Consider the equilibrium CO(g) + 3H_2(g) ightleftharpoons CH_4(g) + H_2O(g) If the pressure applied over the system increases by two fold at constant temperature then: (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below:

Solution & Explanation

### Core Logic Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration (c = n/V) for both reactants and products instantly. Statement (B) is correct: The reaction has Delta n_g = 2 - 4 = -2. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant (K) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Confirms that equilibrium constant remains unchanged. ### Pattern Recognition Always remember: pressure changes shift positions but NEVER alter the value of the equilibrium constant K_c or K_p. Only temperature changes can change K. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

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Equilibrium

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Q76 jee_main_2024_27_jan_morning Salt Hydrolysis
Given below are two statements: Statement (I): Aqueous solution of ammonium carbonate is basic. Statement (II): Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on K_a and K_b value of acid and the base forming it. In the light of the above statements, choose the most appropriate answer from the options given below :
  • A. Both Statement I and Statement II are correct
  • B. Statement I is correct but Statement II is incorrect
  • C. Both Statement I and Statement II are incorrect
  • D. Statement I is incorrect but Statement II is correct

Solution

### Related Formula pH of a weak acid-weak base salt system: textpH = 7 + frac12(textpK_a - textpK_b) ### Core Logic Ammonium carbonate, (textNH_4)_2textCO_3, is formed from a weak acid (textH_2textCO_3, K_a approx 4.3 times 10^-7) and weak base (textNH_4textOH, K_b approx 1.8 times 10^-5). Since K_b > K_a, the aqueous medium accumulates an excess of hydroxyl particles over hydronium, forming a basic system (textpH > 7). Both statements are structurally accurate descriptions. ### Chapter Mix Class 11 Chemistry: Equilibrium
Q84 jee_main_2024_29_jan_morning Kp and Kc Relationship
For the reaction mathrmN_2mathrmO_4(mathrmg) rightleftharpoons 2mathrmNO_2(mathrmg) mathrmK_p = 0.492 atm at 300mathrmK . mathrmK_c for the reaction at same temperature is \_\_\_\_\_\_ times 10^-2 . (textGiven: R = 0.082 text L atm mathrmmol^-1 textK^-1)
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula K_p = K_c cdot (RT)^Delta n_g ### Core Logic For the given gaseous equilibrium reaction: N_2O_4(g) rightleftharpoons 2NO_2(g) First, find the change in the number of moles of gas (Delta n_g): Delta n_g = n_p - n_r = 2 - 1 = 1 ### Step 1: Calculation Substitute the given values into the K_p - K_c relationship: K_p = 0.492 R = 0.082 T = 300text K 0.492 = K_c cdot (0.082 times 300)^1 K_c = frac0.4920.082 times 300 K_c = frac0.49224.6 K_c = 0.02 Converting to the requested format (x times 10^-2): K_c = 2 times 10^-2 So, the value is 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q85 jee_main_2024_30_january_evening Buffer Solutions
The pH of an aqueous solution containing 1M benzoic acid (pK_a = 4.20) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ______ mL.
Numerical Answer. Answer: 100 to 100

Solution

### Related Formula Henderson-Hasselbalch Equation for Acidic Buffers: mathrmpH = pK_a + log left( frac[textSalt][textAcid] right) ### Core Logic Let the volume of 1M Benzoic acid be V_a mL and the volume of 1M Sodium benzoate be V_s mL. Total volume = V_s + V_a = 300\,textmL. Millimoles of acid = 1 times V_a = V_a Millimoles of salt = 1 times V_s = V_s Applying Henderson's Equation: 4.5 = 4.2 + log left(fracV_sV_aright) ### Step 1: Calculate Volume Ratio log left(fracV_sV_aright) = 4.5 - 4.2 = 0.3 Since log 2 approx 0.3, we have: fracV_sV_a = 2 V_s = 2 V_a ### Step 2: Substitute and Solve We know V_s + V_a = 300 Substituting V_s = 2 V_a: 2 V_a + V_a = 300 3 V_a = 300 V_a = 100 \, textmL ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q82 jee_main_2024_30_jan_morning Solubility Product
The pH at which Mg(OH)_2 [K_sp=1times 10^-11] begins to precipitate from a solution containing 0.10text M Mg^2+ ions is
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula K_sp = [Mg^2+][OH^-]^2 pOH = -log[OH^-] pH + pOH = 14 ### Core Logic Precipitation begins just when the ionic product equals the solubility product (Q_sp = K_sp). ### Step 1: Calculating required [OH-] [Mg^2+][OH^-]^2 = 10^-11 Given [Mg^2+] = 0.10 text M 0.10 times [OH^-]^2 = 10^-11 [OH^-]^2 = 10^-10 [OH^-] = 10^-5 text M ### Step 2: Finding pH pOH = -log(10^-5) = 5 pH = 14 - pOH pH = 14 - 5 = 9 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q71 jee_main_2024_31_jan_evening Equilibrium Constants (Kp and Kc)
A_(g) rightleftharpoons B_(g) + fracC2_(g) The correct relationship between K_P, alpha and equilibrium pressure P is
  • A. K_P = fracalpha^frac12P^frac12(2 + alpha)^frac12
  • B. K_P = fracalpha^frac32P^frac12(2 + alpha)^frac12(1 - alpha)
  • C. K_P = fracalpha^frac12P^frac32(2 + alpha)^frac32
  • D. K_P = fracalpha^frac12P^frac12(2 + alpha)^frac32

Solution

### Related Formula K_P = fracP_B cdot (P_C)^frac12P_A where P_i is the partial pressure of component i. ### Step 1: Setting up the ICE Table For the reaction A_(g) rightleftharpoons B_(g) + frac12 C_(g) Let initial moles of A = 1. At equilibrium: Moles of A = 1 - alpha Moles of B = alpha Moles of C = fracalpha2 Total moles at equilibrium = (1 - alpha) + alpha + fracalpha2 = 1 + fracalpha2 = frac2 + alpha2 ### Step 2: Calculating Partial Pressures Using mole fraction times Total Pressure (P): P_A = frac1 - alpha1 + fracalpha2 cdot P P_B = fracalpha1 + fracalpha2 cdot P P_C = fracfracalpha21 + fracalpha2 cdot P ### Step 3: Calculating Kp K_P = fracP_B cdot (P_C)^frac12P_A K_P = fracleft( fracalpha1 + alpha/2 P right) cdot left( fracalpha/21 + alpha/2 P right)^1/2frac1 - alpha1 + alpha/2 P K_P = fracalpha cdot (alpha/2)^1/2 cdot P^3/2(1 + alpha/2)^3/2 cdot frac1 + alpha/2(1 - alpha) P K_P = fracalpha^3/2 cdot P^1/2sqrt2 cdot (1 + alpha/2)^1/2 cdot (1 - alpha) Since 1 + alpha/2 = frac2+alpha2, the sqrt2 in denominator cancels out perfectly leaving: K_P = fracalpha^frac32 P^frac12(2 + alpha)^frac12(1 - alpha) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

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