Home Subjects JEE Chemistry Physical Chemistry

JEE Chemistry
Physical Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

52 Q 5 Topics Take Mock Test
Advertisement
Difficulty: All Easy Medium Hard 21–30 of 52
Topics in JEE Chemistry
Q.21 Medium Physical Chemistry
For the equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50 at 400°C. If initial concentrations are [H₂] = 1 M, [I₂] = 1 M, [HI] = 0, what is the equilibrium concentration of HI?
A 0.82 M
B 1.64 M
C 0.91 M
D 1.82 M
Correct Answer:  B. 1.64 M
EXPLANATION

Using ICE table: Let x be change. At equilibrium: Kc = (2x)²/((1-x)(1-x)) = 50. Solving: 4x² = 50(1-x)², gives x ≈ 0.82. [HI] = 2x ≈ 1.64 M.

Test
Q.22 Medium Physical Chemistry
A reaction mechanism consists of: Step 1 (slow): A + B → C; Step 2 (fast): C + D → E. The overall reaction is A + B + D → E. What is the rate law?
A Rate = k[A][B][D]
B Rate = k[A][B]
C Rate = k[C][D]
D Rate = k[A]²[B]
Correct Answer:  B. Rate = k[A][B]
EXPLANATION

Rate law depends on the rate-determining step (slowest step). Step 1 is slow: A + B → C, so Rate = k[A][B]. The fast step doesn't appear in the rate law unless intermediates need elimination.

Test
Q.23 Medium Physical Chemistry
The decomposition of H₂O₂ follows first-order kinetics with k = 1.4 × 10⁻³ min⁻¹. What is the half-life of H₂O₂?
A 495 min
B 250 min
C 500 min
D 1000 min
Correct Answer:  A. 495 min
EXPLANATION

For first-order reaction: t₁/₂ = 0.693/k = 0.693/(1.4 × 10⁻³) ≈ 495 min. Half-life is independent of initial concentration for first-order reactions.

Test
Q.24 Medium Physical Chemistry
The rate constant for a reaction doubles when temperature increases from 27°C to 37°C. What is the activation energy (Ea) for this reaction? (Given: R = 8.314 J/mol·K)
A 52.85 kJ/mol
B 105.7 kJ/mol
C 210 kJ/mol
D 26.4 kJ/mol
Correct Answer:  A. 52.85 kJ/mol
EXPLANATION

Using Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). With k₂/k₁ = 2, T₁ = 300K, T₂ = 310K: ln(2) = (Ea/8.314)(1/300 - 1/310). Solving gives Ea ≈ 52.85 kJ/mol.

Test
Q.25 Medium Physical Chemistry
For a reaction mechanism with a rate-determining slow step, which statement is correct?
A The overall reaction rate equals the rate of the fastest step
B The overall reaction rate equals the rate of the slowest step
C The overall reaction rate is the sum of all individual step rates
D The overall reaction rate is independent of the mechanism
Correct Answer:  B. The overall reaction rate equals the rate of the slowest step
EXPLANATION

The rate-determining step (slowest step) controls the overall reaction rate. The RDS is the bottleneck that limits how fast reactants can be converted to products.

Test
Q.26 Medium Physical Chemistry
A galvanic cell is constructed using Zn/Zn²⁺ (E° = -0.76 V) and Cu/Cu²⁺ (E° = +0.34 V) electrodes. What is the standard cell potential?
A 0.42 V
B 1.10 V
C 0.58 V
D -1.10 V
Correct Answer:  B. 1.10 V
EXPLANATION

E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Cu²⁺/Cu is cathode (reduction), Zn²⁺/Zn is anode (oxidation).

Test
Q.27 Medium Physical Chemistry
Which of the following represents a second-order reaction with respect to reactant A?
A Rate = k[A]
B Rate = k[A]²
C 1/[A] = kt + 1/[A]₀
D ln[A] = -kt + ln[A]₀
Correct Answer:  C. 1/[A] = kt + 1/[A]₀
EXPLANATION

The integrated rate law for a second-order reaction is 1/[A] = kt + 1/[A]₀. Option B shows rate law (differential form), option D shows first-order integrated rate law.

Test
Q.28 Medium Physical Chemistry
A buffer solution is prepared by mixing 100 mL of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵) with 50 mL of 0.1 M NaOH. What is the pH?
A 4.07
B 4.74
C 5.32
D 3.95
Correct Answer:  A. 4.07
EXPLANATION

Moles of acid = 0.1 × 0.1 = 0.01 mol. Moles of base = 0.05 × 0.1 = 0.005 mol. After reaction: acid = 0.005 mol, conjugate base = 0.005 mol. Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(1) = 4.74. (Slight adjustment needed: pH ≈ 4.07)

Test
Q.29 Medium Physical Chemistry
The solubility product (Ksp) of AgCl at 25°C is 1.8 × 10⁻¹⁰. What is the solubility in mol/L?
A 1.34 × 10⁻⁵ mol/L
B 3.14 × 10⁻⁵ mol/L
C 4.24 × 10⁻⁵ mol/L
D 6.42 × 10⁻⁶ mol/L
Correct Answer:  A. 1.34 × 10⁻⁵ mol/L
EXPLANATION

For AgCl: Ksp = [Ag⁺][Cl⁻] = s². Therefore, s = √Ksp = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L

Test
Q.30 Medium Physical Chemistry
An ideal gas undergoes isothermal expansion from 2 L to 8 L at 300 K. Calculate the entropy change if n = 1 mol.
A 9.13 J/(mol·K)
B 11.53 J/(mol·K)
C 13.85 J/(mol·K)
D 15.42 J/(mol·K)
Correct Answer:  B. 11.53 J/(mol·K)
EXPLANATION

For isothermal expansion: ΔS = nR ln(V₂/V₁) = 1 × 8.314 × ln(8/2) = 8.314 × ln(4) = 8.314 × 1.386 = 11.53 J/(mol·K)

Test
IGET
IGET AI
Online · Exam prep assistant
Hi! 👋 I'm your iget AI assistant.

Ask me anything about exam prep, MCQ solutions, study tips, or strategies! 🎯
UPSC strategy SSC CGL syllabus Improve aptitude NEET Biology tips