Previous year questions: Equilibrium

Equilibrium30 min readLesson 8 of 8

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks) At temperature T, the equilibrium constant of A2(g) + B2(g) giving C(g) is 2.7 x 10^-5. What is the equilibrium constant of (1/3)A2(g) + (1/3)B2(g) giving (1/3)C(g) at the same temperature? Answer: When an equation is multiplied by a factor, the equilibrium constant is raised to the power of that factor. Halving an equation takes the square root of K, and reversing it takes the reciprocal. Here the target equation is one third of the original, so the new constant is the cube root of 2.7 x 10^-5. Since 2.7 x 10^-5 equals 27 x 10^-6 and the cube root of 27 is 3 while the cube root of 10^-6 is 10^-2, the new constant is 3 x 10^-2. The answer is option D in the official key.

K^{'} = K^{1/3} = (2.7 \times 1 0^{- 5})^{1/3} = 3 \times 1 0^{- 2}

Equilibrium constant for a scaled equation

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks) The pH of a solution obtained by mixing 5 mL of 0.1 M NH4OH with 250 mL of 0.1 M NH4Cl is x multiplied by 10^-2. Find the nearest integer x. (pKb of NH4OH = 4.74) Answer: The mixture is a basic buffer, so use pOH = pKb + log([salt]/[base]). The amounts are 5 mL x 0.1 M = 0.5 mmol of base and 250 mL x 0.1 M = 25 mmol of salt, and the volumes cancel in the ratio, so the salt to base ratio is 50. The logarithm of 50 is 1.70, so pOH = 4.74 + 1.70 = 6.44. Then pH = 14 - 6.44 = 7.56, which is 756 x 10^-2. The value of x is 756.

pO H = 4.74 + lo g \frac{25}{0.5} = 4.74 + 1.70 = 6.44 p H = 14 - 6.44 = 7.56 = 756 \times 1 0^{- 2}

Basic buffer from the salt to base ratio

Q3 (JEE Main 2026, 4 Apr Shift 2, 4 marks) 20 mL of an acetic acid solution requires 28.4 mL of 0.1 M NaOH for neutralisation. A solution X is prepared by mixing 20 mL of the same acid with 14.2 mL of 0.1 M NaOH. What is the pH of X? (pKa of acetic acid = 4.75) Answer: First find the acid concentration. The moles of base used are 28.4 mL x 0.1 M = 2.84 mmol, which equals the acid moles in 20 mL, so the acid is 0.142 M. In solution X, 20 mL of this acid gives 2.84 mmol of acid and 14.2 mL of 0.1 M NaOH gives 1.42 mmol of base. The base neutralises exactly half the acid, leaving 1.42 mmol of acid and forming 1.42 mmol of acetate. The mixture is a buffer with equal concentrations of acid and salt, so pH equals pKa, which is 4.75. The correct option is 4.75.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks) The reaction A(g) giving B(g) + C(g) starts with an amount a of A. At equilibrium the amount of A remaining is (a - x) at a total pressure p. Write the equilibrium constant Kp. Answer: At equilibrium the moles are (a - x) of A, x of B and x of C, so the total is (a + x). The partial pressures follow from the mole fractions: pA is (a - x)p/(a + x) and both pB and pC are xp/(a + x). Kp is pB times pC divided by pA. Substituting gives x squared times p squared divided by (a + x) squared, divided by (a - x)p/(a + x), which simplifies to x squared p divided by (a + x)(a - x), that is x squared p over (a squared - x squared). This is the expression in option B.

K_{p} = \frac{p _{B} p _{C}}{p _{A}} = \frac{x ^{2} p}{( a + x ) ( a - x )} = \frac{x ^{2} p}{a ^{2} - x ^{2}}

Kp for A giving B + C from extent x and total pressure p

Q5 (JEE Main 2026, 2 Apr Shift 1, 4 marks) The solubility products of Ag2CrO4 and AgBr are 32x and 4y respectively at 298 K. Find the ratio of the molarity of Ag2CrO4 to the molarity of AgBr in their saturated solutions. Answer: For Ag2CrO4 the dissociation gives two silver ions and one chromate ion, so Ksp equals 4 times the cube of the solubility. Setting 4s cubed equal to 32x gives s cubed equal to 8x, so the solubility is 2 times the cube root of x. For AgBr the dissociation gives one silver ion and one bromide ion, so Ksp equals the square of the solubility. Setting s squared equal to 4y gives the solubility 2 times the square root of y. The ratio is therefore 2x^(1/3) divided by 2y^(1/2), which is x^(1/3) divided by the square root of y. This matches the last option.

s_{1} = (\frac{32 x}{4})^{1/3} = 2 x^{1/3} s_{2} = 4 y = 2 y \frac{s _{1}}{s _{2}} = \frac{x ^{1/3}}{y}

Solubilities of Ag2CrO4 and AgBr and their ratio

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks) Consider the reactions 2xy(g) giving x2(g) + y2(g) with K1 = 2.5 x 10^5 and xy(g) + (1/2)z2(g) giving xyz(g) with K2 = 5 x 10^-3. Find K3 for the equilibrium (1/2)x2(g) + (1/2)y2(g) + (1/2)z2(g) giving xyz(g). Answer: Build the target reaction from the two given reactions. Reverse the first reaction to get x2 + y2 giving 2xy; its constant becomes 1/K1. Halve it to get (1/2)x2 + (1/2)y2 giving xy; the constant becomes the square root of 1/K1. Add the second reaction as written, xy + (1/2)z2 giving xyz, with constant K2. The sum is the target reaction, so K3 is K2 times the reciprocal square root of K1. With K1 equal to 2.5 x 10^5, the square root is 500, so K3 is 5 x 10^-3 divided by 500, which is 1.0 x 10^-5. The correct option is 1.0 x 10^-5.

K_{3} = K_{2} (\frac{1}{K _{1}})^{1/2} = \frac{5 \times 1 0 ^{- 3}}{500} = 1.0 \times 1 0^{- 5}

Combining equilibrium constants for the target reaction

Q7 (NCERT Exemplar, 3 marks) For the reaction NH4Cl(s) giving NH3(g) + HCl(g), what is the value of delta n for the relation Kp = Kc (RT)^delta n: 1, 0.5, 1.5 or 2? Answer: Count only the gaseous species. The reactant side has no gas because ammonium chloride is a solid, and the product side has two gases, ammonia and hydrogen chloride. The change in gas moles is therefore two minus zero, which is 2. The correct value is 2, and the relation becomes Kp = Kc(RT)^2. Source: NCERT Exemplar Class 11 Chemistry, Unit 7, MCQ 1.

Q8 (NCERT Exemplar, 3 marks) For the reaction H2(g) + I2(g) giving 2HI(g), the standard Gibbs energy change is positive. Is the equilibrium constant zero, greater than one, equal to one, or less than one? Answer: The standard Gibbs energy change and the equilibrium constant are related by delta G standard equal to minus RT ln K. If delta G standard is positive, ln K must be negative, and therefore K is less than one. A negative ln K means the standard state favours the reactants, so at equilibrium the reactants dominate. The answer is K less than one. Source: NCERT Exemplar Class 11 Chemistry, Unit 7, MCQ 2.

Q9 (NCERT Exemplar, 3 marks) Which feature does not belong to equilibria of physical processes: the need for a closed vessel at one fixed temperature, constant values for all measurable properties, every physical process coming to a halt at equilibrium, or opposing processes running at the same rate? Answer: A physical equilibrium is dynamic: the two processes keep running at matching rates, so the measured properties hold steady while change continues at the molecular level. The claim that everything stops is therefore the wrong one. The closed-vessel requirement and the constancy of properties are both genuine features. Source: NCERT Exemplar Class 11 Chemistry, Unit 7, MCQ 3.

Q10 (NCERT Exemplar, 3 marks) Which statement is incorrect: in an insulated flask with ice and water the masses do not change, adding oxalic acid to a solution of iron(III) nitrate and potassium thiocyanate deepens the red colour, adding a catalyst does not change the equilibrium constant, or the equilibrium constant of a reaction with negative delta H decreases as temperature rises? Answer: The last two statements are correct. A catalyst changes only the rate, so K is unchanged, and an exothermic reaction has a smaller K at higher temperature. In the ice-water system the two phases stay in equilibrium, so the masses remain constant, which is also correct. The incorrect statement is the oxalic acid one. Oxalic acid removes iron(III) ions by forming a complex with them, which shifts the thiocyanate equilibrium and decreases the red colour rather than deepening it. The answer is the second statement. Source: NCERT Exemplar Class 11 Chemistry, Unit 7, MCQ 5.

in hinglish

PYQ mein equilibrium constants, buffers, weak acids aur Ksp sabse zyada poochhe jaate hain. Equation scale karne par K ki power badalti hai: halve karo to square root, reverse karo to reciprocal, one third karo to cube root. Buffer ke sawaal mein Henderson equation pH = pKa + log([salt]/[acid]) lagao, aur basic buffer ke liye pOH = pKb + log([salt]/[base]). Titration ke half-neutralisation point par pH = pKa hota hai. Kp expression A ⇌ B + C ke liye x²p/(a² - x²) hota hai. Ksp ke liye dissociation equation se solubility nikaalo; AB ke liye s², AB2 ke liye 4s³. K3 nikaalne ke liye target reaction ko given reactions ka combination likho aur unke K multiply/divide karo. Exemplar questions mein Δn (sirf gases), ΔG° vs K ka sign, dynamic equilibrium aur common ion effect ke concepts poochhe jaate hain. Numerical mein neat expression, approximation aur units likho.