Chapter 12 Chemical Equilibrium Solutions

Maharashtra State Board 11th Chemistry Solutions Chapter 12 Chemical Equilibrium

1. Choose the correct option

Question A.
The equilibrium, H₂O_(l)⇌ H⁺_(aq)+ OH^–_(aq)is
a. dynamic
b. static
c. physical
d. mechanical
Answer:
a. dynamic

Question B.
For the equilibrium, A ⇌ 2B + Heat, the number of ‘A’ molecules increases if
a. volume is increased
b. temperature is increased
c. catalyst is added
d. concentration of B is decreased
Answer:
b. temperature is increased

Question C.
For the equilibrium Cl_2(g)+ 2NO_(g)⇌ 2NOCl_(g)the concentration of NOCl will increase if the equilibrium is disturbed by ………..
a. adding Cl₂
b. removing NO
c. adding NOCl
d. removal of Cl₂
Answer:
a. adding Cl₂

Question D.
The relation between K_cand K_pfor the reaction A_(g)+ B_(g)⇌ 2C_(g)+ D_(g)is
a. K_c= K_p/RT
b. K_p= Kc²
c. K_c=
d. K_p/K_c= 1
Answer:
a. K_c= K_p/RT

Question E.
When volume of the equilibrium reaction C_(s)+ H₂O_(g)⇌ CO_(g)+ H_2(g)is increased at constant temperature the equilibrium will
a. shift from left to right
b. shift from right to left
c. be unaltered
d. can not be predicted
Answer:
a. shift from left to right

2. Answer the following

Question A.
State Law of Mass action.
Answer:
Law of mass action: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.

Question B.
Write an expression for equilibrium constant with respect to concerntration.
Answer:
For a reversible chemical reaction at equilibrium, aA + bB ⇌ cD + dD
Equilibrium constant (K_c) =

Question C.
Derive mathematically value of K_pfor A_(g)+ B_(g)⇌ C_(g)+ D_(g).
Answer:
When the concentrations of reactants and products in gaseous reactions are expressed in terms of their partial pressure, then the equilibrium constant is represented as K_p.
∴ For the reaction,
A_(g)+ B_(g)⇌ C_(g)+ D_(g)
the equilibrium constant (K_C) can be expressed using partial pressure as: K_p=
Where P_A, P_B, P_Cand P_Dare equilibrium partial pressures of A, B, C and D respectively.

Question D.
Write expressions of K_Cfor following chemical reactions
i. 2SO_2(g)+ O_2(g)⇌ 2SO_3(g)
ii. N₂O_4(g)⇌ 2NO_2(g)
Answer:
i. 2SO_2(g)+ O_2(g)⇌ 2SO_3(g)
K_c=

ii. N₂O_4(g)⇌ 2NO_2(g)
K_c=

Question E.
Mention various applications of equilibrium constant.
Answer:
Various applications of equilibrium constant:

Question F.
How does the change of pressure affect the value of equilibrium constant ?
Answer:
The change of pressure does not affect the value of equilibrium constant.

Question J.
Differentiate irreversible and reversible reaction.
Answer:
Irreversible reaction:

Reversible reaction:

Question K.
Write suitable conditions of concentration, temperature and pressure used during manufacture of ammonia by Haber process.
Answer:
i. Concentration: Addition of H₂or N₂both favours forward reaction. This increases the yield of NH₃.
ii. Temperature: The formation NH₃is exothermic. Hence, low temperature should favour the formation of NH₃. However, at low temperatures, the rate of reaction is small. At high temperatures, the reaction occurs rapidly but decomposition of NH₃occurs. Hence, optimum temperature of about 773 K is used.
iii. Pressure: The forward reaction is favoured with high pressure as it proceeds with decrease in number of moles. At high pressure, the catalyst becomes inefficient. Therefore, optimum pressure needs to be used. The optimum pressure is about 250 atm.

Question L.
Relate the terms reversible reactions and dynamic equilibrium.
Answer:

Thus, if the reaction is not reversible then it cannot attain dynamic equilibrium.

Question M.
For the equilibrium.

state the effect of
a. Addition of Ba²⁺ion.
b. Removal of SO₄^2-ion
c. Addition of BaSO_4(s)
on the equilibrium.
Answer:
a. Addition of Ba²⁺ion will favour the reverse reaction, (that is, equilibrium shifts from right to left). This increases the amount of BaSO₄.
b. Removal of ion will favour the forward reaction, (that is, equilibrium shifts from left to right). This decreases the amount of BaSO₄.
c. Addition of BaSO_4(s)will not affect the equilibrium as the equilibrium constant expression does not include pure solids.

3. Explain :

Question A.
Dynamic nature of chemical equilibrium with suitable example.
Answer:
Dynamic nature of chemical equilibrium:
i. Consider a chemical reaction: A ⇌ B.
K_c= [B]/[A]
At equilibrium, the ratio of concentration of the product to that of the concentration of the reactant is constant and this is equal to K_c.

ii. At this stage reaction takes place in both the directions with same speed although the reaction appears to have stopped. Thus, the chemical equilibrium is dynamic in nature. Dynamic means moving and at a microscopic level, the system is in motion.

iii. For example, in the reaction between H₂and I₂to form HI, the colour of the reaction mixture becomes constant because the concentrations of H₂, I₂and HI become constant at equilibrium.
H₂+ I₂⇌ 2HI
Thus, when equilibrium is reached, the reaction appears to have stopped. However, this is not the case. The reaction is still going on in the forward and backward direction but the rate of forward reaction is equal to the rate of backward reaction. Hence, chemical equilibrium is dynamic in nature and not static.

Question B.
Relation between K_cand K_p.
Answer:
Consider a general reversible reaction:
aA_(g)+ bB_(g)⇌ cC_(g)+ dD_(g)
The equilibrium constant (K_p) in terms of partial pressure is given by equation:
K_p= …………(1)
For a mixture of ideal gases, the partial pressure of each component is directly proportional to its concentration at constant temperature.
For component A,
P_AV = n_ART
P_A= × RT
is molar concentration of A in mol dm^-3V
∴ P_A= [A]RT where, [A] =
Similarly, for other components, P_B= [B]RT, P_C= [C]RT, P_D= [D]RT
Now substituting equations for P_A, P_B, P_C, P_Din equation (1), we get

where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation.
R = 0.08206 L atm K^-1mol^-1
[Note: While calculating the value of K_p, pressure should be expressed in bar, because standard state of pressure is 1 bar. 1 pascal (Pa) = 1 N m^-2and 1 bar = 10⁵Pa]

Question C.
State and explain Le Chatelier’s principle with reference to
1. change in temperature
2. change in concerntration.
Answer:
Statement: When a system at equilibrium is subjected to a change in any of the factors determining the equilibrium conditions of a system, system will respond in such a way as to minimize the effect of change.

1. Change in temperature:

2. Change in concentration:

Question D.
a. Reversible reaction
b. Rate of reaction
Answer:
a. Reversible reaction:
i. Reactions which do not go to completion and occur in both the directions simultaneously are called reversible reactions.
ii. Reversible reactions proceed in both directions. The direction from reactants to products is the forward reaction, whereas the opposite reaction from products to reactants is called the reverse or backward reaction.
iii. A reversible reaction is denoted by drawing in between the reactants and product a double arrow, one pointing in the forward direction and other in the reverse direction (⇌ or ⇄).
ii. At high temperature in an open container, the CO₂gas formed will escape away. Therefore, it is not possible to obtain back
e.g. a. H_2(g)+ I_2(g)⇌ 2HI_(g)
b. CH₃COOH_(aq)+ H₂O_(l)⇌ CH₃COO^–_(aq)+ H₃O⁺_(aq)

b. Rate of reaction:
Rate of a chemical reaction:
i. The rate of a chemical reaction can be determined by measuring the extent to which the concentration of a reactant decreases in the given time interval, or extent to which the concentration of a product increases in the given time interval.
ii. Mathematically, the rate of reaction is expressed as:
Rate =
where, d[reactant] and d[product] are the small decrease or increase in concentration during the small time interval dT.

Question E.
What is the effect of adding chloride on the position of the equilibrium ?
AgCl_(s)⇌ Ag⁺_(aq)+ Cl^–_(aq)
Answer:
Addition of Cl ion will favour the reverse reaction, (that is, equilibrium shift from right to left) This increases the amount of AgCl.

11th Chemistry Digest Chapter 12 Chemical Equilibrium Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Try this. (Textbook Page No. 174)

Can you tell? (Textbook Page No. 174)

(Textbook Page No. 174)

Internet my friend (Textbook Page No. 175)

The binding of oxygen to haemoglobin is a reversible reaction.
Hb + 4O₂⇌ Hb.4O₂
When the oxygen concentration is high (in the lungs), haemoglobin and oxygen combine to form oxyhaemoglobin and the reaction achieves equilibrium. But, when the oxygen concentration is low (in the body tissue), the reverse reaction occurs, that is, oxyhaemoglobin dissociates to haemoglobin and oxygen.
Thus, an equilibrium exists in the formation of oxyhaemoglobin in the human body.

ii. Refrigeration system in equilibrium:
a. Refrigeration system works on the principle of thermal equilibrium i.e., when a cold body comes in contact with a hot body then the heat flows from hot body to cold body until both the bodies attain the same temperature.
b. In the same way, a liquid (called as refrigerant) passes through the various compartments in the refrigerator and eventually lowers the temperature inside the refrigerator. This cycle is briefly described below:
Refrigerant flows through the compressor, which raises the pressure of the refrigerant. Next, the refrigerant flows through the condenser, where it condenses from vapor form to liquid form, giving off heat in the process. The heat given off is what j makes the condenser “hot to the touch.” After the condenser, the refrigerant goes through the expansion valve, where it experiences a pressure drop. Finally, the refrigerant goes to the evaporator. The refrigerant draws heat from the evaporator which causes the refrigerant to vaporize. The evaporator draws heat from the region that is to be cooled. The vaporized refrigerant goes back to the compressor to restart the cycle. In each of the heat transfer process, equilibrium is achieved (that is, heat given off is equivalent to the cooling achieved.)

[Note: Students are expected to collect additional information about equilibrium existing in the formation of oxyhaemoglobin in human body’ and ‘refrigeration system in equilibrium on their own.]

Try this. (Textbook Page No. 176)

Try this. (Textbook Page No. 176)

Do you know? (Textbook Page No. 177)

Observe and discuss. (Textbook Page No. 177)

[Note: For any reversible reaction in a closed system whenever the opposing reactions (forward and reaction) are occurring at different rates, the forward reaction will gradually become slower and the reverse reaction will become faster. Finally, the rates become equal and equilibrium is established.]

Discuss (Textbook Page No. 177)

i. Consider the following dissociation reaction:

The reaction is carried out in a closed vessel starting with hydrogen iodide.
ii. Now, let us start with hydrogen and iodine vapour in a closed container at a certain temperature.
H_2(g)+ I_2(g)⇌ 2HI_(g)
Answer:
i. Starting with hydrogen iodide:
Observations:
a. At first, there is an increase in the intensity of violet colour.
b. After certain time, the increase in the intensity of violet colour stops.
c. When contents in a closed vessel are analyzed at this stage, it is observed that reaction mixture contains the hydrogen iodide, hydrogen and iodine with their concentrations being constant over time.
Inference:
The rate of decomposition of HI becomes equal to the rate of combination of H₂and I₂. At equilibrium, no net change is observed and both reactions continue to occur at equal rates.
Thus, the reaction represents chemical equilibrium.

ii. Starting with hydrogen and iodine:
Observations:
a. At first, there is a decrease in the intensity of violet colour.
b. After certain time, the decrease in the intensity of violet colour stops.
c. When contents in a closed vessel are analyzed at this stage, it is observed that reaction mixture contains hydrogen, iodine and hydrogen iodide with their concentrations being constant over time.
Inference:
The rate of combination of H₂and I₂becomes equal to the rate of decomposition of HI. The reaction attains chemical equilibrium.

Can you recall? (Textbook Page No. 180)

Just think. (Textbook page no. 181)

Can you tell? (Textbook Page No. 183)

ii. For the reaction, K_c= 10¹⁸at 550 K
If the value of K_c>>> 10³, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.

Use your brain power (Textbook Page No. 183)

Internet my friend (Textbook Page No. 183)

Can you tell? (Textbook Page No. 188)

i. If NH₃is added to the equilibrium system (Haber process), in which direction will the equilibrium shift to consume added NH₃to reduce the effect of stress?
ii. In this process, out of the reactions (reverse and forward reaction), which reaction will occur to a greater extent?
iii. What will be the effect on yield of NH₃?
Answer:
i. If NH₃is added to the equilibrium system, the equilibrium will shift from right to left to consume added NH₃to reduce the effect of stress.
ii. If NH₃is added to the equilibrium system, then reverse reaction will occur to greater extent.
iii. If NH₃is added to the equilibrium system, the equilibrium will shift in reverse direction and the yield of NH₃will decrease.

Internet my friend (Textbook Page No. 188)

i. Collect information about Haber process in chemical equilibrium.
ii. Youtube.Freescienceslessons: The Haber process
Answer:
i.
[Note: Students can use the above link as reference and collect information about chemical equilibrium involved in Haber process.]
ii. Students are expected to refer ‘The Haber process ’ on YouTube channel ‘Freescienceslessons’