What is the answer to the Chemical Thermodynamics NEET 2026 protein question?
Option A is correct: this Chemical Thermodynamics NEET 2026 question gives equal equilibrium concentrations, making the equilibrium constant one and the standard Gibbs energy zero. A protein changes reversibly between its native state N and denatured state D:
The two states have equal equilibrium concentrations at the stated temperature. The temperature and standard denaturation enthalpy are:
Find the closest standard entropy change in:
The question bank rates this as medium, with an expected solving time of 60 seconds on its scale. Neither is an official NTA difficulty rating or time limit.
Why is the standard Gibbs energy zero in this question?
The equilibrium constant is one because the concentrations are equal, not simply because the protein is at equilibrium. For this one-to-one reaction, divide the equilibrium concentration of the denatured state by that of the native state:
Substitute the stated equality of concentrations:
The official solution then uses the relation between standard Gibbs energy and the equilibrium constant:
Reaction Gibbs energy and standard reaction Gibbs energy are not interchangeable. The reaction Gibbs energy is zero at equilibrium:
The standard reaction Gibbs energy is zero here because the equilibrium constant equals one:
An equilibrium mixture need not contain equal concentrations of reactant and product. Its equilibrium constant can differ from one, leaving its standard Gibbs energy nonzero.
Keep the concentration-ratio step in your working. Skipping directly from “at equilibrium” to zero standard Gibbs energy would miss the condition that makes this solution valid.
How do you calculate the entropy change and match option A?
Divide the standard enthalpy change by the absolute temperature only after establishing zero standard Gibbs energy. Start with the Gibbs equation and substitute that condition:
Move the entropy term to the other side, then divide by temperature:
Convert the temperature exactly as in the official solution:
Substitute the enthalpy with its unit:
The final answer is option A:
Dividing enthalpy per mole by kelvin gives the requested entropy unit. Keep the enthalpy in kilojoules because the options use kilojoules:
Converting the result into joules changes both the numerical value and the energy unit:
That is not 2000 in the option’s kilojoule-based units. Option B therefore does not represent the calculated entropy.
The enthalpy-over-temperature expression is not a universal entropy formula. It follows here from zero standard Gibbs energy; otherwise, the Gibbs-energy term must remain in the calculation.
How does using Celsius produce option D?
Dividing the enthalpy by the Celsius reading reproduces option D. The incorrect numerical calculation is:
The error is at temperature substitution, not in rearranging the Gibbs equation. The temperature multiplying entropy must be absolute temperature, expressed in kelvin:
The correct denominator is:
The Celsius reading of 60 cannot replace it. Celsius has a shifted zero point, so using it directly in this product gives the wrong result even when the algebra is correct.
Use this execution rule: convert temperature before substituting, then check the requested energy unit before selecting an option. Write the kelvin value before dividing, rather than trying to correct the temperature scale afterwards.
Can you apply the same reasoning to three related questions?
Use the concentration ratio to establish standard Gibbs energy, then check the reaction direction before calculating entropy. These three original practice questions are based on the supplied problem, not additional verified NEET PYQs.
What is the standard Gibbs energy when the equilibrium concentrations are equal?
Question 1: It is zero for the stated one-to-one equilibrium. Equal native and denatured concentrations give an equilibrium constant of one:
Include the concentration ratio in your justification. “It is at equilibrium” alone does not establish zero standard Gibbs energy.
What is the standard entropy change for the reverse process?
Question 2: At the same temperature, the reverse process has the opposite standard entropy change. Reversing the reaction exchanges its initial and final states, changing the sign but not the magnitude:
What changes if the denatured protein has the higher equilibrium concentration?
Question 3: The standard Gibbs energy is negative, so the entropy shortcut no longer follows. For the same forward, one-to-one reaction:
The mixture is still at equilibrium, but its standard Gibbs energy is not zero. Retain that term when rearranging:
Practise reproducing the original solution against the question bank’s 60-second target. Use this sequence:
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Frequently asked questions
What is the answer to the Chemical Thermodynamics NEET 2026 protein question?
Option A is correct: the standard entropy change is 2.0 kJ K⁻¹ mol⁻¹. Equal equilibrium concentrations give K = 1 and ΔG° = 0, so ΔS° = ΔH°/T. Using T = 60 + 273 = 333 K gives ΔS° = 666/333 = 2.0 kJ K⁻¹ mol⁻¹.
Is standard Gibbs energy always zero at equilibrium?
No. Reaction Gibbs energy, ΔG, is zero at equilibrium, but standard reaction Gibbs energy, ΔG°, is zero only when K = 1. In this one-to-one protein equilibrium, equal native and denatured concentrations establish K = 1.
Why is the protein entropy answer 2.0 and not 2000?
The options ask for entropy in kJ K⁻¹ mol⁻¹, so 666 kJ mol⁻¹ divided by 333 K gives 2.0 in the requested units. This equals 2000 J K⁻¹ mol⁻¹, not 2000 kJ K⁻¹ mol⁻¹. Option B therefore uses the wrong numerical value for the stated unit.
Why can't I use 60°C in the Gibbs equation?
The temperature in ΔG° = ΔH° − TΔS° must be absolute temperature in kelvin. Using 60 instead of 333 gives 666/60 = 11.1, reproducing the incorrect option D. Convert the temperature before substituting.