KPK G10 Chemistry Notes Chapter9 (Chemical Equilibrium)

This 10th class chemistry has been prepared according to the syllabus of all KPK G10 Boards. Apart from kpk textbook, 10 class chemistry notes are not followed in other boards. These are kpk Board, Peshawar Board, Mardan Board, swat Board, Abbottabad Board, di khan Board, and swabi Board.

Contents show

10th Class Chemistry, ch 9, Reversible Reaction & Dynamic

Lastly, we do our best to make these notes useful to you. But if you notice a mistake, any suggestions for further correction are invited. And if you know that our efforts help you, share it with your peers because “sharing is caring”.

G10 Chemistry Notes Chapter9
G10 Chemistry Notes Chapter9

KPK Class 10 Chemistry Notes Chapter 9 short questions

Q.1) Define chemical equilibrium with two examples.

Answer: 
Chemical equilibrium: A reversible reaction is in a chemical equilibrium state when the rate of forwarding reaction equals the rate of reverse reaction and the concentrations of its products and reactants remain constant.
Examples:
i. 2SO2(s) + O2(g) ⇌ 2SO3(g)
ii. H2(s) +  I2(g) ⇌ 2HI(g)

Q.2) How would you identify that dynamic equilibrium is established? 

Answer: 
The dynamic equilibrium is an equilibrium in which the two chemical processes continue at an equal rate in opposite direction. At the beginning of a reaction, the concentrations of reactants are at their maximum. That means that the rate of the forward reaction is very fast and the rate of reverse reaction is negligible. As the reactants react, their concentrations fall, thus the rates of the forward reaction falls as time goes on and reverse reaction speeds up. Eventually, the rates of the two reactions will become equal. At this point there won’t be any further change in the amounts of reactants and products in the mixture. That is when the equilibrium is established.
                                                         Rate of Forward Reaction = Rate of Reverse Reaction

Read more: List of Chemistry Class 9 Notes All Chapters 2021 | all boards

Q.3) Compare the different macroscopic characteristics of forward and reverse reactions?     

Answer: 
Comparison of the macroscopic characteristics of forward and reverse reactions:

Forward ReactionReverse Reaction
It is a reaction in which reactants react to form products. It is a reaction in which products react to form reaction .
It takes place from left to right. It takes place from right to left  
At initial stage, the rate of forward reaction is very fastIn the beginning, rate of reverse reaction is negligible. 
It slows down gradually. It speeds up gradually. 

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

Answer: 
We can determine the direction of reaction with help of equilubrum constant expression.
Kc = [products] / [Reactants]
The direction of chemical reaction at any particular time can be predicted by means of [products] / [reactants] ratio. the value of [products] / [reactants] ratio leads to one of the following three possibilities.
a. When ratio of [products] / [reactants] is less than Kc. The system is not at equilibrium and more products are required to reach the equilibrium. So, the reaction will move in the forward direaction.
b. When ratio of [products] / [reactants] is greater than Kc. The system is not at equilibrium and more reactants are required to reach equilibrium. The reaction will move in the reverse direction.
c. When ratio of [products] / [reactants] is equal to Kc. The system is at equilibrium. 

Matric part 2 Chemistry Chapter 9 Chemical Equilibrium

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

Answer: 
Relationship between active masses of the reactants and the rate of reaction has been described by law of mass action. The law states that the rate of reaction is directly proportional to the product of the active masses of the reacting substances. Active mass is usually considered as the molar concentrations expressed as square brackets. For example for the following reaction:
A + B ⇌ C + D
Rate of forward reaction = R(Forward) ∝ [A][B]                          (R(Forward) = Rf)
Rate of reverse reaction = R(Reverse) ∝ [C][D]                           (R(Reverse) = Rr)

Read more: 9 Class Islamiat Notes Chapters 1 to chapters 9 in English medium

Q.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 the equilibrium constant for the reaction at this temperature? 

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

Answer: 
Data:
The equilibrium concentrations are:
[N2] = 0.602 mol.dm-3
[H2] = 0.420 mol.dm-3
[NH3] = 0.113 mol.dm-3
Solution:
Reaction equation
  N2(g)     +   3H2(g)     ⇌     2NH3(g)
The equilibrium constant expression for this reaction is:
Kc = [NH3]2 / [N2][H2]3
Now put the equilibrium concentration values in the equilibrium expression:
Kc = [0.113]2 / [0.602][0.420]3
Kc = 0.286 mol-2dm-6

Q.7) State conditions necessary for chemical equilibrium. 

Answer: 
Following are the conditions necessary for chemical equilibrium:
1. Concentration of reactants and products
2. Temperature of the system
3. Pressure of the system
4. Volume of the system
5. Catalyst, if used in the system remains unchanged.

Q.8) Write equilibrium constant expression for the following reactions

i)           N2(g)         +    3H2(g)      ⇌     2NH3(g)   
 ii)           2H2(g)      +    O2(g)         ⇌     2H2O(g)   
 iii)           4NH3(g)  +    5O2(g)      ⇌     4NO(g)   +   6H2O(g)     

Answer: 
i) N2(g)         +    3H2(g)      ⇌     2NH3(g)
Rate of forward reaction
Rf = kf [N2] [H2]3
Rate of reverse reaction
Rr = kr [NH3]2
Equilibrium constant expression
Kc = [NH3]2 / [N2][H2]3
ii) 2H2(g)      +    O2(g)         ⇌     2H2O(g)  
Rate of forward reaction
Rf = kf [H2]2 [O2]
Rate of reverse reaction
Rr = kr [H2O]2
Equilibrium constant expression
Kc = [H2O]2 / [H2]2 [O2]
iii) 4NH3(g)  +    5O2(g)      ⇌     4NO(g)   +   6H2O(g)
Rate of forward reaction
Rf = kf [NH3]4 [O2]5
Rate of reverse reaction
Rr = kr [NO]4 [H2O]6
Equilibrium constant expression
Kc = [NO]4 [H2O]6 / [NH3]4 [O2]5

Chemical Equilibrium , Free Chemistry kpk Class 10

Q.9) A reaction between gaseous sulphur dioxide and oxygen gas to produce gaseous sulphur trioxide takes place at 600 °C. At this temperature, the concentration of SO2 is found to be 1.50 mol/dm3, the concentration of O2 is 1.25 mol/dm3 and the concentration of SO3 3.50 mol/dm3. Using the balanced chemical equation, calculate the equilibrium constant for this system. 

Answer: 
Data:
Concentration of SO2 = 1.50 mol. dm-3
Concentration of O2 = 1.25 mol. dm-3
Concentration of SO3 = 3.50 mol. dm-3
Equilibrium constant (Kc) = ?
Solution:
Balanced chemical equation,
         2SO2(g)     +      O2(g) ⇌ 2SO3(g)  
Kc Expression
Kc =  [SO3]2  / [SO2]2[O2]
Insert the concentration values along the units
Units of Kc = [3.50 mol. dm-3]2 / [1.50 mol. dm-3]2 x [1.25 mol. dm-3]
= 12.25 [mole.dm-3]2 / (2.25 x 1.25 ) [mol. dm-3]3
= 12.25 / 2.8125 [mol. dm-3]-1
Kc = 4.36 mol-1 dm3 

Q.10) Describe the effect of temperature on equilibrium constant. 

Answer: 
The value of equilibrium constant is dependent on the temperature of the system. At higher temperature, the particles have more kinetic energy, they move at a faster speed, and there will be more successful collisions resulting in the formation of more products. Whereas, the low temperature, in the reaction is slowed down and the equilibrium constant changes accordingly. 

Read more: Biology class 10 notes All chapters 10 to chapters 18

class 10 chemistry chapter 1 chemical equilibrium exercise answers

Long Questions

Q.1) SO3(g) decomposes to form SO2(g) and O2(g). For this reaction write, 

(i) Chemical equation (ii) Kc expression, and (iii) Derive the units of Kc for this reaction.

Answer: 
i) Chemical Equation
2SO3(g)         ⇌         2SO2(g)     +      O2(g)
ii) Kc Expression
Kc = [SO2]2[O2] / [SO3]2
iii) Derive the units of Kc
Kc = [SO2]2[O2] / [SO3]2
Units of Kc = [mole.dm-3]2 x [mole.dm-3] / [mole.dm-3]2
= [mole.dm-3]2 x [mole.dm-3]2  x [mole.dm-3]
Units of Kc = mole.dm3     

Q.2) (a) Describe the equilibrium state with the help of a graph and an example. 

ntration of Answer: 
When we think of the term equilibrium, the first word that usually comes to mind is “balance”. However, the balance may be achieved in a variety of ways. Thus, when the rate of the forward reaction is the same as the rate of the reverse reaction, the composition of the reaction mixture remains constant, it is called a chemical equilibrium state.
A(g) + B(g)  ⇌   C(g)   + D(g)
Initially, the concentration of reactants A and B is taken the same and is maximum and the concentration of products C and D is zero. As the forward reaction proceeds the concentration of reactants (A and B) decreases and the concentration of products (C and D) increases simultaneously. With the passage of time, the forward reaction slows down and reverse reaction speeds up. After some time, the rate of the forward reaction and reverse reaction becomes equal; at this state, the concentration of reactants and products becomes constant and attain equilibrium. Eventually, both reactions (forward and reverse) attain the same rate; it is called dynamic equilibrium state.
                         Rate of the forward reaction = Rate of the reverse reaction
It is represented graphically in the figure below;

chemical equilibrium class 10 notes

Graph showing the rate of forward and reverse reaction and equilibrium state

The examples of equilibrium state are:
N2(g) + 3H2(g)  ⇌   2NH3(g)
2HgO(g)          ⇌     2Hg(g) + O2(g)

Q.2) (b) Define the law of mass action. 

Answer: 
Law of mass action was put forward by Guldberg and Waage in 1869. It shows a relation between the concentration of reactants and products of a chemical reaction at equilibrium. This law states that the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances, with each mass raised to a power equal to their number of moles in the chemical equation. An active mass is usually considered as the molar concentration with units of mole.dm-3, expressed in square brackets.
For example, consider a general reversible reaction:
aA + bB ⇌ cC + dD
According to the law of mass action:
Rate of forward reaction ∝ [A]a[B]b
Rf = kf [A]a[B]b
kf is the proportionality constant for the forward reaction.
Rate of reverse reaction ∝ [C]c[D]d
Rr = kr [C]c[D]d
kr is the proportionality constant for the reverse reaction.

Read more: Physics Class 10 Notes Chapters 10 to chapters 18

Q.3) (a) Derive an expression for the equilibrium constant and explain its units. 

Answer: 
The equilibrium constant is the ratio of the product of the concentration of the products to the product of the concentration of the reactants at equilibrium state.
The numerical value of Kc for a particular equilibrium system is obtained experimentally. The chemists examine the equilibrium mixture and determine the concentration of all substances. When we know the balanced chemical equation of a reaction we can write the equilibrium constant (Kc). While writing the equilibrium constant, the products side is written as the numerator and the reactants as the denominator. Thus we can calculate the numerical value and unit of Kc, by putting the equilibrium values of the products side at the numerator and the reactants side at the denominator and the coefficient is raised to the power of that substance in the chemical equation. 

kpk 2

The unit of Kc depends on the equilibrium constant expression for the given reaction.
i. Kc has no units for a reaction in which the number of moles of reactants and products are equal. This is because concentration units cancel out in the expression of Kc, for example, in the given reaction we have,
H2(g) + I2(g)  ⇌   2HI(g)
Kc = [HI]2 / [H2][I2]
Units of Kc = [mole.dm-3]2 / [mole.dm-3][mole.dm-3] = No Units
ii. Kc has units for a reaction in which the number of moles of products is greater than reactants in a balanced chemical equation. For example, in the given reaction we have,
N2O4(g)   ⇌  2NO2(g)    
Kc = [NO2]2 / [N2O4]
Units of Kc = [mole.dm-3]2 / [mole.dm-3] = mole.dm-3
iii. Similarly, Kc has units for a reaction in which the number of moles of products is less than the reactants in a balanced chemical equation. For example, in the given reaction we have,
N2(g) + 3H2(g)  ⇌   2NH3(g)
Units of Kc = [mole.dm-3]2 /  [mole.dm-3] [mole.dm-3]3 = 1 / [mole.dm-3]2

Q.3) (b) How can you predict the direction of reaction for the Kc value?

Answer: 
We can determine the direction of reaction with the help of equilibrium constant expression.
Kc = [products] / [Reactants]
The direction of chemical reaction at any particular time can be predicted by means of [products] / [reactants] ratio. the value of [products] / [reactants] ratio leads to one of the following three possibilities.
a. When ratio of [products] / [reactants] is less than Kc. The system is not at equilibrium and more products are required to reach the equilibrium. So, the reaction will move in the forward direction.
b. When ratio of [products] / [reactants] is greater than Kc. The system is not at equilibrium and more reactants are required to reach equilibrium. The reaction will move in the reverse direction.
c. When ratio of [products] / [reactants]is equal to Kc. The system is at equilibrium. 

Read more: Physics numericals for class 10 chapter 18 pdf

Q.4) (a) Kc has different units in different reactions. Prove it with suitable examples. 

Answer: 
The unit of Kc depends on the equilibrium constant expression for the given reaction.
i. Kc has no units for a reaction in which the number of moles of reactants and products are equal. This is because concentration units cancel out in the expression of Kc, for example, in the given reaction we have,
H2(g) + I2(g)  ⇌   2HI(g)
Kc = [HI]2 / [H2][I2]
Units of Kc = [mole.dm-3]2 / [mole.dm-3][mole.dm-3] = No Units
ii. Kc has units for a reaction in which the number of moles of products is greater than reactants in a balanced chemical equation. For example, in the given reaction we have,
N2O4(g)   ⇌  2NO2(g)    
Kc = [NO2]2 / [N2O4]
Units of Kc = [mole.dm-3]2 / [mole.dm-3] = mole.dm-3
iii. Similarly, Kc has units for a reaction in which the number of moles of products is less than the reactants in a balanced chemical equation. For example, in the given reaction we have,
N2(g) + 3H2(g)  ⇌   2NH3(g)
Units of Kc = [mole.dm-3]2 /  [mole.dm-3] [mole.dm-3]3 = 1 / [mole.dm-3]2

Q.4) (b) How can you predict the extent of reaction from the Kc value.  

Answer: 
The extent to which a chemical reaction may proceed can be determined with the help of equilibrium constant expression.
Consider the general reaction,
                              aA + bB ⇌ cC + dD
Kc  = [C]c[D]d / [A]a[B]b
The extent of reaction depends upon the magnitude of ‘Kc‘ so when,
i. Kc is very small:
When the concentration of [A] and [B] is large and that of [C] and [D] is small, the equilibrium mixture will contain the reactants mainly and only a small amount of products will be present. It reflects that the reaction does not proceed appreciably in the forward direction.
ii. Kc is very large:
When the concentration of [A] and [B] is small and that of [C] and [D] is large, the equilibrium mixture will consist almost entirely of products and only a small amount of the reactants will be present. This indicates that the reaction is completed in the forward direction.
iii. Kc is neither very large nor very small:
The concentration of the products and reactants will be very close and hence, Kc is close to 1.0. Thus, the equilibrium mixture will contain an appreciable amount of products and reactants. 

Q.5) (a) Kc expression for a reaction is given below, 

Kc = [Cl2]2[H2O]2 / [HCl]4[O2]
For this reaction write,
i. Reactants and products                                 ii. Derive the units of Kc 

Answer: 
i. Reactants and products
For Kc the products side is written as the numerator and the reactants as the denominator
Kc = [Cl2]2[H2O]2 / [HCl]4[O2]
Products = Chlorine (Cl2) and Water (H2O)
Reactants = Hydrochloric acid (HCl) and Oxygen (O2)
ii. Derive the units of Kc 
Units of Kc = [mole.dm-3]2 [mole.dm-3]2 /  [mole.dm-3]4 [mole.dm-3]
= [mole.dm-3]4  /  [mole.dm-3]5
Units of Kc = 1 / [mole.dm-3]

Q.5) (b) Explain the importance of equilibrium constant, support your answer with example and reasons. 

Answer: 
The value of equilibrium constant is specific and remains constant at a particular temperature. The value of Kc helps us to predict.
1. Direction of Reaction:
We can determine the direction of reaction with the help of equilibrium constant expression.
Kc = [products] / [Reactants]
The direction of chemical reaction at any particular time can be predicted by means of [products] / [reactants] ratio. the value of [products] / [reactants] ratio leads to one of the following three possibilities.
a. When ratio of [products] / [reactants] is less than Kc. The system is not at equilibrium and more products are required to reach the equilibrium. So, the reaction will move in the forward direction.
b. When ratio of [products] / [reactants] is greater than Kc. The system is not at equilibrium and more reactants are required to reach equilibrium. The reaction will move in the reverse direction.
c. When ratio of [products] / [reactants]is equal to Kc. The system is at equilibrium.
2. The Extent of a Chemical Reaction:
The extent to which a chemical reaction may proceed can be determined with the help of equilibrium constant expression.
Consider the general reaction,
aA + bB ⇌ cC + dD
Kc  = [C]c[D]d / [A]a[B]b
The extent of reaction depends upon the magnitude of ‘Kc‘ so when,
i. Kc is very small:
When the concentration of [A] and [B] is large and that of [C] and [D] is small, the equilibrium mixture will contain the reactants mainly and only a small amount of products will be present. It reflects that the reaction does not proceed appreciably in the forward direction.
ii. Kc is very large:
When the concentration of [A] and [B] is small and that of [C] and [D] is large, the equilibrium mixture will consist almost entirely of products and only a small amount of the reactants will be present. This indicates that the reaction is completed in the forward direction.
iii. Kc is neither very large nor very small:
The concentration of the products and reactants will be very close and hence, Kc is close to 1.0. Thus, the equilibrium mixture will contain an appreciable amount of products and reactants.
3. The Effect of External Conditions on the Position of Equilibrium
When a system reaches equilibrium it will remain in the same state indefinitely, if the conditions do not change. However, the equilibrium state of a system is disturbed if external conditions are changed, e.g. change of pressure, temperature and concentrations of reactants and products alter the position of the equilibrium. Whenever, the equilibrium is disturbed by changes in the external conditions, the system always tends to restore equilibrium.

Leave a Reply