What was the NEET 2025 chemical kinetics question on first-order isomerization of A to B?
Molecule A isomerises to B by first-order kinetics at exactly 1000 K. The energy barrier relative to reactant is 191.48 kJ mol⁻¹ with frequency factor 10²⁰. The time for 50 % of A molecules to change into B must be expressed in picoseconds as nearest integer. R is given as 8.314 J K⁻¹ mol⁻¹. The answer is 69.
How do you solve the NEET 2025 first-order isomerization numerical using the official method?
The official solution yields 69.
For first-order reaction
The rate constant comes from the Arrhenius equation
Convert Ea to 191.48 × 10³ J mol⁻¹ then
Divide by 2.303: 23.031 / 2.303 ≈ 10. Therefore
Then
s⁻¹.
Half-life:
s.
Since 1 s = 10^{12} ps therefore
ps rounded to nearest integer 69.
Every step matches the verified NEET 2025 official solution. Keep coefficients and exponents on separate lines.
What unit conversion slip produces the distractor 6.93 in this chemical kinetics NEET 2025 question?
Students correctly obtain k = 10^{10} s^{-1} and t½ = 6.93 × 10^{-11} s but then record the coefficient 6.93 directly as the answer in picoseconds. They omit the multiplication by 10^{12}.
This is a pure unit-conversion method error after the exponential arithmetic is finished. Correct discipline is to keep the full scientific notation (6.93 × 10^{-11}) intact until the final conversion step 10^{-11} × 10^{12} = 10^{1} is written explicitly.
Why does the exponent conversion from e to base-10 make this Arrhenius first-order question hard-tier?
The exponent conversion is the real test because 23.031 is deliberately chosen so Ea/RT divided by 2.303 yields exactly 10. This lets e^(-x) convert cleanly to 10^{-10}.
Comparison: using R = 8.314 × 10^{-3} kJ and keeping Ea in kJ gives identical 23.031 ratio. Half-life formula remains independent of initial concentration, a key first-order property tested here.
What practice numericals reuse the same first-order and Arrhenius concepts from chemical kinetics?
Question 1: A first-order reaction has A = 4 × 10^{13} s^{-1} and Ea = 98.5 kJ mol^{-1} at 500 K; find the half-life in seconds (use same e-to-10 method).
Question 2: Rate constant of a first-order process doubles when temperature rises from 300 K to 310 K; calculate Ea in kJ mol^{-1} using log form of Arrhenius.
Question 3: For an isomerisation following first-order kinetics with t½ = 2.0 × 10^{-3} s, what is the value of k and the percentage remaining after 10^{-2} s?
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What practical tips let you solve such chemical kinetics numericals in under 90 seconds?
Always write Ea/RT first and check if the number is close to an integer multiple of 2.303 before approximating.
Keep track of powers of ten on a separate line from the mantissa (6.93 and 10^{-11}).
Memorise 1 s = 10^{12} ps, 10^9 ns, 10^6 μs so conversion is automatic.
Never substitute R = 2 cal or any non-SI value in NEET 2025 paper.
How does this hard numerical fit the NEET 2025 pattern?
Chemistry carries 45 compulsory questions (180 marks) with no Section B. This hard-tier numerical tests integration of two topics in one 4-mark question.
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Frequently asked questions
What was the NEET 2025 chemical kinetics question on first-order isomerization?
Molecule A isomerises to B by first-order kinetics at exactly 1000 K with Ea of 191.48 kJ mol⁻¹ and frequency factor 10²⁰. The question asks for the time for 50% of A molecules to change into B expressed in picoseconds as nearest integer. The official answer is 69.
How do you solve the NEET 2025 first-order isomerization numerical?
Apply t½ = 0.693/k for the first-order reaction. Use Arrhenius equation k = A e^{-Ea/RT} with Ea converted to 191480 J mol⁻¹ to get Ea/RT = 23.031. This divides by 2.303 to give exactly 10 so e^{-23.031} = 10^{-10}, yielding k = 10^{10} s^{-1}. Then t½ = 6.93 × 10^{-11} s which equals 69.3 ps and rounds to 69.
What is the common mistake in chemical kinetics NEET 2025 picosecond question?
Students correctly calculate k = 10^{10} s^{-1} and t½ = 6.93 × 10^{-11} s but forget to multiply by 10^{12} to convert seconds to picoseconds. They directly enter the coefficient 6.93 as the answer. Always retain full scientific notation until the final explicit unit conversion step 10^{-11} × 10^{12} = 10.
Why does Ea/RT divided by 2.303 give exactly 10 in NEET 2025 chemical kinetics?
The value 191.48 kJ mol⁻¹ is deliberately chosen so Ea/RT equals 23.031 at 1000 K with R = 8.314. Dividing 23.031 by 2.303 yields exactly 10, allowing clean conversion of e^{-23.031} directly to 10^{-10}. This tests precise handling of the Arrhenius exponential without approximation errors.
What are practice questions for chemical kinetics first order NEET 2025?
Practice with A = 4×10^{13} s^{-1}, Ea = 98.5 kJ mol^{-1} at 500 K to find half-life. Another is finding Ea when k doubles from 300 K to 310 K using log form of Arrhenius. A third gives t½ = 2.0×10^{-3} s and asks for k plus percentage remaining after 10^{-2} s. Solve these using the same e-to-base-10 method.