Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. and k are constants having appropriate units.
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JEE Advanced 2022 · Paper 1, Chemistry question 15
LIST-I
- Iunder all possible initial concentrations of X
- IIwhere initial concentrations of X are much less than
- IIIwhere initial concentrations of X are much higher than
- IVwhere initial concentration of X is much higher than
LIST-II
- PPlot of half-life against initial concentration: straight line with positive slope and intercept (figure).

- QPlot of half-life against initial concentration: horizontal line (figure).

- RPlot of rate against initial concentration: rises and levels off (figure).

- SPlot of [X] against time: straight falling line (figure).
![[X] against time: a straight line falling with constant slope.](/figures/ja-2022-p1/ja-2022-p1-che-q15-f4.png)
- TPlot of ln[X] against time: straight falling line (figure).
![ln[X] against time: a straight line falling with constant slope.](/figures/ja-2022-p1/ja-2022-p1-che-q15-f5.png)
Answer and worked solution
Official answer: A
Let .
(I) The full rate law. Integrating gives
At half-life, . This is linear in with a positive intercept → P.
(II) . The rate becomes : first order, constant half-life → Q.
(III) . The rate becomes : zero order, so falls linearly with time → S.
(IV) with . The rate becomes : first order, so is linear in time → T.