Chemical Kinetics
Punjab Board · Class 12 · Chemistry
Most important questions from Chemical Kinetics for Punjab Board Class 12 Chemistry board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
For the reaction: Rate = k[A]^½[B]², what is the overall order of the reaction?
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2.5
Step 1: The overall order of a reaction is the sum of all the exponents of the concentration terms in the rate law. Step 2: Order with respect to A = ½. Step 3: Order with respect to B = 2. Step 4: Overall order = ½ + 2 = 2.5. Step 5: Option B (1.5) incorrectly adds ½ + 1; Option C (2) ignores the ½ power; Option D (3) adds 1 + 2, ignoring the fractional power for A.
The molecularity of a reaction is defined as:
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The number of reacting species that collide simultaneously in an elementary reaction
Step 1: Molecularity refers specifically to elementary reactions. Step 2: It counts the number of atoms, ions, or molecules that must collide simultaneously for the reaction to occur. Step 3: It can be 1 (unimolecular), 2 (bimolecular), or rarely 3 (trimolecular). Step 4: Option B describes 'order of reaction', not molecularity — this is a very common confusion. Step 5: Options C and D are completely unrelated definitions. Molecularity is a theoretical concept applicable only to elementary reactions.
Which of the following is the correct integrated rate equation for a zero order reaction?
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[R] = [R]₀ - kt
Step 1: For a zero order reaction, Rate = -d[R]/dt = k. Step 2: Integrating both sides: [R] = -kt + constant. Step 3: At t = 0, [R] = [R]₀, so the constant = [R]₀. Step 4: Therefore, [R] = [R]₀ - kt. This is a straight line equation where [R] vs t gives slope = -k. Step 5: Options B, C, and D all describe first order reactions — ln[R] vs t plot and the exponential decay form are characteristics of first order, not zero order reactions.
For a first order reaction, the half-life (t₁/₂) is given by:
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t₁/₂ = 0.693 / k
Step 1: For a first order reaction, k = (2.303/t) log([R]₀/[R]). Step 2: At half-life, [R] = [R]₀/2. Step 3: So k = (2.303/t₁/₂) log([R]₀ / [R]₀/2) = (2.303/t₁/₂) log 2. Step 4: Since log 2 = 0.301, k = (2.303 × 0.301)/t₁/₂ = 0.693/t₁/₂. Rearranging: t₁/₂ = 0.693/k. Step 5: Option B ([R]₀/2k) is the half-life formula for ZERO order reactions. Option C inverts the formula. Option D uses 2.303 instead of 0.693.
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