Chemical Kinetics JEE 2025: Pseudo-First-Order Rate Error

What is the answer to the chemical kinetics JEE 2025 rate-error question?

The chemical kinetics JEE 2025 answer is 04.17% because the error is measured against the true rate, not the approximate rate. This question comes from JEE Advanced 2025, Paper 2, Chemistry. It is a numerical-answer question, not an MCQ.

The reaction and its measured rate law are:

A + R \to Product, r = k [A] [R]

Initially, the concentration of R is ten times that of A. The pseudo-first-order approximation treats R as unchanged, fixing the effective rate constant at: $k^{'} = k [R]_{0}$

The task is to find the percentage relative rate error after 40% of A has reacted. Both rates must be calculated at that same conversion.

The question bank tags this problem as medium difficulty, with a 120-second target. Neither is an official exam classification.

How much of each reactant remains after 40% completion?

A retains 60% of its initial concentration, while R retains 96%. The reaction consumes equal amounts of A and R because their stoichiometric ratio is one to one. Equal amounts consumed do not mean equal percentages consumed when the starting concentrations differ.

Follow the official solution by defining the initial concentrations:

[A]_{0} = A_{0}, [R]_{0} = 10 A_{0}

At 40% completion, the concentration of A consumed is:

Δ [A]_{consumed} = 0.40 A_{0}

The same concentration of R is consumed:

Δ [R]_{consumed} = 0.40 A_{0}

Subtract that amount from each starting concentration:

[A] = A_{0} - 0.40 A_{0} = 0.60 A_{0}

[R] = 10 A_{0} - 0.40 A_{0} = 9.60 A_{0}

The percentage decreases therefore differ:

A depletion = \frac{0.40 A_{0}}{A_{0}} \times 100 = 40 %

R depletion = \frac{0.40 A_{0}}{10 A_{0}} \times 100 = 4 %

Do not reduce R by 40%. Stoichiometry fixes the amount consumed; the initial concentration determines what percentage that amount represents.

How do you calculate the true and pseudo-first-order rates?

Use the remaining A concentration in both rates, but use different R concentrations. The true rate uses the R actually left. The approximation uses its initial concentration, so it gives a higher rate.

The measured law is first order in each reactant and therefore second order overall. Substituting the remaining concentrations gives:

r = k [A] [R] = k (0.60 A_{0}) (9.60 A_{0}) = 5.76 k A_{0}^{2}

For the pseudo-first-order calculation, freeze R at its initial concentration:

[R]_{approx} = [R]_{0} = 10 A_{0}

This gives a nominally constant effective rate constant: $k^{'} = k [R]_{0} = 10 k A_{0}$

The approximate rate is then first order in A:

r^{'} = k^{'} [A] = (10 k A_{0}) (0.60 A_{0}) = 6 k A_{0}^{2}

The two rates at the same conversion are:

\underset{true rate}{\underset{⏟}{r = 5.76 k A_{0}^{2}}} \underset{approximate rate}{\underset{⏟}{r^{'} = 6 k A_{0}^{2}}}

The approximation freezes R, not A. Both calculations use the same remaining A concentration, so their difference isolates the error caused by treating R as constant. The approximate rate exceeds the true rate because it uses more R than actually remains.

Which rate belongs in the relative-error denominator?

Divide by the true rate. Relative error measures the approximation’s departure from the actual value, using that actual value as the reference. Since the approximation overestimates the rate here, the difference is positive.

Percentage relative error = \frac{r^{'} - r}{r} \times 100

Substitute both rates:

= \frac{6 k A_{0}^{2} - 5.76 k A_{0}^{2}}{5.76 k A_{0}^{2}} \times 100

First subtract in the numerator, then cancel the common factor:

= \frac{0.24 k A_{0}^{2}}{5.76 k A_{0}^{2}} \times 100 = \frac{0.24}{5.76} \times 100

$= 4.1667 % \approx 4.17 %$ The numerical entry matching the supplied official answer is:

04.17

No absolute initial concentration is needed because the common concentration factor cancels. No integrated rate law is needed either: the question gives the conversion and asks for an instantaneous rate error, not the time taken to reach that conversion.

Why does dividing by the approximate rate give the wrong 4.00% result?

That calculation measures the difference relative to the approximation, not the true rate, producing 4.00% instead of 4.17%. This numerical-answer question has no supplied options. Here, 4.00% is a derived wrong result, not an official distractor.

The incorrect calculation is:

\frac{6 - 5.76}{6} \times 100 = 4.00 %

The denominator represents the approximate rate. Using it changes the reference quantity and answers a different question.

Compare the two valid but different quantities:

R depletion relative to its initial concentration:

\frac{0.40}{10} \times 100 = 4 %

Rate overestimation relative to the true rate:

\frac{0.40}{9.60} \times 100 = 4.1667 %

A 4% decrease in R therefore does not imply a 4% overestimate relative to the true rate. Identify the reference quantity before substituting into any relative-error formula.

How does the error change with more excess reactant or greater conversion?

More excess R reduces the error at fixed conversion; greater conversion increases it at a fixed initial ratio. The answers below are approximately 2.04% and 5.26%, respectively. Both questions are original practice variations, not additional verified PYQs.

What is the error with twentyfold initial R at 40% completion?

Original practice variation 1 gives approximately 2.04% relative rate error. Keep the same reaction and measured rate law, but use these initial concentrations:

[A]_{0} = A_{0}, [R]_{0} = 20 A_{0}

Equal amounts are consumed, giving:

[A] = 0.60 A_{0}, [R] = 19.60 A_{0}

Calculate the true rate and then the approximate rate:

r = k (0.60 A_{0}) (19.60 A_{0}) = 11.76 k A_{0}^{2}

r^{'} = k (0.60 A_{0}) (20 A_{0}) = 12 k A_{0}^{2}

Error = \frac{12 - 11.76}{11.76} \times 100 = \frac{0.24}{11.76} \times 100 \approx 2.04 %

What is the error with tenfold initial R at 50% completion?

Original practice variation 2 gives approximately 5.26% relative rate error. Keep the same reaction and measured rate law, restore the original initial ratio, and consume half the starting A:

[A]_{0} = A_{0}, [R]_{0} = 10 A_{0}

The remaining concentrations and corresponding rates are:

[A] = 0.50 A_{0}, [R] = 10 A_{0} - 0.50 A_{0} = 9.50 A_{0}

r = k (0.50 A_{0}) (9.50 A_{0}) = 4.75 k A_{0}^{2}

r^{'} = k (0.50 A_{0}) (10 A_{0}) = 5 k A_{0}^{2}

Error = \frac{5 - 4.75}{4.75} \times 100 = \frac{0.25}{4.75} \times 100 \approx 5.26 %

Greater excess improves the approximation at fixed conversion; greater conversion worsens it at a fixed initial ratio. Redo both variations with the answers covered: subtract equal amounts, calculate both rates at the same conversion, and divide their difference by the true rate.

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Frequently asked questions

What is the answer to the JEE Advanced 2025 chemical kinetics rate-error question?

The percentage relative rate error is approximately 4.17%, with the supplied official numerical answer written as 04.17. Calculate it as (approximate rate − true rate) / true rate × 100. The pseudo-first-order approximation overestimates the rate because it treats the excess reactant concentration as unchanged.

Why does R decrease by only 4% when 40% of A reacts?

The reaction consumes A and R in a 1:1 ratio, but R starts at ten times the concentration of A. Consuming 40% of the initial A therefore consumes only 4% of the initial R. A retains 60% of its initial concentration, while R retains 96%.

Why is the rate error 4.17% instead of 4%?

Relative rate error uses the true rate as its denominator, giving (6 − 5.76) / 5.76 × 100 = 4.17% after cancelling common factors. Dividing by the approximate rate instead gives (6 − 5.76) / 6 × 100 = 4.00%. That uses a different reference quantity and does not answer the question asked.

Do I need an integrated rate law for the JEE 2025 rate-error question?

No integrated rate law is needed because the question gives the conversion and asks for an instantaneous rate error, not elapsed time. Use stoichiometry to find the remaining concentrations, then calculate the true and approximate rates at that same conversion.