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JEE Main MCQ questions — Mathematics, Physics, Chemistry for engineering entrance.

238 Q 3 Subjects 12th (PCM)
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Q.31 Medium JEE Chemistry Chemical Kinetics
A reaction has activation energy of 50 kJ/mol. If the temperature is increased from 300 K to 310 K, the rate constant increases by a factor of approximately (R = 8.314 J/mol·K):
A 1.5
B 2.0
C 2.5
D 3.0
Correct Answer:  B. 2.0
EXPLANATION

Using Arrhenius equation: log(k₂/k₁) = (Ea/2.303R)(T₂-T₁)/(T₁T₂). With Ea = 50,000 J/mol, ΔT = 10 K, this gives log(k₂/k₁) ≈ 0.30, so k₂/k₁ ≈ 2.0

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Q.32 Medium JEE Chemistry Chemical Kinetics
Which of the following is an example of a homogeneous catalyst?
A Fe₃O₄ in ammonia synthesis
B H₂SO₄ in esterification reaction
C Pt surface in hydrogenation
D MnO₂ in decomposition of KMnO₄
Correct Answer:  B. H₂SO₄ in esterification reaction
EXPLANATION

A homogeneous catalyst is in the same phase as reactants. H₂SO₄ (liquid) catalyzes esterification of reactants (liquid), making it homogeneous. Others are heterogeneous catalysts.

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Q.33 Medium JEE Chemistry Chemical Kinetics
If a reaction is first-order with rate constant k = 0.1 min⁻¹, what fraction of the reactant remains after 5 half-lives?
A 1/32
B 1/16
C 1/8
D 1/4
Correct Answer:  A. 1/32
EXPLANATION

After n half-lives, fraction remaining = (1/2)ⁿ. After 5 half-lives: (1/2)⁵ = 1/32.

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Q.34 Medium JEE Chemistry Chemical Kinetics
At 300 K, a reaction has a half-life of 10 minutes. At 310 K, the half-life becomes 5 minutes. What is the approximate value of temperature coefficient (assuming RRT ≈ 2)?
A 1.5
B 2.0
C 2.5
D 3.0
Correct Answer:  B. 2.0
EXPLANATION

For a first-order reaction, if half-life decreases from 10 to 5 minutes (becomes half) with a 10 K increase, this indicates the reaction rate doubles per 10 K, giving a temperature coefficient of 2.

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Q.35 Medium JEE Chemistry Chemical Kinetics
For the reaction A → B, the integrated rate law for zero-order kinetics is:
A [A] = [A]₀ - kt
B ln[A] = ln[A]₀ - kt
C 1/[A] = 1/[A]₀ + kt
D [A]² = [A]₀² - kt
Correct Answer:  A. [A] = [A]₀ - kt
EXPLANATION

For zero-order reaction: d[A]/dt = -k, integrating gives [A] = [A]₀ - kt, which is a linear equation.

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Q.36 Medium JEE Chemistry Chemical Kinetics
According to Arrhenius equation, k = Ae^(-Eₐ/RT), a catalyst increases reaction rate by:
A Increasing the frequency factor A
B Decreasing the activation energy Eₐ
C Increasing the temperature T
D Decreasing the pre-exponential factor
Correct Answer:  B. Decreasing the activation energy Eₐ
EXPLANATION

A catalyst provides an alternative reaction pathway with lower activation energy, thus increasing the rate constant k without affecting A or T.

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Q.37 Medium JEE Chemistry Chemical Kinetics
Which of the following graphs represents a first-order reaction?
A Plot of [A] vs t is linear
B Plot of ln[A] vs t is linear
C Plot of 1/[A] vs t is linear
D Plot of [A]² vs t is linear
Correct Answer:  B. Plot of ln[A] vs t is linear
EXPLANATION

For a first-order reaction, ln[A] = ln[A]₀ - kt, so a plot of ln[A] vs t gives a straight line with slope -k.

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Q.38 Medium JEE Chemistry Chemical Kinetics
The mechanism of a reaction is: Step 1: A + B → C (slow), Step 2: C + D → E + F (fast). What is the overall reaction and the rate law?
A A + B + D → E + F; rate = k[A][B]
B A + B + D → E + F; rate = k[C][D]
C A + B + C + D → E + F; rate = k[A][B][D]
D A + B → E + F; rate = k[A]²[B]
Correct Answer:  A. A + B + D → E + F; rate = k[A][B]
EXPLANATION

The overall reaction is obtained by adding all steps and canceling intermediates: A + B + D → E + F. Rate law is determined by the slow step: rate = k[A][B]

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Q.39 Medium JEE Chemistry Chemical Kinetics
For a first-order reaction, if the initial concentration is [A]₀ = 0.5 M and after 30 seconds it becomes 0.25 M, what is the rate constant?
A 0.0231 s⁻¹
B 0.0115 s⁻¹
C 0.0462 s⁻¹
D 0.0577 s⁻¹
Correct Answer:  A. 0.0231 s⁻¹
EXPLANATION

Using ln([A]₀/[A]ₜ) = kt, ln(0.5/0.25) = k × 30, ln(2) = k × 30, k = 0.693/30 = 0.0231 s⁻¹

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Q.40 Medium JEE Chemistry Electrochemistry
A galvanic cell is constructed using Zn|Zn²⁺ and Cu|Cu²⁺ half-cells. If the concentration of Zn²⁺ is increased from 1 M to 10 M at 25°C, how does this affect the cell potential? (E°cell = 1.1 V)
A Decreases by 0.0296 V
B Increases by 0.0296 V
C Increases by 0.059 V
D Remains unchanged
Correct Answer:  A. Decreases by 0.0296 V
EXPLANATION

Using Nernst equation: Ecell = E°cell - (0.059/n)log(Q). Increasing [Zn²⁺] increases Q, making the log term positive, which decreases Ecell. ΔE = -(0.059/2)log(10) = -0.0295 ≈ -0.0296 V.

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