Samacheer Kalvi Class 11 Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium

Textbook Evaluation:

I. Choose the best answer:

Question 1. If the rate constants for the forward and reversible reaction are \( 0.8 \times 10^{-5} \) and \( 1.6 \times 10^{-4} \) respectively, the value of the equilibrium constant is,
(a) 20
(b) \( 0.2 \times 10^{-1} \)
(c) 0.05
(d) None of the options
Answer: (a) 20
In simple words: The equilibrium constant is found by dividing the rate constant of the forward reaction by the rate constant of the reverse reaction. When you do this calculation, the result is 20. This value helps us understand how far the reaction goes towards products.

🎯 Exam Tip: Remember the formula for the equilibrium constant (K) in terms of rate constants: K = \( \frac{k_f}{k_r} \), where \( k_f \) is the forward rate constant and \( k_r \) is the reverse rate constant. Pay attention to units and scientific notation.

Question 2. At a given temperature and pressure, the equilibrium constant values for the equilibria
\( 3A_2 + B_2 + 2Cr \xrightarrow{K_1} 2A_3BC \)
and
\( A_3BC \xrightarrow{K_2} \frac{3}{2} [A_2] + \frac{1}{2} B_2 + C \)
The relation between \( K_1 \) and \( K_2 \) is
(a) \( K_1 = \frac{1}{\sqrt{K_2}} \)
(b) \( K_2 = K_1^{-1 / 2} \)
(c) \( K_1^2 = 2K_2 \)
(d) \( \frac{K_1}{2} = K_2 \)
Answer: (b) \( K_2 = K_1^{-1 / 2} \)
In simple words: The second reaction is the reverse of the first reaction, and it's also multiplied by a factor of 1/2. So, the equilibrium constant for the second reaction, \( K_2 \), is the inverse of the square root of \( K_1 \). This means \( K_2 \) is equal to \( K_1 \) raised to the power of negative one-half.

🎯 Exam Tip: When a reaction is reversed, its equilibrium constant becomes the inverse (1/K) of the original. If the stoichiometric coefficients are multiplied by a factor (n), the new equilibrium constant becomes \( K^n \).

Question 3. The equilibrium constant for a reaction at room temperature is \( K_1 \) and that at 700 K is \( K_2 \). If \( K_1 > K_2 \), then
(a) The forward reaction is exothermic
(b) The forward reaction is endothermic
(c) The system does not attain equilibrium
(d) The reverse reaction is exothermic
Answer: (a) The forward reaction is exothermic
In simple words: If the equilibrium constant decreases when the temperature increases (from room temperature to 700 K), it means that the forward reaction gives off heat. This type of reaction is called exothermic. When heat is added, the reaction shifts to the left, reducing the product amount.

🎯 Exam Tip: For exothermic reactions, increasing the temperature decreases the equilibrium constant (K), favoring reactants. For endothermic reactions, increasing the temperature increases K, favoring products.

Question 4. The formation of ammonia from \( N_2(g) \) and \( H_2(g) \) is a reversible reaction \( 2NO(g) + O_2(g) \xrightarrow{K_2} 2NO_2(g) + Heat \). What is the effect of increase of temperature on this equilibrium reaction
(a) equilibrium is unaltered
(b) formation of ammonia is favoured
(c) equilibrium is shifted to the left
(d) reaction rate does not change
Answer: (c) equilibrium is shifted to the left
In simple words: The reaction releases heat, meaning it is an exothermic reaction. If you increase the temperature, it's like adding more heat. To reduce this extra heat, the reaction will move backward, away from making more products, shifting to the left.

🎯 Exam Tip: For exothermic reactions, adding heat (increasing temperature) shifts the equilibrium to the reactant side. For endothermic reactions, adding heat shifts it to the product side. This is according to Le Chatelier's Principle.

Question 5. Solubility of carbon dioxide gas in cold water can be increased by
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of the options
Answer: (a) increase in pressure
In simple words: To get more carbon dioxide gas to dissolve in cold water, you need to push down on the gas. This extra push, or higher pressure, forces more gas into the water. Think of how soda cans stay fizzy under pressure.

🎯 Exam Tip: The solubility of gases in liquids increases with increasing pressure (Henry's Law) and decreases with increasing temperature.

Question 6. Which one of the following is an incorrect statement?
(a) for a system at equilibrium, Q is always less than the equilibrium constant
(b) equilibrium can be attained from either side of the reaction
(c) the presence of catalyst affects both the forward reaction and reverse reaction to the same extent
(d) Equilibrium constant varied with temperature
Answer: (a) for a system at equilibrium, Q is always less than the equilibrium constant
In simple words: At equilibrium, the reaction quotient (Q) is exactly equal to the equilibrium constant (K). It is not less than K. All the other statements about equilibrium are correct, like how reactions can start from either side or how catalysts speed up both forward and reverse reactions equally.

🎯 Exam Tip: A key characteristic of equilibrium is that Q = K. If Q < K, the reaction moves forward to reach equilibrium. If Q > K, the reaction moves backward.

Question 7. \( K_1 \) and \( K_2 \) are the equilibrium constants for the reactions respectively.
\( N_2(g) + O_2(g) \xrightarrow{K_1} 2NO(g) \)
\( 2NO(g) + O_2(g) \xrightarrow{K_2} 2NO_2(g) \)
What is the equilibrium constant for the reaction \( NO_2(g) \rightleftharpoons \frac{1}{2} N_2(g) + O_2(g) \)
(a) \( \frac{1}{\sqrt{K_1 K_2}} \)
(b) \( (K_1 = K_2)^{1/2} \)
(c) \( \frac{1}{2 K_1 K_2} \)
(d) \( (\frac{1}{K_1 K_2})^{3 / 2} \)
Answer: (a) \( \frac{1}{\sqrt{K_1 K_2}} \)
In simple words: To find the new equilibrium constant, we combine the first two reactions. The target reaction is the reverse of the sum of the first two reactions, and then all coefficients are divided by two. This means we multiply \( K_1 \) and \( K_2 \), take the inverse of that product, and then take the square root.

🎯 Exam Tip: When adding reactions, multiply their equilibrium constants. When reversing a reaction, invert its constant (1/K). When multiplying coefficients by 'n', raise the constant to the power 'n'. For this problem, \( NO_2(g) \to NO(g) + \frac{1}{2}O_2(g) \) has \( K'_2 = \frac{1}{\sqrt{K_2}} \). And \( NO(g) \to \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \) has \( K'_1 = \frac{1}{\sqrt{K_1}} \). Adding these two reversed and halved reactions yields the target, so the new K is \( K'_1 \times K'_2 = \frac{1}{\sqrt{K_1 K_2}} \).

Question 8. In the equilibrium, \( 2A(g) \rightleftharpoons 2B(g) + C_2(g) \) the equilibrium concentrations of A, B and \( C_2 \) at 400K are \( 1 \times 10^{-4}M \), \( 2.0 \times 10^{-3} M \), \( 1.5 \times 10^{-4} M \) respectively. The value of \( K_c \) for the equilibrium at 400 K is
(a) 0.06
(b) 0.09
(c) 0.62
(d) \( 3 \times 10^{-2} \)
Answer: (d) \( 3 \times 10^{-2} \)
In simple words: To calculate the equilibrium constant \( K_c \), you multiply the concentrations of the products raised to their powers and divide by the concentrations of the reactants raised to their powers. For this reaction, it is \( \frac{[B]^2[C_2]}{[A]^2} \). Plugging in the given numbers gives us \( 3 \times 10^{-2} \).

🎯 Exam Tip: Always write out the correct \( K_c \) expression before substituting values. Make sure to raise each concentration to the power of its stoichiometric coefficient from the balanced equation.

Question 9. An equilibrium constant of \( 3.2 \times 10^{-6} \) for a reaction means, the equilibrium is
(a) largely towards forward direction
(b) largely towards reverse direction
(c) never established
(d) none of the options
Answer: (b) largely towards reverse direction
In simple words: A very small equilibrium constant, like \( 3.2 \times 10^{-6} \), tells us that at equilibrium, there are much more reactants than products. This means the reaction strongly favors the reverse direction, moving towards the starting materials.

🎯 Exam Tip: A small K value (K < 1) means the reaction favors reactants, while a large K value (K > 1) means it favors products. A K value close to 1 indicates significant amounts of both reactants and products at equilibrium.

Question 10. \( \frac{K_C}{K_P} \) for the reaction, \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) is
(a) \( \frac{1}{RT} \)
(b) \( \sqrt{RT} \)
(c) RT
(d) \( (RT)^2 \)
Answer: (d) \( (RT)^2 \)
In simple words: The relationship between \( K_p \) and \( K_c \) is \( K_p = K_c (RT)^{\Delta n_g} \). First, we calculate \( \Delta n_g \), which is the difference in moles of gaseous products minus moles of gaseous reactants. For this reaction, \( \Delta n_g = 2 - (1+3) = 2 - 4 = -2 \). So, \( K_p = K_c (RT)^{-2} \). If we rearrange this to find \( \frac{K_c}{K_p} \), we get \( (RT)^2 \).

🎯 Exam Tip: Always remember the formula \( K_p = K_c (RT)^{\Delta n_g} \). The most common error is miscalculating \( \Delta n_g \) or getting the exponent sign wrong. Remember to include only gaseous species in the calculation of \( \Delta n_g \).

Question 11. For the reaction \( AB (g) \rightleftharpoons A(g) + B(g) \), at equilibrium, AB is 20 % dissociated at a total pressure of P, the equilibrium constant \( K_p \) is related to the total pressure by the expression
(a) \( P = 24 K_p \)
(b) \( P = 8 K_p \)
(c) \( 24 P = K_p \)
(d) none of the options
Answer: (a) \( P = 24 K_p \)
In simple words: When AB breaks down by 20%, we can find the partial pressures of A, B, and the remaining AB in terms of the total pressure P. Then, we use these partial pressures in the \( K_p \) formula. The calculation shows that \( K_p \) is equal to \( \frac{P}{24} \), which can be rewritten as \( P = 24 K_p \).

🎯 Exam Tip: For dissociation problems, set up an ICE (Initial, Change, Equilibrium) table to determine the moles of each species at equilibrium. Express partial pressures using mole fractions and total pressure, then substitute into the \( K_p \) expression.

Question 12. In which of the following equilibrium, \( K_p \) and \( K_c \) are not equal?
(a) \( 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \)
(b) \( SO_2(g) + NO_2 \rightleftharpoons SO_3(g) + NO(g) \)
(c) \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \)
(d) \( PCl_5 \rightleftharpoons PCl_3(g) + Cl_2(g) \)
Answer: (d) \( PCl_5 \rightleftharpoons PCl_3(g) + Cl_2(g) \)
In simple words: \( K_p \) and \( K_c \) are only equal when the number of gas moles on the product side is the same as on the reactant side (\( \Delta n_g = 0 \)). In reaction (d), there is one mole of gas on the reactant side (\( PCl_5 \)) and two moles of gas on the product side (\( PCl_3 + Cl_2 \)), so \( \Delta n_g = 1 \). This means \( K_p \) and \( K_c \) will not be equal.

🎯 Exam Tip: \( K_p = K_c \) only when \( \Delta n_g = 0 \). To check this, sum the stoichiometric coefficients of gaseous products and subtract the sum of stoichiometric coefficients of gaseous reactants. If the result is zero, \( K_p = K_c \).

Question 13. If x is the fraction of \( PCl_5 \) dissociated at equilibrium in the reaction \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) then starting with 0.5 mole of \( PCl_5 \), the total number of moles of reactants and products at equilibrium is
(a) \( 0.5 - x \)
(b) \( x + 0.5 \)
(c) \( 2x + 0.5 \)
(d) \( x + 1 \)
Answer: (b) \( x + 0.5 \)
In simple words: If you start with 0.5 moles of \( PCl_5 \) and x is the part that breaks down, then at equilibrium, you'll have \( 0.5 - x \) moles of \( PCl_5 \), \( x \) moles of \( PCl_3 \), and \( x \) moles of \( Cl_2 \). Adding these up gives \( (0.5 - x) + x + x \), which simplifies to \( 0.5 + x \).

🎯 Exam Tip: Use an ICE table (Initial, Change, Equilibrium) to clearly track the moles of each species. This helps avoid errors in calculating total moles at equilibrium. Remember to consider the starting amount and the fraction dissociated.

Question 14. The values of \( K_{p1} \) and \( K_{p2} \) for the reactions, \( X \rightleftharpoons Y + Z \), \( A \rightleftharpoons 2B \) are in the ratio 9 : 1 if degree of dissociation of X and A be equal then total pressure at equilibrium \( P_1 \), and \( P_2 \) are in the ratio
(a) 36:1
(b) 1:1
(c) 3:1
(d) 1:9
Answer: (a) 36:1
In simple words: For reaction X, the expression for \( K_{p1} \) is \( \frac{\alpha^2 P_1}{1-\alpha^2} \). For reaction A, the expression for \( K_{p2} \) is \( \frac{4\alpha^2 P_2}{1-\alpha^2} \). Given that \( \frac{K_{p1}}{K_{p2}} = \frac{9}{1} \) and the degree of dissociation \( \alpha \) is the same, we can divide the \( K_p \) expressions. After simplifying, we find that \( \frac{P_1}{P_2} = \frac{9}{4} \times 4 = 36 \). So the ratio is 36:1.

🎯 Exam Tip: Ensure you correctly derive the \( K_p \) expressions for each reaction, especially considering the \( \alpha \) (degree of dissociation) and total pressure. Pay close attention to the stoichiometry for each reaction, as it affects the exponent of \( \alpha \) and the overall factor of P.

Question 15. In the reaction \( Fe(OH)_3(S) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq) \), if the concentration of \( OH^- \) ions is decreased by 1/4 times, then the equilibrium concentration of \( Fe^{3+} \) will
(a) not changed
(b) also decreased by 1/4 times
(c) increase by 4 times
(d) increase by 64 times
Answer: (d) increase by 64 times
In simple words: The equilibrium constant expression is \( K_{sp} = [Fe^{3+}][OH^-]^3 \). If the \( OH^- \) concentration is reduced to \( \frac{1}{4} \) of its original value, then to keep \( K_{sp} \) constant, the \( Fe^{3+} \) concentration must increase by a factor of \( 4^3 \), which is 64. This happens to balance the equation according to the constant value.

🎯 Exam Tip: Remember Le Chatelier's principle and the solubility product expression. Changes in concentration of one ion affect the concentration of the other to maintain the \( K_{sp} \) constant. Make sure to apply the stoichiometric coefficient as an exponent correctly.

Question 16. Consider the reaction where \( K_p = 0.5 \) at a particular temperature \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \). If the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true.
(a) more \( PCl_3 \) will be produced
(b) more \( Cl_2 \) will be produced
(c) more \( PCl_5 \) will be produced
(d) none of the options
Answer: (c) more \( PCl_5 \) will be produced
In simple words: To figure out which way the reaction moves, we calculate the reaction quotient \( Q_p \). If the partial pressure of each gas is 1 atm, then \( Q_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{1 \times 1}{1} = 1 \). Since \( Q_p = 1 \) and \( K_p = 0.5 \), \( Q_p > K_p \). This means the reaction will shift to the left, favoring the formation of more \( PCl_5 \) until equilibrium is reached.

🎯 Exam Tip: Compare the reaction quotient (Q) with the equilibrium constant (K) to predict the direction of a shift: if Q < K, the reaction shifts right; if Q > K, it shifts left; if Q = K, it is at equilibrium.

Question 17. Equimolar concentrations of \( H_2 \) and \( I_2 \) are heated to equilibrium in a 1 litre flask. What percentage of the initial concentration of \( H_2 \) has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
(a) 33%
(b) 66%
(c) \( (33)^2 \% \)
(d) 16.5%
Answer: (a) 33%
In simple words: If the rate constants for the forward and reverse reactions are equal, it means the equilibrium constant (K) is 1. When K=1 for \( H_2 + I_2 \rightleftharpoons 2HI \) and you start with equal amounts of \( H_2 \) and \( I_2 \), it turns out that 33% of the initial \( H_2 \) will have reacted to form HI at equilibrium. This is a special case of a balanced reaction.

🎯 Exam Tip: When \( k_f = k_r \), the equilibrium constant K is 1. For the reaction \( H_2 + I_2 \rightleftharpoons 2HI \), set up an ICE table, calculate K, and solve for 'x' (the reacted amount) to find the percentage. This leads to \( x = 1/3 \), or 33.33%.

Question 18. In a chemical equilibrium, the rate constant for the forward reaction is \( 2.5 \times 10^{-2} \), and the equilibrium constant is 50. The rate constant for the reverse reaction is,
(a) 11.5
(b) 50
(c) \( 2 \times 10^2 \)
(d) \( 2 \times 10^{-3} \)
Answer: (d) \( 2 \times 10^{-3} \)
In simple words: We know that the equilibrium constant (K) is the forward rate constant (\( k_f \)) divided by the reverse rate constant (\( k_r \)). If \( K = 50 \) and \( k_f = 2.5 \times 10^{-2} \), we can find \( k_r \) by rearranging the formula to \( k_r = \frac{k_f}{K} \). This calculation gives us \( 2 \times 10^{-3} \).

🎯 Exam Tip: Always remember the fundamental relationship \( K = \frac{k_f}{k_r} \). This allows you to find any one of the three variables if the other two are known. Double-check your calculations with scientific notation.

Question 19. Which of the following is not a general characteristic of equilibrium involving physical process
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remain constant
Answer: (c) All the physical processes stop at equilibrium
In simple words: At equilibrium, physical processes do not stop; they continue to happen but at equal rates in opposite directions. For example, in a liquid-vapor equilibrium, molecules still evaporate and condense, but the total amount of liquid and vapor stays the same. The equilibrium is dynamic, not static.

🎯 Exam Tip: A key concept in equilibrium is its dynamic nature: processes continue at the molecular level, but macroscopically, there are no observable changes. Physical processes, like phase changes, also follow this rule.

Question 20. For the formation of Two moles of \( SO_3(g) \) from \( SO_2 \) and \( O_2 \), the equilibrium constant is \( K_1 \). The equilibrium constant for the dissociation of one mole of \( SO_3 \)
(a) \( \frac{1}{K_1} \)
(b) \( K_1^2 \)
(c) \( \left(\frac{1}{K_1}\right)^{1 / 2} \)
(d) \( \frac{K_1}{2} \)
Answer: (c) \( \left(\frac{1}{K_1}\right)^{1 / 2} \)
In simple words: The first reaction is \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \) with constant \( K_1 \). The second reaction, dissociation of one mole of \( SO_3 \), is \( SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g) \). This is the reverse of the first reaction and also halved. So, the equilibrium constant for the second reaction is \( \sqrt{\frac{1}{K_1}} \) or \( (\frac{1}{K_1})^{1/2} \).

🎯 Exam Tip: Remember these rules for manipulating equilibrium constants: reversing a reaction inverts K (1/K), and dividing all coefficients by 'n' means taking the \( n^{th} \) root of K (or \( K^{1/n} \)).

Question 21. Match the equilibria with the corresponding conditions:
(i) Liquid \( \rightleftharpoons \) Vapour
(ii) Solid \( \rightleftharpoons \) Liquid
(iii) Solid \( \rightleftharpoons \) Vapour
(iv) Solute(s) \( \rightleftharpoons \) Solute (Solution)
1) Melting point
2) Saturated solution
3) Boiling point
4) Sublimation point

🎯 Exam Tip: Understand the basic definitions of phase transitions and solubility. Boiling point is liquid-vapor equilibrium, melting point is solid-liquid, sublimation point is solid-vapor, and saturated solution is solute-solution equilibrium.

Question 22. Consider the following reversible reaction at equilibrium, \( A + B \rightleftharpoons C \). If the concentrations of reactants A and B are doubled, then the equilibrium constant will
(a) be doubled
(b) become one fourth
(c) be halved
(d) remain the same
Answer: (d) remain the same
In simple words: The equilibrium constant, K, depends only on temperature for a specific reaction. It does not change if the concentrations of reactants or products are changed. If you change the concentrations, the reaction might shift to restore equilibrium, but the value of K itself stays the same.

🎯 Exam Tip: The equilibrium constant is specific to a reaction at a given temperature. It does not change with concentration, pressure (if \( \Delta n_g = 0 \)), or the addition of a catalyst. Only temperature can change the value of K.

Question 23. \( [Co(H_2O)_6]^{2+} (aq) (pink) + 4Cl^- (aq) \rightleftharpoons [CoCl_4]^{2-}(aq) (blue) + 6 H_2O (l) \)
In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in color. On the basis of this information, which one of the following is true?
(a) \( \Delta H > 0 \) for the forward reaction
(b) \( \Delta H = 0 \) for the reverse reaction
(c) \( \Delta H < 0 \) for the forward reaction
(d) Sign of the \( \Delta H \) cannot be predicted based on this information
Answer: (a) \( \Delta H > 0 \) for the forward reaction
In simple words: The reaction is blue at room temperature. When cooled, it turns pink. This means that cooling shifts the equilibrium to the left, favoring the pink reactant \( [Co(H_2O)_6]^{2+} \). According to Le Chatelier's principle, a shift to the left upon cooling means the reverse reaction is exothermic, and the forward reaction is endothermic. An endothermic reaction has a positive \( \Delta H \).

🎯 Exam Tip: Use Le Chatelier's principle: if cooling (removing heat) shifts equilibrium to the left, the forward reaction absorbs heat (endothermic, \( \Delta H > 0 \)). If cooling shifts to the right, the forward reaction releases heat (exothermic, \( \Delta H < 0 \)).

Question 24. The equilibrium constants of the following reactions are:
\( N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1 \)
\( N_2 + O_2 \rightleftharpoons 2NO; K_2 \)
\( H_2 + 1/2O_2 \rightleftharpoons H_2O; K_3 \)
The equilibrium constant (K) for the reaction;
\( 2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O \), will be
(a) \( K_2^3 K_3 / K_1 \)
(b) \( K_2^3 K_3^2 / K_1 \)
(c) \( K_2 K_3^3 / K_1 \)
(d) \( K_2 K_3 / K_1 \)
Answer: (c) \( K_2 K_3^3 / K_1 \)
In simple words: To get the target reaction, we need to reverse the first reaction (\( \frac{1}{K_1} \)), use the second reaction as is (\( K_2 \)), and multiply the third reaction by 3 (\( K_3^3 \)). Then we combine these. So, the overall equilibrium constant K is \( \frac{1}{K_1} \times K_2 \times K_3^3 \), which simplifies to \( \frac{K_2 K_3^3}{K_1} \).

🎯 Exam Tip: When manipulating equilibrium reactions, remember: reversing a reaction means inverting its K (1/K); multiplying coefficients by a factor 'n' means raising K to the power 'n' (\( K^n \)); and adding reactions means multiplying their respective K values.

Question 25. A 20 litre container at 400 K contains \( CO_2 (g) \) at pressure 0.4 atm and an excess (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of \( CO_2 \) attains its maximum value will be: Given that: \( SrCO_3(S) \rightleftharpoons SrO + CO_2 (g) [K_p = 1.6 \text{ atm}] \)
(a) 2 litre
(b) 5 litre
(c) 10 litre
(d) 4 litre
Answer: (b) 5 litre
In simple words: For the given reaction, the equilibrium constant \( K_p \) is simply equal to the partial pressure of \( CO_2 \). We are given \( K_p = 1.6 \text{ atm} \). So, the maximum pressure \( P_{CO_2} \) can be is 1.6 atm. We use Boyle's law (\( P_1 V_1 = P_2 V_2 \)). We start with 0.4 atm in 20 liters. So, \( 0.4 \times 20 = 1.6 \times V_2 \). Solving for \( V_2 \) gives 5 liters.

🎯 Exam Tip: For reactions involving solids and gases, only the partial pressures of gases are included in \( K_p \). Remember to identify the maximum possible pressure from \( K_p \) and then use gas laws (like Boyle's Law) for volume-pressure relationships.

II. Write brief answer to the following questions:

Question 26. If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer: The equilibrium state is called dynamic because, at this point, the forward and backward reactions are still happening. Even though the overall amounts of reactants and products stay constant, the individual molecules are constantly changing from reactants to products and back again. The rate of the forward reaction becomes equal to the rate of the backward reaction. This ongoing activity at the molecular level makes it dynamic.
In simple words: Equilibrium is dynamic because reactions keep going in both directions, but they balance each other out. So, the total amounts don't change, but molecules are always moving.

🎯 Exam Tip: Emphasize that "dynamic" means continuous microscopic activity even when macroscopic properties appear constant. This is a fundamental concept of chemical equilibrium.

Question 27. For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.
Answer: The equilibrium constant (\( K_c \)) has a constant value at a specific temperature. The reaction quotient (Q), however, is not always constant. Q is calculated using the concentrations of reactants and products at any given moment. The value of Q changes as the reaction proceeds towards equilibrium. Only when the system reaches equilibrium, the value of Q becomes equal to \( K_c \) and then it remains constant. Until then, Q keeps changing. So, \( K_c \) is constant, and Q is constant only at equilibrium.
In simple words: The equilibrium constant (K) stays the same for a reaction at a set temperature. The reaction quotient (Q) changes as the reaction moves along. Q is only constant when the reaction reaches equilibrium, becoming equal to K.

🎯 Exam Tip: Differentiate between the equilibrium constant (K) and the reaction quotient (Q). K is a fixed value at a given temperature, while Q is a variable that indicates the current state of a reaction relative to equilibrium.

Question 28. What is the relation between \( K_p \) and \( K_c \)? Given one example for which \( K_p \) is equal to \( K_c \).
Answer: The relationship between \( K_p \) (equilibrium constant in terms of partial pressures) and \( K_c \) (equilibrium constant in terms of molar concentrations) is given by the equation: \( K_p = K_c (RT)^{\Delta n_g} \). Here, R is the ideal gas constant, T is the absolute temperature in Kelvin, and \( \Delta n_g \) is the change in the number of moles of gaseous substances in the reaction. \( \Delta n_g \) is calculated as (moles of gaseous products) - (moles of gaseous reactants). For \( K_p \) to be equal to \( K_c \), \( \Delta n_g \) must be zero.
An example of a reaction where \( K_p = K_c \) is: \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \). In this reaction, there are 2 moles of gaseous products and \( 1+1=2 \) moles of gaseous reactants, so \( \Delta n_g = 2 - 2 = 0 \). Therefore, \( K_p = K_c \).
In simple words: \( K_p \) and \( K_c \) are related by the formula \( K_p = K_c (RT)^{\Delta n_g} \). They are the same only if the number of gas molecules doesn't change from the left side to the right side of the reaction. For example, in the reaction of hydrogen and iodine making hydrogen iodide, the number of gas molecules stays the same, so \( K_p \) equals \( K_c \).

🎯 Exam Tip: Clearly define all terms in the \( K_p = K_c (RT)^{\Delta n_g} \) equation and correctly calculate \( \Delta n_g \). Remember that only gaseous species count for \( \Delta n_g \).

Question 29. For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is \( K_c \) larger or smaller than \( K_p \).
Answer: For a gaseous homogeneous reaction where the number of moles of products is greater than the number of moles of reactants, \( \Delta n_g \) (change in moles of gas) will be positive. Using the relationship \( K_p = K_c (RT)^{\Delta n_g} \), if \( \Delta n_g \) is positive, then \( K_p \) will be larger than \( K_c \) because (RT) is a positive value greater than 1 at normal temperatures. Therefore, \( K_c \) will be smaller than \( K_p \).
In simple words: If a reaction makes more gas molecules than it starts with, then \( \Delta n_g \) is positive. This means \( K_p \) will be bigger than \( K_c \). So, \( K_c \) is smaller than \( K_p \).

🎯 Exam Tip: Understand how the sign of \( \Delta n_g \) affects the relationship between \( K_p \) and \( K_c \). If \( \Delta n_g > 0 \), then \( K_p > K_c \). If \( \Delta n_g < 0 \), then \( K_p < K_c \). If \( \Delta n_g = 0 \), then \( K_p = K_c \).

Question 30. When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant, in which direction does the reaction proceed to reach equilibrium?
Answer: When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), it means that the concentrations of products are too high and the concentrations of reactants are too low compared to what they would be at equilibrium. To reach equilibrium, the reaction will proceed in the reverse direction. This shift helps to decrease the product concentrations and increase the reactant concentrations until Q becomes equal to K.
In simple words: If Q is bigger than K, there are too many products. So, the reaction will go backward, making more reactants, until Q becomes equal to K.

🎯 Exam Tip: Think of Q as a measure of the current state of the reaction. If Q is too large (more products), the reaction shifts left. If Q is too small (more reactants), it shifts right. If Q equals K, it's already balanced.

Question 31. For the reaction, \( A_2(g) + B_2(g) \rightleftharpoons 2AB(g); \Delta H \) is -ve. the following molecular scenes represent different reaction mixtures. (A-green, B-blue)