Zero-Order Reaction NEET 2026: Why the Answer Is 0.75 min

What is the correct answer to the zero-order reaction NEET 2026 question?

B, 0.75 min is the answer to the zero-order reaction NEET 2026 question. The coefficient 2 enters through the reaction-rate definition: the given constant describes the normalised reaction rate, so A disappears at twice that rate.

This Chemistry MCQ belongs to Chemical Kinetics. Its medium difficulty rating comes from the question bank, not an official NEET exam classification.

A undergoes the zero-order conversion below. Calculate the time by which three-quarters of the initial A has reacted.

2 A \to B, k = 1.0 m o l L^{- 1} m i n^{- 1}, [A]_{0} = 2 M

The supplied options are:

Why should you write the rate equation before the concentration equation?

The official solution treats the given constant as the stoichiometrically normalised reaction rate, not A’s disappearance rate. This definition determines how the coefficient 2 enters the concentration equation. Zero order means the reaction rate is independent of A’s concentration while the stated law applies: $Rate = k$

For the coefficient of 2 in the supplied equation:

Rate = - \frac{1}{2} \frac{d [A]}{d t} = k

Multiplying by 2 gives:

- \frac{d [A]}{d t} = 2 k = 2.0 m o l L^{- 1} m i n^{- 1}

A therefore disappears twice as fast as the normalised reaction rate. To obtain the concentration equation, separate the differentials: $d [A] = - 2 k d t$

Integrate between the initial and final concentrations, and between zero and elapsed time:

\int_{[A]_{0}}^{[A]_{t}} d [A] = - 2 k \int_{0}^{t} d t

$[A]_{t} - [A]_{0} = - 2 k t$ Rearranging:

[A]_{t} = [A]_{0} - 2 k t

The coefficient 2 fixes the relationship between the two rates. It does not make the reaction second order. Reaction order concerns concentration dependence in the rate law, not the coefficient in the balanced equation.

How do you convert 75% completion into concentration remaining?

At 75% completion, 75% of the initial A has been consumed and 25% remains. Calculate the concentration remaining separately so that you do not substitute the consumed amount in its place.

Fraction remaining = 1 - \frac{75}{100} = 0.25

[A]_{t} = 0.25 \times 2 = 0.50 M

The concentration consumed is:

[A]_{0} - [A]_{t} = 2.00 - 0.50 = 1.50 M

0.50 M is remaining A; 1.50 M is consumed A. The remaining concentration belongs on the left of the integrated equation. The consumed concentration is the drop used when calculating time directly.

How does substitution give 0.75 min, and do the units agree?

Substitution gives option B, 0.75 min, equivalent to 45 s. The concentration drop is divided by A’s disappearance rate, which is twice the supplied reaction-rate constant. The resulting unit is minutes.

Start with the integrated equation: $[A]_{t} = [A]_{0} - 2 k t$

Using concentrations in mol per litre and time in minutes: $0.5 = 2 - 2 (1) t$ $2 t = 2 - 0.5 = 1.5$

t = \frac{1.5}{2} = 0.75 m i n = 45 s

The dimensions reduce correctly:

\frac{m o l L^{- 1}}{m o l L^{- 1} m i n^{- 1}} = m i n

Check the arithmetic directly by dividing the concentration consumed by A’s constant disappearance rate:

t = \frac{1.50 M}{2.0 M m i n^{- 1}} = 0.75 m i n

Both routes track the loss of A. The calculated 45 s is the reaction duration, not a suggested time limit for solving the MCQ.

Why does option A give 1.5 min instead?

Option A results from treating the supplied reaction-rate constant as A’s disappearance-rate constant. The error happens before substitution, not during division: the calculation leaves out the relationship between the two rates.

The incorrect calculation for this question is:

t = \frac{[A]_{0} - [A]_{t}}{k} = \frac{2 - 0.5}{1} = 1.5 m i n

The familiar concentration equation is valid when its constant is defined for A’s disappearance:

- \frac{d [A]}{d t} = k_{A}

[A]_{t} = [A]_{0} - k_{A} t

Here, the two constants are related by: $k_{A} = 2 k$

There is no conflict between the integrated equations. They use constants defined for different rates, but both constants have identical units:

[k_{A}] = [k] = m o l L^{- 1} m i n^{- 1}

Dimensional analysis cannot detect the missing factor 2. It checks units, not which rate the constant describes.

Define the reaction rate, relate it to the species being tracked, and only then use the integrated equation. Do not insert a factor 2 merely because you see a coefficient 2.

How do you solve related Chemical Kinetics questions using the same rate definition?

Use A’s disappearance rate, twice the given reaction rate, in each calculation below. These are author-created practice questions, not additional NEET PYQs. They use the same normalised rate convention as the original question.

Question 1: What is the half-life with the original reaction, initial concentration and rate constant?

At half-life, half the initial A remains. Divide the concentration consumed by A’s disappearance rate:

[A]_{t} = \frac{2}{2} = 1 M

t_{1 / 2} = \frac{2 - 1}{2 \times 1} = 0.50 m i n

Question 2: With the original data, what is A’s concentration after 0.25 min?

Subtract the concentration consumed from the initial concentration. Then use that loss to calculate percentage completion:

[A]_{t} = 2 - 2 (1) (0.25) = 1.50 M

\frac{2.00 - 1.50}{2.00} \times 100 % = 25 %

Question 3: If the initial concentration doubles to 4 M and the reaction-rate constant stays unchanged, how long does 75% completion take?

One-quarter of the new initial concentration remains. The rate is unchanged, but more A must react:

[A]_{t} = 0.25 \times 4 = 1 M

t = \frac{4 - 1}{2 \times 1} = 1.50 m i n

The third answer is twice the original time because twice as much A must disappear at the same rate. Concentration-independent rate does not mean concentration-independent time for a fixed percentage completion.

All three calculations stop while A remains and assume the stated zero-order law still applies. Before your next substitution, write what the given constant measures: normalised reaction rate or reactant disappearance rate.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

Keep going with Capacitor Networks NEET 2026: Why P Stores One-Sixth.

Frequently asked questions

What is the answer to the zero-order reaction NEET 2026 question?

The answer is option B, 0.75 min, or 45 s. For 2A → B, the supplied k = 1.0 mol L⁻¹ min⁻¹ describes the normalised reaction rate, so A disappears at 2k. At 75% completion, 1.50 M of the initial 2 M is consumed, giving t = 1.50/2.0 = 0.75 min.

Why is 1.5 min wrong in this zero-order reaction question?

The 1.5 min result treats the supplied normalised reaction-rate constant as A's disappearance-rate constant. For 2A → B, A disappears at twice the normalised reaction rate, so the concentration drop must be divided by 2k. Both rate constants have the same units, so dimensional analysis cannot detect this missing factor.

How much A remains at 75% completion if the initial concentration is 2 M?

At 75% completion, 25% of the initial A remains. The remaining concentration is 0.25 × 2 = 0.50 M, while the concentration consumed is 1.50 M. Use 0.50 M as the final concentration, not the consumed amount.

Does the coefficient 2 in 2A → B make the reaction second order?

No: reaction order describes concentration dependence in the rate law, not the coefficient in the balanced equation. For the stated zero-order law, the normalised reaction rate equals k and is independent of A's concentration while that law applies. The coefficient 2 instead gives the relationship −d[A]/dt = 2k.