The equilibrium constant (K_p) for the thermal decomposition of water vapor: textH_2textO(g) rightleftharpoons textH_2text(g) + frac12textO_2text(g) quad (Delta G^circ = 92.34 text kJ mol^-1) is evaluated as 8.0 times 10^-3 at 2300 text K under a total pressure of 1 text bar. Under these specific conditions, the degree of dissociation (alpha) of water is _________ times 10^-2 (as the nearest integer value). [Assume alpha ll 1].

Numerical Answer Type:
Enter a numerical value Answer: 5 to 5 +4 marks

Solution & Explanation

### Related Formula Gas phase dissociation equilibrium setup: textH_2textO(g) rightleftharpoons textH_2text(g) + frac12textO_2text(g) Partial pressure equilibrium expression: K_p = fracP_textH_2 cdot (P_textO_2)^1/2P_textH_2textO ### Execution Step 1: Set up the mole distribution table at equilibrium assuming 1 initial mole: * textH_2textO = 1 - alpha * textH_2 = alpha * textO_2 = fracalpha2 Step 2: Calculate the total moles (n_T) at equilibrium: n_T = (1 - alpha) + alpha + fracalpha2 = 1 + fracalpha2 Given alpha ll 1, we can approximate n_T approx 1. Step 3: Express the partial pressures using total pressure P = 1 text bar: P_textH_2textO = frac1-alpha1 cdot P approx 1 cdot 1 = 1 P_textH_2 = alpha cdot P = alpha P_textO_2 = fracalpha2 cdot P = fracalpha2 Step 4: Substitute these partial pressures into the K_p expression: K_p = fracalpha cdot left(fracalpha2right)^1/21 = fracalpha^3/2sqrt2 Step 5: Equate to the given value of K_p = 8.0 times 10^-3 and solve for alpha: 8.0 times 10^-3 = fracalpha^3/2sqrt2 implies alpha^3/2 = 8sqrt2 times 10^-3 Cube both sides to clear fractional exponents: alpha^3 = left(8sqrt2 times 10^-3right)^2 = 128 times 10^-6 alpha = sqrt[3]128 times 10^-2 approx 5.03 times 10^-2 Matching the target template alpha = 5.03 times 10^-2, the integer value is **5**. ### Pattern Recognition When alpha ll 1, the total mole expression simplifies to 1, and the denominator (1-alpha) drops out. This simplifies the expression to K_p propto alpha^1 + Delta n_g, allowing you to quickly isolate alpha via standard powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium Class 11 Chemistry: Chemical Thermodynamics

Reference Study Guides

Chemical Equilibrium

More Chemical Equilibrium Previous-Year Questions

Q71 jee_main_2026_21_jan_evening Ionic Equilibrium and Polyprotic Acids
The first and second ionization constants of textH_2textX are 2.5 times 10^-8 and 1.0 times 10^-13 respectively. The concentration of textX^2- in 0.1text M textH_2textX solution is \_ times 10^-15text M. (Nearest Integer)
Numerical Answer. Answer: 100 to 100

Solution

### Core Logic For a weak diprotic acid textH_2textX, the concentration of textX^2- in a solution of concentration C is approximately equal to the second dissociation constant K_a_2 under standard weak acid approximation conditions ([textX^2-] approx K_a_2 = 1.0 times 10^-13 text M). Converting to the required format: 1.0 times 10^-13 = 100 times 10^-15 text M. ### Step 1: Final Calculation Thus, the integer value is 100. ### Pattern Recognition Sees: polyprotic acid dissociation equilibrium calculation for secondary anion concentration. Trap: Assuming complete dissociation for the second step. ### Chapter Mix Class 11 Chemistry: Equilibrium
Q63 jee_main_2026_22_january_evening Buffer Solution pH Calculation
Which of the following mixture gives a buffer solution with pH = 9.25 ? Given: pK_b(textNH_4textOH) = 4.75
  • A. 0.2textM textNH_4textOH (0.4text L) + 0.1textM textHCl (1text L)
  • B. 0.2textM textNH_4textOH (0.5text L) + 0.1textM textHCl (0.5text L)
  • C. 0.5textM textNH_4textOH (0.2text L) + 0.2textM textHCl (0.5text L)
  • D. 0.4textM textNH_4textOH (1text L) + 0.1textM textHCl (1text L)

Solution

### Related Formula textpOH = 14 - textpH = 14 - 9.25 = 4.75 textpOH = pK_b + log frac[textSalt][textBase] ### Core Logic Step 1: Determine condition for textpOH = 4.75: 4.75 = 4.75 + log frac[textSalt][textBase] implies log frac[textSalt][textBase] = 0 implies [textSalt] = [textBase] Step 2: Test Option (2): textInitial millimoles of textNH_4textOH = 0.2 times 500 = 100text mmol textInitial millimoles of textHCl = 0.1 times 500 = 50text mmol textMillimoles of textNH_4textCl formed = 50text mmol textMillimoles of textNH_4textOH remaining = 100 - 50 = 50text mmol Since [textSalt] = [textBase], textpOH = pK_b = 4.75 and textpH = 9.25. ### Pattern Recognition Sees: Basic buffer when textpH = 14 - pK_b. Shortcut: Requires millimoles of remaining weak base to equal millimoles of formed conjugate salt (i.e. acid amount must be exactly half of base amount). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q70 jee_main_2026_23_january_morning Relation between Kp and Kc
Consider the general reaction given below at 400 K xA(g) rightleftharpoons yB(g) The values of K_p and K_c are studied under the same condition of temperature but variation in x and y. (i) K_p = 85.87 and K_c = 2.586 appropriate units (ii) K_p = 0.862 and K_c = 28.62 appropriate units The value of x and y in (i) and (ii) respectively are:
  • A. text(i) 3, 1 ; (ii) 3, 1
  • B. text(i) 4, 1 ; (ii) 4, 1
  • C. text(i) 1, 3 ; (ii) 2, 1
  • D. text(i) 1, 2 ; (ii) 2, 1

Solution

### Related Formula K_p = K_c (RT)^Delta n_g ### Core Logic For the reaction xA(g) rightleftharpoons yB(g), Delta n_g = y - x. We use R = 0.0821 text L atm K^-1 textmol^-1 and T = 400 text K. RT = 0.0821 times 400 = 32.84. ### Step 1: Case (i) Analysis In case (i), K_p = 85.87 and K_c = 2.586. Since K_p > K_c, Delta n_g > 0, implying y - x > 0. 85.87 = 2.586 times (32.84)^y-x 33.20 approx (32.84)^y-x Solving gives y - x = 1. Looking at options for (i), "1, 2" (where x=1, y=2) fits perfectly because 2 - 1 = 1. ### Step 2: Case (ii) Analysis In case (ii), K_p = 0.862 and K_c = 28.62. Since K_p < K_c, Delta n_g < 0, implying y - x < 0. Looking at options for (ii), "2, 1" (where x=2, y=1) gives y - x = -1. K_p = K_c times (32.84)^-1 implies 0.862 = 28.62 / 32.84 approx 0.87 This matches correctly. ### Pattern Recognition Just check the signs! K_p > K_c Rightarrow y > x. K_p < K_c Rightarrow y < x. Only option (4) satisfies both conditions simultaneously without even needing full calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q72 jee_main_2026_23_january_morning Degree of Dissociation and Kp
For the following gas phase equilibrium reaction at constant temperature, NH_3(g) rightleftharpoons frac12N_2(g) + frac32H_2(g) If the total pressure is sqrt3 atm and the pressure equilibrium constant (K_p) is 9 atm, then the degree of dissociation is given as (x times 10^-2)^-1/2. The value of x is ____ (Nearest integer)
Numerical Answer. Answer: 125 to 125

Solution

### Related Formula K_p = frac(P_N_2)^1/2 (P_H_2)^3/2(P_NH_3) ### Core Logic Let the degree of dissociation of NH_3 be alpha. Initial moles: NH_3 = 1 Equilibrium moles: NH_3 = 1 - alpha, N_2 = fracalpha2, H_2 = frac3alpha2 Total moles = 1 - alpha + fracalpha2 + frac3alpha2 = 1 + alpha ### Step 1: Setting up Partial Pressures Using mole fractions and total pressure P_T = sqrt3: P_NH_3 = frac1-alpha1+alpha P_T P_N_2 = fracalpha/21+alpha P_T P_H_2 = frac3alpha/21+alpha P_T ### Step 2: Solving for alpha K_p = frac (fracalpha/21+alpha P_T)^1/2 cdot (frac3alpha/21+alpha P_T)^3/2 frac1-alpha1+alpha P_T K_p = frac (alpha/2)^1/2 (3alpha/2)^3/2 1-alpha cdot fracP_T^2P_T (1+alpha) cdot (1+alpha) Wait, simplify the P_T term: K_p = frac (alpha/2)^1/2 (3alpha/2)^3/2 1-alpha cdot left(fracP_T1+alpharight) Substitute K_p = 9 and P_T = sqrt3: 9 = frac (alpha/2)^1/2 (3alpha/2)^3/2 (1-alpha) cdot fracsqrt31+alpha 9 = frac sqrt27 cdot (alpha/2)^2 (1-alpha)(1+alpha) cdot sqrt3 9 = frac 9 cdot (alpha^2/4) 1 - alpha^2 1 = frac alpha^2/4 1 - alpha^2 1 - alpha^2 = fracalpha^24 1 = frac5alpha^24 alpha^2 = frac45 = 0.8 alpha = (0.8)^1/2 ### Step 3: Matching the given format We are given alpha = (x times 10^-2)^-1/2. Equating both forms: (0.8)^1/2 = (x times 10^-2)^-1/2 Invert both sides: x times 10^-2 = frac10.8 x times 10^-2 = 1.25 x = 125 ### Pattern Recognition When doing fractional stoichiometric power K_p calculations, grouping the alpha terms early simplifies algebra massively. 3^3/2 times 3^1/2 = 9, which perfectly cancels the K_p value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q71 jee_main_2026_23_january_evening Chemical Equilibrium and Le Chatelier's Principle
X_2(g) + Y_2(g) rightleftharpoons 2Z(g) X_2(g) and Y_2(g) are added to a 1 text L flask and it is found that the system attains the above equilibrium at T(K) with the number of moles of X_2(g), Y_2(g) and Z(g) being 3, 3 and 9 mol respectively (equilibrium moles). Under this conditions of equilibrium, 10 mol of Z(g) is added to the flask and the temperature is maintained at T(K). Then the number of moles of Z(g) in the flask when the new equilibrium is established is ____. (Nearest integer).
Numerical Answer. Answer: 15 to 15

Solution

### Related Formula K_c = frac[Z]^2[X_2][Y_2] ### Core Logic For the reaction X_2(g) + Y_2(g) rightleftharpoons 2Z(g): First, calculate the equilibrium constant K_c using the initial equilibrium concentrations (moles in 1 L volume). [X_2] = 3 text M [Y_2] = 3 text M [Z] = 9 text M K_c = frac(9)^2(3)(3) = frac819 = 9 ### Step 1: Shift due to addition of Z Now, 10 moles of Z(g) are added. The new initial moles are: X_2 = 3 Y_2 = 3 Z = 9 + 10 = 19 The reaction quotient Q_c > K_c, so the reaction shifts backward. Let the change in X_2 and Y_2 be +x moles, then Z changes by -2x moles. At new equilibrium: X_2 = 3 + x Y_2 = 3 + x Z = 19 - 2x ### Step 2: Calculate New Equilibrium K_c = frac(19 - 2x)^2(3 + x)(3 + x) = 9 Take the square root of both sides (concentrations must be positive): frac19 - 2x3 + x = 3 19 - 2x = 3(3 + x) 19 - 2x = 9 + 3x 10 = 5x implies x = 2 The number of moles of Z(g) at the new equilibrium is: textMoles of Z = 19 - 2x = 19 - 2(2) = 19 - 4 = 15 ### Pattern Recognition If a reaction's K_c evaluates to a perfect square and Delta n_g = 0, adding products sets up a neat quadratic that can be directly resolved by taking the root of both sides, heavily speeding up the algebraic work. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)