.These reactions go to completion.

Ii.They is represented by a single arrow sign

Iii.No equilibrium state is established in irreversible reactions. .These reactions proceed in one direction only.

A reversible reaction has the following characteristics:

“Those chemical reactions in which the reactants react to form the “Those chemical reactions which proceed both in forward and -Those chemical reactions in which reactants are changed into ‘Those chemical reactions which proceed in forward direction only are called irreversible reactions. – OR

Zn (s) + 2HCl (as) ►ZnCl2 (s) + H2 (g)

2H2 (g) + 02 (g) (I)

2Na (s) + Cl2 (g) (s)

C(s) + 02 (g) CO2-. (g)

Examples:

Following are a few examples of reversible reunions i, Reaction of nitrogen with hydrogen:

N2 (9) + 3H2 (9) 2NH3 (g)

At the start, reaction goes in forward direction but as soon as some amount of ammonia is formed, the ammonia molecules dissociate to nitrogen and hydrogen. Thus the reaction is reversed.

Some other examples are:

Ii.Reaction of sulphur dioxide with oxygen:

2S02 (g) + 02(g) 2S03 (g)

Ii.Reaction of hydrogen with iodine:

112(g) + I2 (g) 2Hl (g)

Ii.Reaction of nitrogen with oxygen: N2 (g) + 02(g) 2NO 3

Q2. What is meant by chemical equilibrium/ Dynamic equilibrium? Explain with examples.

Ans. Chemical equilibrium/Dynamic equilibrium:

History:

Chemical equilibrium (Dynamic equilibrium) was first discover-ed by a French chemist Claude Louis Berthelot in 1803_

Definition:

The state of a reversible chemical reaction in which the of forward reaction becomes equal to the rate of reverse reaction is cs e: chemical equilibrium/Dynamic equilibrium: OR

The state of a reversible chemical reaction a: 73-7i- are continuously changing to products and the – _ changing to the reactants back but the Concentra: 3-S

Products remain constant is called chemical e

Equilibrium.”

Explanation:

Consider a general chemical reaction in which reactant A reacts with reactant B in gaseous state, in a closed container. To form products C

And D.

A (g) + B (g) C (g) + D (9)

The detail about this general reversible reaction is given below;

i. Concentrations at the start:

At the start of the reaction the concentrations of A and B are maximum while that of C and D are zero.

ii. Concentrations after some time:

With the passage of time the concentrations of C and D gradually increase while that of A and B gradually decreases.

iii. Variation of rate of forward and reverse reactions: Change in concentrations of reactants or products per unit time is called rate of reaction.

In the beginning, the rate of forward reaction is maximum while rate of reverse reaction is almost zero but as the time passes and sufficient amounts of C and D are formed then the rate of backward reaction increases while that of forward reaction decreases until rate of forward reaction becomes equal to the rate of reverse reaction. Thus a state of dynamic equilibrium is established.

iv. Concentration at equilibrium state:

The concentrations of reactants or products remain constant at equilibrium state. These concentrations are called equilibrium concentrations.

v. Graphical representation:

Reactants [A] or [B] Equilibrium

Products (C] or [D]

Examples:

i. Change of a liquid to gas in a closed container:

Consider a closed container which is partially filled with a liquid at a given temperature. At the start. Evaporation starts and the vapour molecules are collected at the liquid surface. At the passage of time the collected gas molecules over the liquid surface converts to liquid back condensation starts). In the beginning rate of evaporation is faster than the rate of condensation but with the passage of time the rate of evaporation becomes equal to the rate of condensation and thus a state of dynamic equilibrium is established.

Liquid Evaporation

– Gases

Condensation

Equilibrium state is also established in the following chemical reactions: i. Reaction of sulphur dioxide with oxygen: 2S02 (9)4- 02(9) 2SO2. (g)

For example:

Consider the following general chemical equation.

A+13–•C+D Then according to law of mass action:

R a [AJ [E3]

r = k [A][B]

Where,

r = rate of reaction

= me 3′ concentrations of reacting substances While. k = rate constant of the reaction

For a more general equation law of mass action can be written as:

AA + bB cC 4″ dD

r fAriBt

r = klAr [B]t’

-=re a and b = number of moles of reacting substances in the balanced equation.

Q3. Derive an expression for equilibrium constant.

Ans. Derivation of the expression for the equilibrium constant: Consider a general reversible reaction in which two reactants A and B reacts to form two products C and D. A + B C + D

Law of mass action for forward reaction:

Let the molar concentrations of both reactants be represented by [A] and [B]. Then according to law of mass action the rate of forward reaction (r1) is directly proportional to the product of active masses (molar concentrations) of reactants A and B. By applying law of mass action to forward reaction we get:

RI a [A][B]

r1 = kf [A][B]

Where r1 represents rate of forward reaction while kf is the rate constant of forward reaction.

Law of mass action for reverse reaction:

As the above reaction is reversible therefore, when sufficient C and D are formed they react to form the reactant A and B again. Applying taw

01 MSS Ation for reverse reaction:

R a [C][D]

r2 = k. [C][D]

Where r2 represents rate of reverse reaction while kt is the rate constant of reverse reaction.

At equilibrium:

The rate of forward reaction = the rate of reverse reaction

= r2

Putting the values of RI and r2 we get:

– Kf [A][B] = k,[q[D]

On rearranging:

Kf. [C][D]

Kr [A][B]

[C][D] Ki

Kc – [A] [B] where K =

Kr

For more general reaction K, is written as: AA + bB cC + dD

K, _ [CY [D]j [Ar [B]}

Where a, b, c and d represents the number of mdes ‘Torn e chemical equation.

Note: If the number of moles is one. Then it is not written Equilibrium constant ash K is called equilibrium constant. The eat_ ‘barium CC-:•z7=-7 defined as “The ratio of the product of the active masses of the *: the product of active masses of reactants is called eq…..-_- – s7F.

A large value of K means that at :-. — Concentration is greater than the reactants and vice versa The equilibrium constant is independent of pressure. C_ _- and catalyst and its value is constant for a particular reaction_

Activity:

Write equilibrium constant expressions for the following reversible reactions:

i. N2 (g) + 3H2 (g) 2NH3 (g)

ii. PCI3 (g) + Cl2 (g) PCI5 (g)

Solution:

As we know that: n umber of moles

K [P

Product of products concentrations] m her of moles

Imo) dm-31′ _ 1 1

[Mol d, m-’] 3 tmol tmof din-61

= moI-2dm6

Case-2: It may have no units if the number of moles of reactants and products are equal in the balanced chemical equation.

Example:

H2 (g) + 12 (g) 2HI (g)

K, for this reaction is:

[HI]: -6brril-

Kc = P23s21 – No units

[H:][12]

Therefore, it is concluded that equilibrium c-:-.7-Stan*. ^as different units for different chemical reactions.

Q4.Write an equilibrium constant expression for the reaction of nitrogen and hydrogen to form ammonia at 150-200atm pressure and 400°C to 500°C while iron is used as catalyst.

Ans: uilibrium Constant Expression First we write the chemical equation for the reaction of nitrogen with hydrogen to form ammonia:

N2 (9) 3H2 (g) 2NH3 (g)

Now IC for this reaction is:

[NI -1,]: K, –

[N2][H2F

Example 9.3:

In the equilibrium mixture. The concentration of hydrogen and ionone is 0.04 moles per dm3 each while that of hydrogen iodide is 0.08 moles per dm3. Find lc for the following reaction:

Hz (g) + 12 (g) 2H1 (g)

Given:

Chemical equation = H2 (g) + 12 (g) 2HI (g)

Concentrations of hydrogen = 0.04mol/dm3

Concentrations of iodine = 0.04mol/dm3

Concentration of hydrogen iodide = 0.08 mol/dm3

Required:

Value of Kc

i. Kc – LVH’12

[A.AH2r

Ii.Kc-

[PC/,]

I Pa

Q6.What is equilibrium constant? Explain its units.

Ans. Equilibrium constant:

The ratio of the product of the active masses of the products to the product of active masses of reactants is called equilibrium constant.” Denotation:

It is denoted by K., where the subscript c indicates the equilibrium concentrations of various species in term of mole/dm3or liter.

For example consider the following general reaction

A+B

Then K, for this reaction will be:

[CP]

K, =

[A][B]

For more general reaction K, is written as: aA + bB cC + dD

[D],’Kc – [D]

[A]a [B]b

Where [A],[B],[C] and[D] represents molar concentration of reactants and products while amebic and d represents the number of moles from the balanced chemical equation.

Units of equilibrium constant:

The unit of equilibrium constant depends on the equilibrium constant expression. We have two cases regarding the units of equilibrium constant:

Case-1:

It may have units if the number of moles of reactants and products are unequal in the balanced chemical equation.

Example:

N2 (g) + 3H2 (g) 2NH3 (g)

For this reaction is:

= [N11,] 2 [tnol dm-‘] =

[N,][H, r 1mol dm-‘][mol dm-‘r

Solution:

Writing equilibrium constant expression for the above reaction:

K, = [11112

[1121112]

Putting the values we get:

[0.08mo/ dm-3 f

K: =

Result:

Value of Kc = 4

Eyamnlo 9.4!

CO (g) + H20 (g) CO2 (g) + H2 (g)

For the above reaction at certain high temperature the concentration of CO is 0.0600 mol/dm3, H2O is 0.120 mol/dm3, CO2 is 0.150 mol/dm3 and H2 is 0.300 mol/dm3. Calculate the value of the equilibrium constant.

Given:

Chemical equation = CO (g) + H20 (g)i CO2(g) + H2(g)

Concentrations of CO = 0.0600 mol/dm

Concentration of H2O = 0.120 mol/dm3

Concentration of CO2 = 0.150 mol/dm3

Concentration of H2 = 0.300 mol/dm3

Required:

To calculate the value. Of 1<c

Solution:

Writing equilibrium constant expression for the above reaction:

Kc = [COz][H2] [CO] [H2 0]

Putting the values we get:

[0.150 mol.dm-1[0.300mol.dm-3]

[0.0600 mol.dm-3][0.120mol.dm-3]

K [0.150, r…..Ifilft7ii–./–n-][0.300!n011-ar’r1- ]

C

[0.0600 Moldier-] [0.120n011-drir]

, [0.150][0.300]

[0.0600][0.120] 0.045

0.0072

Kc = 6.25

Result:

Value of K- = 6.25

Practice Problem 9.4

An equilibrium mixture of N2, 02 and NO gases at 1500K is determined to consist of 6.4×10-3molldm3 of N2, 1.7×10.3mol/dm3 of 02 and 1.1×10-5madm3 of NO. What is the equilibrium constant for the given system?

Given:

Concentrations of NO =1.1 dmolldm3

Concentration of N2 = 6.4×10-3mol/dm3 Concentration of 02 = 1.7×10-3mo1/dm3 required:

To calculate the value of K,

Solution:

Chemical equation for this equilibrium mixture is: N2 (g) + 02(9) 2N0 (g)

Writing equilibrium constant expression for the above reaction:

P V0]-

KC –

[N2][02] Putting the values we get:

X 10-5 mol.dm-312

6.4 x 10 x10-3 mol.dml

[1.21x le° mol 2 .dm I

C

[10.88 x10- ion/ 2 .dm-6]

[1.21 x10-1° –

Kc –

1.088x 10-5

1.21×10-1°x10′

K 1.088

Kc=1.112×10-5

Kc= 6.25

Result:

Value of K, = 1.112×10-5

Q.7. what is the importance of equilibrium constant? OR

What are the applications of equilibrium constant?

Ans. Importance/applications of equilibrium constant:

The value of equilibrium constant is specific and remains constant at a particular temperature. The equilibrium constant can be used to predict:

i. The direction of chemical reaction

ii. The extent of chemical reaction

iii. The effect of change in conditions upon a chemical reaction in equilibrium state.

The detail is given below:

1. Prediction of direction of reaction:

The direction of reaction can be predicted by comparing the value of reaction quotient (Q,) with that of K.

Q0 is the ratio of initial molar concentration of products to initial molar concentration of reactants i.e.

_ [Initial concentration 01 products]

[Initial concentration reactants]

By comparing the value of Q0 with K, we have three possibilities: i.Qc<K,

If the value of 0 is less than K0 then the reaction will proceed in forward direction to form products.

Ieq,>Kc:

If the value of () cis greater than K, then the reaction will go in the reverse direction to form reactants.

= Kc:

If the value of Q, is equal to K0 then the reaction is at equilibrium i.e. rate of forward reaction is equal to the rate of reverse reaction.

2. Extent of chemical reaction:

The value of K, tells us about the extent of reaction from which quantities of reactants or products can also be predicted. There are three possibilities:

i. K, value very small:

If K, value is very small then the forward reaction will not occur to an appreciable extent and the reverse reaction (from right to left) will go nearly to completion. Thus we have large amount of reactants_ as compared to products.

For example:

K, value for the following reaction is 1 x10-3° at 25 °C which is very small N2 (g) + 02 (g) 2NO (g)

The equilibrium mixture of above reaction will contain large amounts of N2 and 02 and trace amount of NO.

Kc value very large:

If K, value is very large then the reverse reaction will not occur to an appreciable extent and the forward reaction (from left to right) will go nearly to completion. Thus we have large amount of products as compared to reactants.

For example:

K, value for the following reaction is 1 x1038 at 25 °C which is very large. 201 (g) 012 (g)

Thus the equilibrium mixture of above reaction will consist of almost entirely all chlorine molecules. The chlorine atoms will be in negligible amount.

iii. Kc value neither very small nor very large (moderate)

If K, value is neither very small nor very large then neither the forward nor the reverse reaction goes to completion. Thus both reactants and products will be present in neither very large nor very small amounts. For example:

The K, value for the following reaction is 0.36 at 25 °C which is moderate.

N204 (g) 2NO2 (9)

The equilibrium mixture of above reaction will contain both N204 and NO2 in moderate amounts.

3. The effect of change in external conditions:

Once a system has attained the equilibrium then it is possible to change the ratio of products to reactants by changing the external conditions i.e. concentration, pressure, temperature. Changing these conditions will disturb the equilibrium. We can deduce the direction in which equilibrium will shift when one of these changes. This can be done by applying Le-Chatelier’s principle stated by French chemist Henri Louis Le Chatelier in 1888.

Le-Chatelier’s principle:

This principle states that if a system at equilibrium is disturbed by some change, the system will shift so as to counteract the effect of the change. (OR)

If the equilibrium of a system is disturb through anybody. It resists changing its position.

We can determine the extent of shift by working with the equilibrium constant Kc.

Q.8. Write down the conditions necessary for equilibrium.

Ans: Conditions necessary for equilibrium:

If a system attains equilibrium then it will remain at equilibrium until its conditions are changed. When the conditions are disturbed the equilibrium will also be disturbed.

Followings are the conditions necessary for chemical equilibrium:

i. Closed container:

Equilibrium state can only be attained in closed container. It cannot be attained in open container because in open container the gaseous reactants and products will escape due to which there will be no possibility of equilibrium

ii. Constant concentrations:

When equilibrium state is attained by a reaction in a closed container then the concentrations of various species in the reaction become constant. These concentrations are called equilibrium concentrations.

iii. No change in catalyst:

If the catalyst used at the equilibrium state is unchanged the system will remain at equilibrium.

iv. Constant Volume:

For equilibrium to remain, constant volumes are also necessary.

v. Constant Pressure:

Constant pressure is also necessary for equilibrium state, to remain unchanged.

Society, technology and science

Nitrogen and oxygen are present in our atmosphere which is used in the industrial preparation of nitric acid by Birkland-Eyde process. This process has the following Steps:

Step-1:

In this step air is passed through an electric arc to produce nitric oxide:

2NON2 + •-02

Step-2:

The nitric oxide is further oxidized to NO2 in the presence of oxygen:

2N0-O. 2NO2

Step-3:

The NO2 is dissolved in water in absorption tower t produce nitric acid:

3NO2 + H20 i 2HNO3 + NO

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Multiple Choice Questions 1

Q.1: Choose the correct answers from the given options:

I.At dynamic equilibrium

a. The reverse reaction stops

b. The forward reaction stops

c. Both forward and reverse reactions stop

d. Both forward and reverse reactions continue at the same rate.

ii. A reversible reaction has the following characteristics except:

a. They proceed in both directions

b. They never complete

c. Products do not form reactants again

d. They are represented by

Ii.They equilibrium constant expression for the given reaction is:

N2 (9) + 02 (9) 2N0 (9)

A.

[2A T][20] b. [1’x’]2[0]2

C.

21[02] d. [N012

[2N0] [2N0] [N0]2 IN MO 21

iv. The equilibrium constant K, is:

a. The sum of the two reactants

b. The difference of two rate constants

c. The ratio of the two rate constants

d. The product of the two rate constants

v. When the value of K, is very small it shows that:

a. Reaction will go in the forward direction

b. Reaction will go in reverse direction

c. Reaction is equilibrium

d. Equilibrium will never establish

vi. For which of the following reaction, the will have no unit:

a. N2 (g) +02 (g) 2N0 (g)

b. N2 (g) + 3H2 (g) 2NH3 (g)

c. CO (g) + 3H2 (g) CH4 (g) + H20 (g)

d. 4NH3 (g) + 502(g) 4N0 (g) + 6H20(g)

vii. When the value of IC is very large it shows that:

a. Reaction will go in the forward direction

b. Reaction will go in reverse direction

c. Reaction is equilibrium

d. Equilibrium will never establish

viii. Active mass means:

a. The total mass of reactants

b. The total mass of products

c. The total mass of reactants and products

d. Concentrations of reactants and products in mole per dm3 in a dilt.ite solution

Ix. For a reversible reaction: [, c]4 The equation will be

[A]. [13]-

a.4C (g) 3A (g) + 2B (g)

b.4C (g) A3 (g) + B2 (g)

c.A3 (g) + B2 (g) 04(g)

d. 3A (g) + 2B (g) 4C

x. The reaction between PCI3 and 012 will produce PCI5

PCI3 (g) + Cl2 (g) PCI5 (g) the units of K, for this reaction are a mol.dm-3 b. mo1-1.dm-3 c. mo1-1.dm3

Solution:

Kc- [PCI4]

[PCI3][C12] [Nro/. dm •][1_

1

– Rn ol-1.dm3

[rnol.dm’] d. mol.dm3

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1. Define chemical equilibrium with two examples?

Ans. Chemical Equilibrium:

History:

Chemical equilibrium was first discovered by a French chemist Claude Louis Berthelot in 1803.

Definition:

“The state of a reversible chemical reaction in which the rate of forward reaction becomes equal to the rate of reverse reaction is called chemical equilibrium.”(OR)

“The state of a reversible chemical reaction at which the reactants are continuously changing to products and the products are continuously changing to the reactants back but the concentrations of reactants and products remain constant is called chemical equilibrium.”

Two Examples:

In the following two examples equilibrium state will be established in closed container.

i. Reaction of sulphur dioxide with oxygen:

2S02 (g) + 02(g) 2S03 (g)

ii. Reaction of hydrogen with iodine: H2 (g) + 12 (g) 2H1 (g)

2. How would you identify that dynamic equilibrium has been established?

Ans. Identification of Dynamic Equilibrium:

In order to identify the equilibrium state of a reversible chemical reaction, the following methods are used:

i. Physical Methods such as refractometery, polarimetry, spectrophotometry. Chemicals methods such as titration etc.

In both methods, the equilibrium state of a reversible chemical reaction is identified by knowing the concentration of reactants and products at regular interval of time.

When the concentration of the reactants and products become constant, then reaction is said to be in equilibrium.

3. Compare the different macroscopic properties of forward and reverse reactions?

Ans. Comparison between the macroscopic properties of forward and reverse reactions:

It is given in the table below:

Forward Reaction Reverse Reaction

i.

It is a reaction in which

Reactants react to form
products.

, It is a reaction in which

I products react to form
reactants.

It takes place from left to right.

i It takes place from right to left.

At initial stage, the rate of

Forward reaction is very fast.

At initial stage, the rate of

Reverse reaction is very
negligible.

It slows down gradually.

It speeds up gradually.

Example:

Forward reaction

H2 (g) + 12 (g) 2H1 (g)

Reverse

reaction

4. What is required to predict the direction of a chemical reaction?

Ans. Requirement for predicting direction of a chemical reaction:

The value of K, and 0, is required for predicting the direction of a chemical .reaction. It is explained below:

Prediction of direction of reaction:

The direction of reaction can be predicted by comparing the value of reaction quotient (Q,) with that of K0.

Q, is the ratio of initial molar concentration of products to initial molar concentration of reactants

QC _ concentration of prodiaiNi

[Initial concentration reactants]

By comparing the value of Q, with K0 we have three possibilities: C1, <Ke:

If the value of Q, is less than K0 then the reaction will proceed in forward direction to form products

QC>Kc:

If the value of Q, is greater than K0 then the reaction will go in the reverse direction to form reactants.

QC = Kc

If the value of Q0 is equal to K0 then the reaction is at equilibrium i.e. rate of forward reaction is equal to the rate of reverse reaction.

5. Relate the active mass with rate of a chemical reaction

Ans.Relation of active mass with rate of a chemical equation:

The active mass is related with rate of a chemical equation by law of mass action stated by two Norwegian. Chemists, Goldberg and Peter Waage in 1864. According to law of mass action:

Statement:

The rate or speed of a chemical reaction is directly proportional to the products of active masses of reacting substances raised to a power the number moles of the reacting substances from the balanced chemical equation.”

Active Masses:

Active masses mean molar concentrations i.e. number of moles per dma. Rate of Reaction:

Rate. Of reaction means change in concentrations of reacting substances per unit time.

Mathematical Relation:

Consider the following general chemical equation:

A + C D •

Then according to law of mass action;

r of[A][B]

r = k [A] [B

Where,

r = rate of reaction

[A] And [B] = molar concentrations of reacting substances while k = rate constant of the reaction

For a more general equation law of mass action can be written as: aA + bB + dD

R a [A]”[B]b

r =k [A]a [B]h

Where,

A and b = number of moles of reacting substances in the balanced chemical equation

6. At equilibrium a mixture of N2, H2 and NH3 gas at 500°C is determined to consist of 0.602 mol/dm3 of N2, 0.420 mol/dm3 of H2 and 0.113 mol/dm3 of NH3. What is equilibrium constant for the reaction at this temperature?

Ans.Given:

Concentration of N2 = 0.602 mol/dm3

Concentration of H2 = 0.420 mol/dm3

Concentration of NH3 = 0.113 mol/dm3

Required:

To calculate the equilibrium constant (Ks).

Solution:

The chemical equation for the reaction is:

N2 (g) +3H2 (g) 2NH3 (g)

The equilibrium constant expression for this reaction is:

K – [N1-1,]-

C

[I`12][H2]’

Putting the values in the above equation we get:

[0.113mol/dm3]’

ICS =

[0.602mol/dm3][0.420mo1 dm3]3

0.01277 mo. /2 dm`’r* Ks – 0.04461mo14/dm’2

0.286mo/2 / dm’ Kc –

M0/4 1 dm’

Mol’ .dm’

Fizo/4.dm-‘2

a> Ks = 0.286 mol2 .dm”

Kc = 0.286 mo1-2.dm6

Result:

Equilibrium Constant (Ks) = 0.286 rnol-2.dm6 State conditions necessary for chemical equilibrium.

Ans: For answer see Question no.8

Wrier–expression of Kc for the following reactions:

A) N2 (g) + 3H2 (g) 2NH3 (g)

B) 2H2 (g) 02 (g) 2H20 (g)

C) 4NH3 (g) + 502—‘\–4N0 (g) + 6H20 (g)

Ans.A) Chemical equation:

N2 (g) + 3H2 (g) 2NH3 (g)

Equilibrium constant expression:

K–

[

[N2][H,]

B) Chemical equation:

2H2 (g) + 02 (g) —\2H2o (g) Equilibrium constant expression:

K, – [H, 0]-

[H,]- [07]

C) Chemical equation:

4N1-13(g)-1- 502 4NO 6H20 (g

Equilibrium constant expression: Ks _ [VO]’favor

Lvfid, [0,] 5

(9.A reaction between gaseous sulphur dioxide and oxygen gas to produce gaseous sulphur tri-oxide takes place at 600°C. At this temperature the concentration of SO2 is found to be 1.50 mol/dm3, the concentration of 02 is 1.25 mol/dm3 and the concentration of SO3 is 3.50 mol/dm3.Using balance chemical equation calculates the equilibrium constant for this system?

Ans.Given:

Concentration of SO2 = 1.50 mol/dm3

Concentration of 02 = 1.25 mol/dm3

Concentration of SO3 = 3.50 mol/dm3

Required:

To calculate the equilibrium constant (Kc)

Solution:

The chemical equation for the reaction is: 2S02 (g) + 02 (g) 2S03 (g)

The equilibrium constant expression for this reaction is:

[S0,]-

KC – [s02]2[cm

Putting the values in the above equation we get: – [3.50 mol dm r

[1.50mol/dm’]'[1.25mol: dm’]

12.25 arts

Kc –

2.25_mo-1-; 11-n6 x 1.25mo/ (int’ 1″).-15

2.8125 mot .drn-3

Ke = 4.36 mori.dm3

Result:

Uilibrium Constant (Ks) = 4.36 mor1.dm3

10. Describe the effect of temperature on equilibrium constant? Ans. Effect of Temperature on Equilibrium Constant:

Temperature affects the equilibrium constant of a system. It affects the equilibrium constant value in the following two ways:

Effect on Exothermic Reaction:

If we increase the temperature of an exothermic reaction, the value of equilibrium constant will decrease.

Effect on Endothermic Reaction:

If we increase the temperature of an endothermic reaction, the value of equilibrium constant will increase.

Example:

Consider the following chemical equation:

H2 ei269) 2HI (g)

The Kc value for the above reaction at different temperatures is given below;

Temperature Kc

500K 160

700K 54.

As the forward reaction is exothermic and backward is endothermic therefore when temperature is increased then according to Le-Chatelier’s principle the backward reaction will proceed and more reactants will be formed thus the value of Kc is decreased.

word image 40

.1. S03 (g) decomposes to form SO,9) and 02(p). For this reaction write,

  1. Chemical equation
  2. Kc expression
  3. Derive the units of Kc for this reaction

Ans (a) Chemical Equation:

2S03 (g) (g) + 02 (g)

AnsMISc expression:

The equilibrium constant expression for this reaction is:

Kc – –

[S0,] 2[0.]

[S0, 12

Ans(c) Units of K:

As we know that the equilibrium constant expression is: So. Flo? 1 Sod2

Putting the values we get:

13.1al-cfni]-

AT= mol/dm3

Q.I (a).Describe the equilibrium state with the help of a graph and an example.

Ans For answers see Question no.2.

Define the law of mass action.

Ans For answer see Question no.4

(e).Derive an expression for the equilibrium constant and explain

See Question no.6

(b).How can you predict the direction of reaction for their<,-value?

Ans: or answer sees Question no.7

A). Kc has different units in different reactions. Prove it with suitable examples.

Ans: For answer see Question no.6

(b). How can you predict the extent of reaction from the Kc value? Ans. F r answer

See Question no.7

a) Expression for a reaction is given below,

[C/:][H, 0]2

[HCI]'[02]

For this reaction write,

i. Reactants and products

ii. Derive the units of Kc

Ans (i).Chemical Equation:

The chemical equation for this expression is: 2H20 + 2C12 4HC1 + 02

Reactants for this reaction are:

H2O and C12

Products:

Products for this reaction are:

HCI and 02

Ans (ii) Units of Kc: As

_ [C12]2[H2Or K, –

[110]4[02]

Putting the values in the above equation we get:

Kc – [mol/ dnq [mol/ [moll dm’]’ [mot/dm’]

[Mol/ dm3

K, —

[Mol/ dm r [mol/dm3]

IM0-11-1: Wf

Lmoli-cfm—Thol/dm3]

Rz> Kc

[Mol/dma]

=> Kc =

b. Explain the importance of equilibrium constant, support your answer with examples and reasons.

Ans: For answer see Question no.7

Additional Multiple Choice Questions

1. A reversible reaction proceeds in the:

(a) Reverse direction. (b) Forward direction.

(c) Forward and reverse direction.

(d) More backward less forward.

2. for the reaction A + B C +D.Kc are equal to:

[A]+ [13]. [A] dB]. [C][D]

(a) (B) [A1+ [D].

(c) (D)

[C]+ [D] [B1+[c] [CND] [11](B1

3. Molecules of chlorine do not decompose into atomic chlorine i.e.: 012 201 this is because Kc of this reaction is:

(a) Very large (b) very small- (c) zero (d) 1.

4. How much reaction is complete when Kc. =-1:

(a) 10% (b) 25% (c) 50% (d) 100%.

5. The escape of molecules from the surface of a liquid is called

(a) Evaporation (b) condensation.

(c) Sublimation (d) calcinations.

6. The conversion of a solid into vapours and vapours into solid without passing it through liquid state is called.

(a) Evaporation (b) condensation.

(c) Sublimation (d) calcinations

7. Reactions which proceed in both directions are called—-reactions.

(a) Reversible (b) irreversible

(c) Physical (d) both a&b.

8. Nitric acid (HNO3) is prepared through process.

(a) Solvay (b) Haber

(c) Birkland Eyed (d) down’s process.

Law of mass action was presented by in 1864.

(a) Goldberg & Waage (b) Boyle

(c) Charles (d) Dalton

10. Law of mass action is related to the of products & reactants.

(a)Types (b) concentration

(c) Alkalinity (d) acidity

11. A reversible reaction proceeds in the

(a) Forward direction (b) Reverse direction

(c) Both A & B (d) No change.

12 The unit of K0 in following system is PCI5 PCI3 + C12.

(a) Mo12/L2 (b) Umole

(c) mol/L2 (d) mol/L

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