Maharashtra Board Class 11 Chemistry Chapter 12 Chemical Equilibrium Solutions

Chemical Equilibrium Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 12 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 12 Exercise Solutions

1. Choose The Correct Option

Question A. The equlilibrium, \( H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)} \) is
(a) dynamic
(b) static
(c) physical
(d) mechanical
Answer: (a) dynamic
In simple words: The equilibrium state of water autoionization is dynamic, meaning both forward and reverse reactions continue at equal rates, not stopping.

🎯 Exam Tip: Understanding the dynamic nature of chemical equilibrium is crucial for solving problems related to reaction rates and concentration changes at equilibrium.

Question B. For the equlibrium, \( A \rightleftharpoons 2B + \text{Heat} \), the number of 'A' molecules increases if
(a) volume is increased
(b) temperature is increased
(c) catalyst is added
(d) concerntration of B is decreased
Answer: (b) temperature is increased
In simple words: Increasing temperature for an exothermic reaction (like A ⇌ 2B + Heat) shifts the equilibrium to the left, favoring the reactants and increasing the amount of 'A'.

🎯 Exam Tip: Le Chatelier's principle is key here; for exothermic reactions, heat acts as a product, so adding heat shifts the equilibrium away from the products.

Question C. For the equilibrium \( Cl_{2(g)} + 2NO_{(g)} \rightleftharpoons 2NOCI_{(g)} \) the concerntration of NOCI will increase if the equlibrium is disturbed by .........
(a) adding Cl2
(b) removing NO
(c) adding NOCI
(d) removal of Cl2
Answer: (a) adding Cl2
In simple words: According to Le Chatelier's principle, adding a reactant like Cl2 will shift the equilibrium to the right, favoring product formation and increasing the concentration of NOCl.

🎯 Exam Tip: Remember that adding reactants pushes the equilibrium towards products, while removing products also favors forward reaction, increasing product yield.

Question D. The relation between Kc and Kp for the reaction \( A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)} + D_{(g)} \) is
(a) \( K_c = K_p/RT \)
(b) \( K_p = K_c^2 \)
(c) \( K_c = \frac{1}{\sqrt{K_p}} \)
(d) \( K_p/K_c = 1 \)
Answer: (a) \( K_c = K_p/RT \)
In simple words: For the given reaction, \(\Delta n = (2+1) - (1+1) = 3 - 2 = 1\). The relationship between Kp and Kc is \( K_p = K_c (RT)^{\Delta n} \), so \( K_p = K_c (RT)^1 \), which means \( K_c = K_p/RT \).

🎯 Exam Tip: Always correctly calculate \(\Delta n\) (moles of gaseous products - moles of gaseous reactants) to establish the relationship between Kp and Kc accurately.

Question E. When volume of the equilibrium reaction \( C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_2_{(g)} \) 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
In simple words: Increasing the volume of the container for a gaseous equilibrium shifts the reaction towards the side with more moles of gas to relieve the pressure, in this case, the product side (2 moles of gas vs 1 mole of gas).

🎯 Exam Tip: When volume increases (pressure decreases), the equilibrium shifts to the side with a greater number of gaseous moles. Solids do not contribute to pressure changes.

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.
In simple words: The speed of a chemical reaction is directly related to how much of the reacting substances are present.

🎯 Exam Tip: Clearly define the proportionality between reaction rate and reactant concentrations. Mentioning "at each instant" emphasizes the dynamic nature.

Question B. Write an expression for equilibrium constant with respect to concerntration.
Answer: For a reversible chemical reaction at equilibrium, \( aA + bB \rightleftharpoons cC + dD \)
Equilibrium constant (\( K_c \)) = \( \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
In simple words: The equilibrium constant Kc is a ratio comparing the product of product concentrations (each raised to their stoichiometric power) to the product of reactant concentrations (each raised to their stoichiometric power) at equilibrium.

🎯 Exam Tip: Ensure the exponents match the stoichiometric coefficients in the balanced chemical equation, and products are in the numerator while reactants are in the denominator.

Question C. Derive mathematically value of Kp for \( A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)} \).
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 Kp.
.. For the reaction,
\( A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)} \)
the equilibrium constant (\( K_c \)) can be expressed using partial pressure as: \( K_p = \frac{P_C \times P_D}{P_A \times P_B} \)
Where \( P_A \), \( P_B \), \( P_C \) and \( P_D \) are equilibrium partial pressures of A, B, C and D respectively.
In simple words: Kp is the equilibrium constant expressed using partial pressures of gaseous reactants and products, calculated as the ratio of product partial pressures to reactant partial pressures, each raised to their stoichiometric coefficients.

🎯 Exam Tip: Always specify that Kp applies to gaseous reactions and involves partial pressures, not concentrations. Ensure correct stoichiometric exponents for each partial pressure term.

Question D. Write expressions of Kc for following chemical reactions
(i) \( 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \)
(ii) \( N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)} \)
Answer:
(i) \( 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \)
\( K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} \)

(ii) \( N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)} \)
\( K_c = \frac{[NO_2]^2}{[N_2O_4]} \)
In simple words: Kc expressions are ratios where the concentrations of products (raised to their stoichiometric coefficients) are divided by the concentrations of reactants (raised to their stoichiometric coefficients) at equilibrium.

🎯 Exam Tip: Double-check the stoichiometric coefficients from the balanced equation to ensure they are correctly used as exponents in the Kc expression. Only gaseous and aqueous species are included.

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

• Prediction of the direction of the reaction
• To know the extent of the reaction
• To calculate equilibrium concentrations
• Link between chemical equilibrium and chemical kinetics

🎯 Exam Tip: Focus on the practical utility of Kc in understanding and manipulating chemical reactions, especially its predictive power regarding reaction direction and yield.

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.
In simple words: The numerical value of the equilibrium constant (Kc or Kp) remains unchanged by pressure variations, though the position of equilibrium might shift according to Le Chatelier's principle.

🎯 Exam Tip: Distinguish between the "value of equilibrium constant" (which is temperature-dependent only) and the "position of equilibrium" (which can be affected by pressure, concentration, etc.).

Question J. Differentiate irreversible and reversible reaction.
Answer: Irreversible reaction:
1. Products are not converted back to reactants.
2. Reaction stops completely and almost goes to completion.
3. It can be carried out in an open or closed vessel.
4. It takes place only in one direction. It is represented by \( \rightarrow \)
5. e.g. \( C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \)
Reversible reaction:
1. Products arc converted back to reactants.
2. Reaction appears to have stopped but does not undergo completion.
3. It is generally carried out in a closed vessel.
4. It takes place in both directions. It is represented by \( \rightleftharpoons \)
5. e.g. \( N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \)
In simple words: Irreversible reactions proceed in one direction until reactants are used up, while reversible reactions proceed in both directions, reaching an equilibrium where reactants and products coexist.

🎯 Exam Tip: Key differences include completion (irreversible) vs. equilibrium (reversible), direction of reaction (one way vs. two ways), and the type of container usually required (open for irreversible, closed for reversible).

Question K. Write suitable conditions of concentration, temperature and pressure used during manufacture of ammonia by Haber process.
Answer:
(i) Concentration: Addition of \( H_2 \) or \( N_2 \) both favours forward reaction. This increases the yield of \( NH_3 \).
(ii) Temperature: The formation \( NH_3 \) is exothermic. Hence, low temperature should favour the formation of \( NH_3 \). However, at low temperatures, the rate of reaction is small. At high temperatures, the reaction occurs rapidly but decomposition of \( NH_3 \) occurs. 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.
In simple words: For Haber process ammonia production, high reactant concentrations, an optimal intermediate temperature (around 773 K), and high pressure (around 250 atm) are used to maximize yield while maintaining a reasonable reaction rate.

🎯 Exam Tip: Relate each condition (concentration, temperature, pressure) back to Le Chatelier's principle and the kinetics of the reaction to justify the "optimum" conditions used in industrial processes.

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

• Reversible reactions are the reactions which do not go to completion and occur in both the directions simultaneously.
• If such a reaction is allowed to take place for a long time, so that the concentrations of the reactants and products do not vary with time, then the reaction will attain equilibrium.
• Since, both the forward and backward reactions continue to take place in opposite directions in the same speed, the equilibrium achieved is dynamic in nature.

🎯 Exam Tip: Emphasize that dynamic equilibrium occurs only in reversible reactions and signifies continuous activity, not a static halt, where rates of opposing processes are equal.

Question M. For the equilibrium. \( BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO_{4(aq)}^{2-} \) state the effect of
(a) Addition of \( Ba^{2+} \) ion.
(b) Removal of \( SO_4^{2-} \) ion
(c) Addition of \( BaSO_{4(s)} \)
on the equilibrium.
Answer:
(a) Addition of \( Ba^{2+} \) ion will favour the reverse reaction, (that is, equilibrium shifts from right to left). This increases the amount of \( BaSO_4 \).
(b) Removal of \( SO_4^{2-} \) ion will favour the forward reaction, (that is, equilibrium shifts from left to right). This decreases the amount of \( BaSO_4 \).
(c) Addition of \( BaSO_{4(s)} \) will not affect the equilibrium as the equilibrium constant expression does not include pure solids.
In simple words: Adding \( Ba^{2+} \) shifts equilibrium left to form more solid \( BaSO_4 \); removing \( SO_4^{2-} \) shifts equilibrium right to dissolve more \( BaSO_4 \); adding more solid \( BaSO_4 \) has no effect on equilibrium since its concentration is constant.

🎯 Exam Tip: Remember that changes in concentration of pure solids or liquids do not shift the position of a heterogeneous equilibrium because their concentrations are considered constant and are not included in the equilibrium constant expression.

3. Explain :

Question A. Dynamic nature of chemical equilibrium with suitable example.
Answer: Dynamic nature of chemical equilibrium:
(i) Consider a chemical reaction: \( A \rightleftharpoons B \).
\( K_c = \frac{[B]}{[A]} \)
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 \( H_2 \) and \( I_2 \) to form HI, the colour of the reaction mixture becomes constant because the concentrations of \( H_2 \), \( I_2 \) and HI become constant at equilibrium.
\( H_2 + I_2 \rightleftharpoons 2HI \)
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.
In simple words: Dynamic equilibrium means that at the macroscopic level, a reaction appears to have stopped, but at the microscopic level, both forward and reverse reactions are continuously occurring at equal rates, like in the formation of HI from H2 and I2 where concentrations become constant.

🎯 Exam Tip: To illustrate dynamic equilibrium effectively, highlight that it's a state of constant activity where forward and reverse reaction rates are precisely balanced, not a cessation of chemical change.

Question B. Relation between Kc and Kp.
Answer: Consider a general reversible reaction:
\( aA_{(g)} + bB_{(g)} \rightleftharpoons cC_{(g)} + dD_{(g)} \)
The equilibrium constant (Kp) in terms of partial pressure is given by equation:
\( K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \) ......(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,
\( P_A V = n_A R T \)
\( P_A = \frac{n_A}{V} \times RT \)
\( \frac{n_A}{V} \) is molar concentration of A in mol dm\(^{-3}\)
\( \therefore P_A = [A]RT \) where, \( [A] = \frac{n_A}{V} \)
Similarly, for other components, \( P_B = [B]RT \), \( P_C = [C]RT \), \( P_D = [D]RT \)
Now substituting equations for \( P_A \), \( P_B \), \( P_C \), \( P_D \) in equation (1), we get
\( K_p = \frac{[C]^c(RT)^c[D]^d(RT)^d}{[A]^a(RT)^a[B]^b(RT)^b} \)

\( K_p = \frac{[C]^c[D]^d (RT)^{c+d}}{[A]^a[B]^b (RT)^{a+b}} \)

\( K_p = \frac{[C]^c[D]^d}{[A]^a[B]^b} \times (RT)^{(c+d)-(a+b)} \)

\( K_p = \frac{[C]^c[D]^d}{[A]^a[B]^b} \times (RT)^{\Delta n} \)
But, \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

\( \therefore K_p = K_c (RT)^{\Delta n} \)
where \( \Delta n \) = (number of moles of gaseous products) - (number of moles of gaseous reactants) in the balanced chemical equation.
R = 0.08206 L atm \( K^{-1} mol^{-1} \)
[Note: While calculating the value of Kp, pressure should be expressed in bar, because standard state of pressure is 1 bar. \( 1 \text{ pascal (Pa)} = 1 \text{ N m}^{-2} \) and \( 1 \text{ bar} = 10^5 \text{ Pa} \)]
In simple words: The relationship between Kp and Kc, \( K_p = K_c (RT)^{\Delta n} \), links the equilibrium constants based on partial pressures and concentrations, where \( \Delta n \) represents the change in the number of moles of gaseous species in the balanced reaction.

🎯 Exam Tip: Ensure precise calculation of \( \Delta n \) and correct application of the ideal gas law to convert concentrations to partial pressures (or vice-versa) for this derivation.

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:

• Consider the equilibrium reaction,
  \( PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)} + 92.5 \text{ kJ} \)
• The forward reaction is exothermic. According to Le Chatelier's principle an increase in temperature shifts the position of equilibrium to the left.
• The reverse reaction is endothermic. An endothermic reaction consumes heat. Therefore, the equilibrium must shift in the reverse direction to use up the added heat (heat energy converted to chemical energy).
• Thus, an increase in temperature favours formation of \( PCl_5 \) while a decrease in temperature favours decomposition of \( PCl_5 \).

• Consider reversible reaction representing production of ammonia (\( NH_3 \)).
  \( N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + \text{Heat} \)
• According to Le Chatelier's principle, when \( H_2 \) or \( N_2 \) is added to equilibrium, the effect of addition of \( H_2 \) or \( N_2 \) or is reduced by shifting the equilibrium from left to right so that the added \( N_2 \) or \( H_2 \) is consumed.
• The forward reaction occurs to a large extent than the reverse reaction until the new equilibrium is established. As a result, the yield of \( NH_3 \) is increases.
• In general, if the concentration of one of the species in equilibrium mixture is increased, the position of equilibrium shifts in the opposite so as to reduce the concentration of this species. However, the equilibrium constant remains unchanged.

🎯 Exam Tip: When explaining Le Chatelier's principle, always explicitly state the disturbance, the system's response to minimize that disturbance, and the resulting shift in equilibrium (left or right, favoring reactants or products).

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 (\( \rightleftharpoons \) or 2).
(ii) At high temperature in an open container, the \( CO_2 \) gas formed will escape away. Therefore, it is not possible to obtain back
e.g. (a) \( H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)} \)
(b) \( CH_3COOH_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3COO^-_{(aq)} + H_3O^+_{(aq)} \)

(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 = \( -\frac{d[Reactant]}{dT} \) = \( \frac{d[Product]}{dT} \)
where, d[reactant] and d[product] are the small decrease or increase in concentration during the small time interval dT.
In simple words: A reversible reaction proceeds in both directions to reach equilibrium, denoted by a double arrow, while the rate of reaction quantifies how quickly reactants are consumed or products are formed over time.

🎯 Exam Tip: When defining reversible reactions, emphasize the simultaneous forward and reverse processes and the use of the equilibrium symbol. For reaction rates, highlight the change in concentration over time and the negative sign for reactant consumption.

Question E. What is the effect of adding chloride on the position of the equilibrium ?
\( AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)} \)
Answer: Addition of \( Cl^- \) ion will favour the reverse reaction, (that is, equilibrium shift from right to left) This increases the amount of AgCl.
In simple words: Adding chloride ions to the equilibrium will shift the reaction to the left, favoring the formation of more solid AgCl, due to the common ion effect.

🎯 Exam Tip: This illustrates the common ion effect, a specific application of Le Chatelier's principle. Adding a product ion (like \( Cl^- \)) shifts the equilibrium towards the reactant side (solid AgCl) to reduce the disturbance.

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

Can You Recall? (Textbook Page No. 174)

Question 1. What are the types of the following changes? Natural waterfall, spreading of smoke from burning incense stick, diffusion of fragrance of flowers.
Answer: Natural waterfall, spreading of smoke from burning incense stick and diffusion of fragrance of flowers are irreversible physical changes.
In simple words: These examples represent irreversible physical changes because they naturally proceed in one direction and cannot easily be reversed to their initial state.

🎯 Exam Tip: When classifying changes, distinguish between physical and chemical, and then between reversible and irreversible. These examples are physical changes that naturally tend towards increased entropy and are not easily undone.

Try This. (Textbook Page No. 174)

Question 1. Dissolve 4 g cobalt chloride in 40 mL water. It forms a reddish pink solution. Add 60 mL concentrated HCl to this. It will turn blue. Take 5 mL of this solution in a test tube and place it in a beaker containing ice water mixture. The colour of solution will become pink. Place the same test tube in a beaker containing water at 90 °C. The colour of the solution turns blue.
Answer: Inference: The colour change of the solution from pink to blue is caused by the chemical reaction. On changing the temperature, the direction of the reaction reverses. This indicates that the chemical reaction is reversible. This activity is an example of a reversible chemical reaction.
In simple words: This experiment demonstrates a reversible chemical reaction where changes in temperature or reactant concentration (adding HCl) shift the equilibrium, causing a visible color change between pink and blue.

🎯 Exam Tip: For practical examples, clearly connect the observable changes (color shifts) to the underlying chemical processes (equilibrium shifts due to temperature or concentration changes) and identify the reaction as reversible.

The reaction can be written as:
\( Co(H_2O)_6^{2+} + 4Cl^-_{(aq)} \xrightleftharpoons[\text{Cool}]{\text{Heat}} CoCl_4^{2-}_{(aq)} + 6H_2O_{(l)} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो परखनली दर्शाता है जो हेक्साहाइड्रोकोबाल्ट(II) आयन (गुलाबी) और टेट्राक्लोरोकोबाल्टेट(II) आयन (नीला) के बीच संतुलन को दर्शाते हैं। एक परखनली ठंडे पानी में रखी है, जो गुलाबी हेक्साहाइड्रोकोबाल्ट(II) आयन का समर्थन करती है, जबकि दूसरी गर्म पानी में है, जो नीले टेट्राक्लोरोकोबाल्टेट(II) आयन का समर्थन करती है, जिससे तापमान के संतुलन पर प्रभाव स्पष्ट होता है।
In simple words: The diagram illustrates a reversible reaction between pink hydrated cobalt ions and blue tetrachlorocobaltate ions, where temperature dictates the equilibrium position, showing more pink in cold water and more blue in hot water.

🎯 Exam Tip: Visualizing equilibrium shifts with color changes helps cement understanding of Le Chatelier's principle. Be prepared to explain how specific disturbances (like temperature) cause the observed color changes.

Can You Tell? (Textbook Page No. 174)

Question 1. What does violet colour of the solution in the activity mentioned in Q.2 indicate?
Answer: In the reaction, the reactant \( Co(H_2O)_6^{2+} \) is pink in colour and the product \( CoCl_4^{2-} \) is blue in colour. When the solution contains both the reactant and product, the resulting solution will appear violet. This indicates that the reaction has attained equilibrium (that is, the reaction proceeds in both the direction with equal rates and is a reversible reaction).
In simple words: The violet color signifies that the system has reached a dynamic equilibrium where both the pink reactant and blue product species are present in constant concentrations, and the forward and reverse reactions are occurring at equal rates.

🎯 Exam Tip: When explaining color changes in equilibrium systems, clearly link the observed mixed color to the simultaneous presence of both reactant and product species at a steady state.

(Textbook Page No. 174)

Question 1. Calcium earbonate when heated strongly, decomposes to form calcium oxide and carbon dioxide.
(i) If this reaction is carried out in a closed container, what will we observe?
(ii) Consider this reaction occurring in an open system or container, what will happen? Can we obtain back calcium carbonate?
Answer: At high temperature in a closed container, we will find that after certain time, some calcium carbonate is present. If we continue the experiment over a longer period of time at the same temperature, the concentrations of calcium carbonate, calcium
oxide and carbon dioxide remain unchanged. The reaction thus appears to have stopped and the system has attained the equilibrium. Actually, the reaction does not stop but proceeds in both the directions with equal rates. In other words, calcium carbonate decomposes to give calcium oxide and carbon dioxide at a particular rate. Exactly at the same rate the calcium oxide and carbon dioxide recombine and form calcium carbonate. Thus, in closed container, reversible reaction occurs.
\( CaCO_{3(s)} \xrightleftharpoons[\text{Backward}]{\text{Forward}} CaO_{(s)} + CO_{2(g)} \)
(ii) At high temperature in an open container, the \( CO_2 \) gas formed will escape away. Therefore, it is not possible to obtain back calcium carbonate. Thus, in an open container, irreversible reaction occurs.
\( CaCO_{3(s)} \xrightarrow{\text{Heat}} CaO_{(s)} + CO_{2(g)} \)
In simple words: In a closed container, calcium carbonate decomposition reaches equilibrium as CO2 cannot escape, making it a reversible reaction. In an open container, CO2 escapes, preventing the reverse reaction and making it an irreversible decomposition.

🎯 Exam Tip: Differentiate between open and closed systems for gaseous reactions. A closed system allows equilibrium to be established, while an open system, where a gaseous product can escape, leads to an irreversible reaction.

Internet My Friend (Textbook Page No. 175)

Question 1.
(i) Equilibrium existing in the formation of oxyhaemoglobin in human body
(ii) Refrigeration system in equilibrium
Answer:
(i) Equilibrium existing in the formation of oxyhaemoglobin in human body:
Oxygen is transported in the body with the assistance of red blood cells. The red blood cells contain a pigment called haemoglobin. Each haemoglobin molecule binds four oxygen molecules to form oxyhaemoglobin. Thus, the oxygen molecules are carried to individual cells in the body tissue where they are released.

The binding of oxygen to haemoglobin is a reversible reaction.
\( Hb + 4O_2 \rightleftharpoons Hb \cdot 4O_2 \)
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.)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक प्रशीतन चक्र को दर्शाता है। इसमें कंप्रेसर, कंडेंसर, विस्तार वाल्व और बाष्पीकरणकर्ता के माध्यम से रेफ्रिजरेंट का प्रवाह दिखाया गया है। कंडेंसर में द्रवीकरण के दौरान गर्मी निकलती है और बाष्पीकरणकर्ता द्वारा अवशोषित की जाती है, जो यह दर्शाता है कि शीतलन के लिए ऊष्मा हस्तांतरण प्रक्रियाएँ कैसे संतुलन प्राप्त करती हैं।
In simple words: Chemical equilibrium governs oxygen binding to hemoglobin, releasing it where needed, while refrigeration relies on thermal equilibrium, moving heat from a cold to a hot reservoir through a cyclical process involving phase changes of a refrigerant.

🎯 Exam Tip: For biological examples like hemoglobin, emphasize how varying concentrations (of oxygen) drive equilibrium shifts. For refrigeration, explain how repeated heat transfer and phase changes establish a continuous cooling equilibrium.

Try This. (Textbook Page No. 176)

Question 1.
(i) Place some iodine crystals in a closed vessel. Observe the change in colour
intensity in it.
(ii) What do you see in the flask after some time?
Answer:
(i) The vessel gets slowly filled up with violet coloured vapour of iodine. After a certain time, the intensity of violet colour becomes stable.
(ii) After sometime, both solid iodine and iodine vapour are present in the closed vessel. Iodine crystals will be seen deposited near the mouth of the flask and violet coloured vapour will be filled in the entire flask. It means solid iodine sublimes to give iodine vapour and the iodine vapour condenses to form solid iodine. The stable intensity of the colour indicates a state of equilibrium between solid and vapour iodine.
\( I_{2(s)} \xrightleftharpoons[\text{condensation}]{\text{sublimation}} I_{2(g)} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बंद बोतल दिखाता है जिसके निचले हिस्से में ठोस आयोडीन क्रिस्टल हैं, और ऊपर के स्थान में बैंगनी आयोडीन वाष्प भरी हुई है। यह ऊर्ध्वपातन और संघनन के माध्यम से ठोस आयोडीन और आयोडीन वाष्प के बीच गतिशील संतुलन को दर्शाता है।
In simple words: When iodine crystals are placed in a closed vessel, they sublime to form violet vapor. Eventually, the intensity of the violet color stabilizes, indicating a dynamic equilibrium between solid iodine and iodine vapor where sublimation and condensation occur at equal rates.

🎯 Exam Tip: This experiment demonstrates sublimation and dynamic equilibrium between solid and gas phases. Focus on how the stable color intensity signifies a balanced rate of opposing physical processes (sublimation and deposition).

Try This. (Textbook Page No. 176)

Question 1.
(i) Dissolve a given amount of sugar in minimum amount of water at room temperature.
(ii) Increase the temperature and dissolve more amount of sugar in the same amount of water to make a thick sugar syrup solution.
(iii) Cool the syrup to the room temperature.
Answer: Observation: Sugar crystals separate out.
Inference: The sugar syrup solution prepared is a saturated solution. Therefore, additional amount of sugar cannot be dissolved in it at room temperature.
In a saturated solution, there exists dynamic equilibrium between the solute molecules in the solid state and in dissolved state.
In simple words: This experiment shows that a saturated sugar solution achieves dynamic equilibrium between solid and dissolved sugar. Heating allows more sugar to dissolve, but upon cooling, the excess sugar recrystallizes, maintaining saturation at the lower temperature.

🎯 Exam Tip: Highlight the concept of a saturated solution as a system in dynamic equilibrium. Explain how solubility (and thus the equilibrium position) is temperature-dependent, leading to recrystallization upon cooling.

Sugar(aq) \( \rightleftharpoons \) Sugar(s)
The rate of dissolution of sugar = The rate of crystallization of sugar.
However, when it is heated, additional amount of sugar can be dissolved in it. But when such a thick sugar syrup is cooled again to room temperature, sugar crystals separate out.
In simple words: In a saturated solution, dynamic equilibrium exists between dissolved and solid sugar, where the rates of dissolution and crystallization are equal. Heating allows more sugar to dissolve, but cooling a thick syrup causes sugar crystals to separate as the equilibrium shifts.

🎯 Exam Tip: Understanding the dynamic nature of equilibrium in saturated solutions is crucial. Focus on how temperature changes affect solubility and crystallization rates, leading to crystal separation upon cooling.

Do You Know? (Textbook Page No. 177)

Question 1. What is a saturated solution?
Answer: A saturated solution is the solution when additional solute cannot be dissolved in it at the given temperature. The concentration of solute in a saturated solution depends on temperature.
In simple words: A saturated solution is one where no more solute can dissolve at a specific temperature, meaning it holds the maximum amount of solute possible at that condition.

🎯 Exam Tip: Define a saturated solution accurately, emphasizing the temperature dependency of solute concentration for full marks.

Observe And Discuss. (Textbook Page No. 177)

Question 1. Colourless N2O4 taken in a closed flask is converted to NO2 (a reddish brown gas).
N2O4(g)
Colourless \( \rightleftharpoons \) 2NO2(g)
Reddish brown

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो गोल पेंदे वाले फ्लास्क को दर्शाता है, जिसमें एक में रंगहीन N2O4 गैस और दूसरे में लाल-भूरे रंग की NO2 गैस है। तीर N2O4 के NO2 में और NO2 के N2O4 में रूपांतरण को दर्शाते हैं, जो इस अभिक्रिया के प्रतिवर्ती स्वरूप को इंगित करता है।
Answer: Observation: Initially, the colourless gas (N2O4) turns to reddish brown (NO2) gas. After sometime, the colour becomes lighter indicating the formation of N2O4 from NO2.
Inference: This indicates that the reaction is reversible. In such reaction, the reactants combine to form the products and the products combine to give the reactants. As soon as the forward reaction produces any NO2, the reverse reaction begins and NO2, starts combining back to N2O4. At equilibrium, the concentrations of N2O4 and NO2 remain unchanged and do not vary with time, because the rate of formation of NO2 is equal to the rate of formation of N2O4.
In simple words: When N2O4 gas is placed in a closed flask, it converts to reddish-brown NO2. Over time, the color stabilizes, indicating that the forward (N2O4 to NO2) and reverse (NO2 to N2O4) reactions are occurring at equal rates, establishing a dynamic chemical equilibrium.

🎯 Exam Tip: Focus on explaining the color changes and the concept of dynamic equilibrium, where opposing reaction rates become equal at equilibrium, not the cessation of reaction.

ℹ️ चित्र व्याख्या (Diagram Explanation): बाईं ओर का चित्र समय के साथ N2O4 से NO2 के रूपांतरण के लिए अग्र और प्रतिगामी अभिक्रिया दरों (सांद्रता के संदर्भ में) का ग्राफ दिखाता है, जहां NO2 की सांद्रता बढ़ती है और N2O4 की घटती है, फिर साम्य पर स्थिर हो जाती है। दाईं ओर का चित्र एक प्रतिवर्ती अभिक्रिया के दौरान अभिक्रिया दरों में परिवर्तन को दर्शाता है, जिसमें अग्र अभिक्रिया दर घटती है और प्रतिगामी अभिक्रिया दर बढ़ती है जब तक कि वे साम्य पर बराबर नहीं हो जातीं।

Discuss (Textbook Page No. 177)

Question 1.
(i) Consider the following dissociation reaction:
2HI(g)
Colourless
gas \( \rightleftharpoons \) H2(g) + I2(g)
Violet
coloured gas
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.
H2(g) + I2(g) \( \rightleftharpoons \) 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 H2 and 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 H2 and I2 becomes equal to the rate of decomposition of HI. The reaction attains chemical equilibrium.
In simple words: Whether starting with hydrogen iodide or with hydrogen and iodine, a reversible reaction will proceed until the rates of the forward and reverse reactions become equal. This leads to a constant concentration of reactants and products, like the stable violet color from iodine vapor, indicating chemical equilibrium.

🎯 Exam Tip: Understand that chemical equilibrium can be approached from either direction (reactants or products). The key is that the macroscopic properties (like color intensity) become constant when the rates of forward and reverse reactions are equal.

Can You Recall? (Textbook Page No. 180)

Question 1. Write ideal gas equation with significance of each term involved in it.
Answer: Ideal gas equation is PV = nRT.
where, P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = universal gas constant
T = Absolute temperature of the gas
In simple words: The ideal gas equation, PV=nRT, describes the relationship between pressure (P), volume (V), number of moles (n), and absolute temperature (T) of an ideal gas, with R being the universal gas constant.

🎯 Exam Tip: Remember the ideal gas equation and clearly define each variable along with its typical units to ensure full understanding.

Just Think. (Textbook Page No. 181)

Question 1. Two processes, which are taking place in opposite directions are in equilibrium. How to write equilibrium constant expressions for heterogeneous equilibrium?
Answer: Equilibrium in a system having more than one phase is called heterogeneous equilibrium.
If ethanol is placed in a conical flask, liquid-vapour equilibrium is established.
C2H5OH(l) \( \rightleftharpoons \) C2H5OH(g)
For a given temperature,
Kc = \( \frac{[C_2H_5OH_{(g)}]}{[C_2H_5OH_{(l)}]} \)
But \( [C_2H_5OH_{(l)}] = 1 \)
\( \therefore \) Kc = \( [C_2H_5OH_{(g)}] \)
Thus, at any given temperature, density is constant irrespective of the amount of liquid, and the term in the denominator is also constant.
(ii) similarly, consider I2(g) \( \rightleftharpoons \) I2(g)
Kc = \( [I_2(g)] \)
(iii) Thus, the expression for equilibrium constant does not contain the concentration of pure solids and pure liquids. That is because for any pure liquid and solid, the concentration is simply its density and this will not change no matter how much solid or liquid is used. Hence, the expression for heterogeneous equilibrium only uses the concentration of gases and dissolved substances (aq.). Solids are pure substances with unchanging concentrations and thus equilibria including solids are simplified.
In simple words: For heterogeneous equilibrium, where reactants and products are in different phases, the equilibrium constant (Kc or Kp) only includes the concentrations or partial pressures of gases and dissolved species, as the concentrations of pure solids and liquids remain constant and are effectively incorporated into the constant itself.

🎯 Exam Tip: When writing equilibrium constant expressions for heterogeneous systems, always exclude pure solids and pure liquids. Only include gaseous and aqueous species in the expression.

Can You Tell? (Textbook Page No. 183)

Question 1. Comment on the extent to which the forward reaction will proceed, from the magnitude of the equilibrium constant for the following reactions:
(i) H2(g) + I2(g) \( \rightleftharpoons \) 2HI(g), Kc = 20 at 550 K
(ii) H2(g) + Cl2(g) \( \rightleftharpoons \) 2HCl(g), Kc = 1018 at 550 K
Answer:
(i) For the reaction, Kc = 20 at 550 K
If the value of Kc is the range of 10-3 to 103, the forward and reverse proceed to equal extents.
Hence, the given reaction will form appreciable concentrations of both reactants and the product at equilibrium.
(ii) For the reaction, Kc = 1018 at 550 K
If the value of Kc > > > 10³, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.
In simple words: A Kc value of 20 indicates that at equilibrium, significant amounts of both reactants and products will be present. A very large Kc value like 10^18 means the forward reaction is highly favored, implying the reaction will almost go to completion, with very little reactant remaining at equilibrium.

🎯 Exam Tip: Remember the general rules: Kc < 10^-3 indicates reactants are favored, Kc > 10^3 indicates products are favored, and Kc between 10^-3 and 10^3 means both reactants and products exist in significant concentrations at equilibrium.

Use Your Brain Power (Textbook Page No. 183)

Question 1. The value of Kc for the dissociation reaction: H2(g) \( \rightleftharpoons \) 2H(g) is 1.2 × 10-42 at 500 K. Does the equilibrium mixture contain mainly hydrogen molecules or hydrogen atoms?
Answer: When the value of Kĉ is very low (that is, Kc < 10-3), then at equilibrium, only a small fraction of the reactants is converted into products.
For the given reaction, Kc <<< 10-3 at 500 K.
Hence, the equilibrium mixture contains mainly hydrogen molecules.
In simple words: Given a very small equilibrium constant (Kc = 1.2 x 10^-42), the reaction strongly favors the reactants. Therefore, the equilibrium mixture will predominantly contain hydrogen molecules (H2) rather than hydrogen atoms (H).

🎯 Exam Tip: A very small Kc value (much less than 1) signifies that the equilibrium lies far to the left, meaning reactants are highly favored and products are formed only in very negligible amounts.

Internet My Friend (Textbook Page No. 183)

Question 1. Collect information about chemical equilibrium.
Answer: https://www.chemguide.co.uk/physical/equilibria/introduction.html
In simple words: Chemical equilibrium is a dynamic state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in reactant and product concentrations.

🎯 Exam Tip: When asked to collect information, summarize the main concepts such as dynamic nature, constant concentrations, and factors affecting equilibrium (Le Chatelier's principle).

Can You Tell? (Textbook Page No. 188)

Question 1.
(i) If NH3 is added to the equilibrium system (Haber process), in which direction will the equilibrium shift to consume added NH3 to 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 NH3 is added to the equilibrium system, the equilibrium will shift from right to left to consume added NH3 to reduce the effect of stress.
(ii) If NH3 is added to the equilibrium system, then reverse reaction will occur to greater extent.
(iii) If NH3 is added to the equilibrium system, the equilibrium will shift in reverse direction and the yield of NH3 will decrease.
In simple words: According to Le Chatelier's principle, adding NH3 to the Haber process equilibrium will cause the reaction to shift to the left, favoring the reverse reaction (decomposition of NH3). This shift will reduce the overall yield of NH3, as the system tries to counteract the added product.

🎯 Exam Tip: Apply Le Chatelier's principle: adding a product shifts equilibrium to the reactant side (left), decreasing the product yield. Conversely, removing a product or adding a reactant shifts it to the product side (right), increasing yield.

Internet My Friend (Textbook Page No. 188)

Question 1.
(i) Collect information about Haber process in chemical equilibrium.
(ii) Youtube. Freescienceslessons: The Haber process
Answer:
(i) https://www.chemguide.co.uk/physical/equilibria/haber.html
(ii) Students are expected to refer 'The Haber process' on YouTube channel 'Freescienceslessons'
In simple words: The Haber process is a crucial industrial method for synthesizing ammonia from nitrogen and hydrogen gases, governed by chemical equilibrium principles, and its efficiency is optimized by controlling temperature, pressure, and catalyst use.

🎯 Exam Tip: Focus on the conditions that maximize ammonia yield in the Haber process (moderate temperature, high pressure, catalyst) and relate them to Le Chatelier's principle for a comprehensive understanding.