JEE Challenger
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ChemMedMain 2026 • 2nd Apr - Shift 2

Identification of Correct Statements for Rate Law and Reaction Order

Consider the reaction aX→bY\text{aX} \rightarrow \text{bY}, for which the rate constant at 30∘C30^\circ\text{C} is 1×10−3 mol−1 L s−11 \times 10^{-3}\text{ mol}^{-1}\text{ L}\text{ s}^{-1}. Which of the following statements are true ?

A. When concentration of 'X' is increased to four times, the rate of reaction becomes 16 times. B. The reaction is a second order reaction. C. The half-life period is independent of the concentration of X. D. Decomposition of N2O5\text{N}_2\text{O}_5 is an example of the above reaction. E. The plot shown below is valid for the above reaction.

Choose the correct answer from the options given below :

Question Diagram 1

Options

A

A and B Only

Correct
B

A, B and C Only

C

A, B, D and E Only

D

C and D Only

Topics & Concepts

Chemical KineticsRate of ReactionOrder of Reaction

Step-by-Step Solution

To determine the correct statements for the given reaction aX→bY\text{aX} \rightarrow \text{bY}, we first analyze the rate constant kk and its units:

Given: Rate constant, k=1×10−3 mol−1Ls−1=1×10−3 M−1s−1\text{Rate constant, } k = 1 \times 10^{-3} \text{ mol}^{-1}\text{L}\text{s}^{-1} = 1 \times 10^{-3} \text{ M}^{-1}\text{s}^{-1}

  1. Order of the Reaction: The general unit for the rate constant of an nthn^{\text{th}}-order reaction is given by: Unit of k=(mol L−1)1−ns−1=mol1−nLn−1s−1\text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{s}^{-1} = \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1}

    Comparing the powers of mol\text{mol} or L\text{L}: 1−n=−1  ⟹  n=21 - n = -1 \implies n = 2 Thus, the reaction is of second order.

    • Therefore, Statement B is TRUE.

  2. Effect of Concentration on Rate: For a second-order reaction, the rate law is expressed as: Rate=k[X]2\text{Rate} = k [\text{X}]^2

    If the concentration of X\text{X} is increased by 44 times, i.e., [X]′=4[X][\text{X}]' = 4[\text{X}], the new rate becomes: Rate′=k(4[X])2=16k[X]2=16×Rate\text{Rate}' = k (4[\text{X}])^2 = 16 k [\text{X}]^2 = 16 \times \text{Rate} Hence, the rate of reaction increases by 1616 times.

    • Therefore, Statement A is TRUE.

  3. Half-life Period: The half-life period (t1/2t_{1/2}) for a second-order reaction is given by: t1/2=1k[X]0t_{1/2} = \frac{1}{k [\text{X}]_0} The half-life is inversely proportional to the initial concentration of X\text{X}, meaning it is dependent on concentration (it is independent of concentration only for a first-order reaction).

    • Therefore, Statement C is FALSE.

  4. Decomposition of N2O5\text{N}_2\text{O}_5: The decomposition of N2O5\text{N}_2\text{O}_5 is a standard example of a first-order reaction, not a second-order reaction.

    • Therefore, Statement D is FALSE.

  5. Graphical Representation: For a first-order reaction, the integrated rate equation is: ln⁡([R]0[R])=kt\ln\left(\frac{[\text{R}]_0}{[\text{R}]}\right) = k t This indicates a straight line passing through the origin when ln⁡([R]0[R])\ln\left(\frac{[\text{R}]_0}{[\text{R}]}\right) is plotted against time (tt). For a second-order reaction, the integrated rate equation is: 1[R]−1[R]0=kt\frac{1}{[\text{R}]} - \frac{1}{[\text{R}]_0} = k t Hence, the given plot of ln⁡([R]0[R])\ln\left(\frac{[\text{R}]_0}{[\text{R}]}\right) vs. time\text{time} is valid for a first-order reaction, not a second-order reaction.

    • Therefore, Statement E is FALSE.

Conclusion: Only statements A and B are correct.

Thus, the correct answer option is A.

Identification of Correct Statements for Rate Law and Reaction Order | Chemistry PYQ Solution - JEE Challenger