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Chemical Kinetics and Nuclear Chemistry question

2024 · 9 Apr · Shift 1 · Q26
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Chemical Kinetics and Nuclear Chemistry question

2024 · 9 Apr · Shift 1 · Q26

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
Given below are two statements : Statement I : The rate law for the reaction A+B→CA+B \rightarrow CA+B→C is rate (r)=k[A]2[B](r)=k[A]^2[B](r)=k[A]2[B]. When the concentration of both A\mathrm{A}A and B\mathrm{B}B is doubled, the reaction rate is increased "xxx" times. Statement II : JEE Main 2024 (Online) 9th April Morning Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 35 English The figure is showing "the variation in concentration against time plot" for a "yyy" order reaction. The Value of x+yx+yx+y is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Find xxx from Statement I

Given rate law: r=k[A]2[B]r = k[A]^2[B]r=k[A]2[B]

If both concentrations are doubled, then [A]→2[A],[B]→2[B][A] \to 2[A], \qquad [B] \to 2[B][A]→2[A],[B]→2[B]

New rate: r′=k(2[A])2(2[B])r' = k(2[A])^2(2[B])r′=k(2[A])2(2[B]) r′=k⋅4[A]2⋅2[B]=8k[A]2[B]r' = k \cdot 4[A]^2 \cdot 2[B] = 8k[A]^2[B]r′=k⋅4[A]2⋅2[B]=8k[A]2[B]

So, r′=8rr' = 8rr′=8r Hence, x=8x=8x=8

  1. Find yyy from Statement II

For standard concentration vs time plots:

  • Zero-order: straight line decrease of concentration with time
  • First-order: exponential decay curve
  • Second-order: curved decay with stronger change in slope

The given plot corresponds to a zero-order reaction, for which concentration decreases linearly with time: [A]=[A]0−kt[A] = [A]_0 - kt[A]=[A]0​−kt

Thus, y=0y=0y=0

  1. Compute x+yx+yx+y

x+y=8+0=8x+y = 8+0 = 8x+y=8+0=8

  1. Compare with stored answer

Stored correct answer = 888

Our derived answer matches it.

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