What was the 2026 NEET question on reversible protein denaturation in chemical kinetics?
The 2026 NEET question described reversible denaturation N ⇌ D with equal equilibrium concentrations of native and denatured forms at exactly 60 °C. Given ΔH° = +666 kJ mol⁻¹ for the process, it asked for ΔS° in kJ K⁻¹ mol⁻¹ closest to one of the four values. The answer is exactly 2.0.
How do you solve this chemical kinetics neet 2026 denaturation question step by step?
At equilibrium with [N] = [D], K = [D]/[N] = 1 for N ⇌ D. ΔG° = –RT ln K therefore equals zero.
From the Gibbs-Helmholtz relation ΔG° = ΔH° – TΔS°, rearrangement gives ΔS° = ΔH° / T. T = 60 + 273 = 333 K. Substitute ΔH° = 666 kJ mol⁻¹:
The value 2 carries the unit kJ K⁻¹ mol⁻¹ and matches option A exactly. Every step follows the verified 2026 official solution.
What mistake turns the correct 2.0 into the distractor 333.0?
The method mistake that produces 333.0 is treating ΔS° as equal to the numerical value of temperature instead of performing division of ΔH by T. After writing ΔG° = 0 = ΔH° – TΔS°, some students omit the division step entirely and copy the 333 K figure straight into the answer box.
This single procedural error lands on the distractor. The correct single division yielding 2.0 avoids the trap.
Why does thermodynamics appear inside the chemical kinetics chapter for NEET?
NCERT explicitly links equilibrium constant K to ΔG° = –RT ln K inside the Chemical Kinetics unit. This makes the question legitimate. The temperature at which [N] = [D] is the point where forward and reverse rates are equal at equilibrium.
Arrhenius factor is not needed here but the same chapter expects mastery of both rate and thermodynamic relations.
Which two numericals test the same first-order and Arrhenius ideas with proper unit handling?
Question 1
A first-order reaction has rate constant 2.5 × 10⁻⁴ s⁻¹ at 300 K. Given Ea = 60 kJ mol⁻¹, find k at 310 K. Use R = 8.314 J mol⁻¹ K⁻¹.
Convert Ea to 60000 J mol⁻¹ first. Apply the Arrhenius relation:
Question 2
For a first-order protein degradation, t½ = 20 min at 37 °C. If ΔH° for activation is 80 kJ mol⁻¹, estimate the temperature at which t½ becomes 5 min.
Convert 37 °C to 310 K and note that k is inversely proportional to t½, so k₂/k₁ = 20/5 = 4. Thus ln(4) = 1.386. Use:
Ea = 80000 J mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹ gives Ea/R ≈ 9622. 1.386 / 9622 ≈ 0.000144. 1/310 ≈ 0.003226, so 1/T₂ = 0.003226 – 0.000144 = 0.003082. T₂ = 1/0.003082 ≈ 324.5 K or 51.5 °C.
Both examples demand exact conversion of temperature and consistent energy units before taking logs or dividing. The same habits are required for the 666/333 step. Search past NEET papers by chapter in the past-paper archive to locate every recent numerical that mixes equilibrium constants with first-order half-life and practise them under timed conditions.
How should you revise hybrid thermo-kinetics numericals for NEET 2027?
Always convert °C to K before dividing thermodynamic quantities. Memorise that K = 1 forces ΔG° = 0 and therefore ΔS° = ΔH°/T at that exact temperature. Write this relation on every similar problem before substituting numbers.
Practice 10 mixed questions daily that combine equilibrium constants with Arrhenius or first-order half-life so the division reflex becomes automatic. When a unit or sign slips during practice, photograph the doubt for a step-by-step solution across Physics, Chemistry and Biology on the NEET JEEnius platform. This verification habit prevents loss of 4 marks on similar hard numericals.
Next step: photograph a doubt on NEET JEEnius AI and photograph any question you are stuck on and get a step-by-step solution across Physics, Chemistry and Biology (20 free a month).
Related on NEET JEEnius AI: Organic Compounds Containing Halogens NEET 2026: Radical HBr Addition.
Frequently asked questions
What was the 2026 NEET question on reversible protein denaturation?
The question described N ⇌ D with equal concentrations of native and denatured forms at 60 °C and ΔH° = +666 kJ mol⁻¹. Students had to find ΔS° which equals exactly 2.0 kJ K⁻¹ mol⁻¹.
How do you calculate ΔS° for the chemical kinetics NEET 2026 denaturation question?
At equilibrium [N] = [D] so K = 1 and ΔG° = 0. From ΔG° = ΔH° – TΔS° it follows that ΔS° = ΔH° / T. With T = 333 K, 666 / 333 = 2.0 kJ K⁻¹ mol⁻¹.
Why is thermodynamics asked in the chemical kinetics chapter for NEET?
NCERT links the equilibrium constant K to ΔG° = –RT ln K inside the Chemical Kinetics unit. The temperature where concentrations are equal is where forward and reverse rates are equal at equilibrium.
What mistake turns the correct 2.0 into 333 in the NEET 2026 question?
After writing ΔG° = 0 = ΔH° – TΔS°, students omit the division and copy the temperature 333 K as the answer for ΔS°. The correct single step is dividing 666 by 333 to obtain 2.0.
How to revise hybrid thermo-kinetics numericals for NEET 2027?
Convert °C to K before every calculation. Memorise that K = 1 forces ΔG° = 0 so ΔS° = ΔH°/T. Practise 10 mixed questions daily on equilibrium constants, Arrhenius equation and first-order half-lives with strict unit handling.