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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jul · Shift 1 · Q21
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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jul · Shift 1 · Q21

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

The reaction between X and Y is first order with respect to X and zero order with respect to Y.

Experiment [X]mol L−1{{[X]} \over {mol\,{L^{ - 1}}}}molL−1[X]​ [Y]mol L−1{{[Y]} \over {mol\,{L^{ - 1}}}}molL−1[Y]​ Initial ratemol L−1 min⁡−1{{Initial\,rate} \over {mol\,{L^{ - 1}}\,{{\min }^{ - 1}}}}molL−1min−1Initialrate​
I 0.1 0.1 2×10−32 \times {10^{ - 3}}2×10−3
I L 0.2 4×10−34 \times {10^{ - 3}}4×10−3
III 0.4 0.4 M×10−3M \times {10^{ - 3}}M×10−3
IV 0.1 0.2 2×10−32 \times {10^{ - 3}}2×10−3

Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)

Numerical answer
View written solutionFree

Correct answer: 40

  1. Write the rate law

Since the reaction is first order in XXX and zero order in YYY, the rate law is

r=k[X]1[Y]0=k[X]r = k[X]^1[Y]^0 = k[X]r=k[X]1[Y]0=k[X]

So, the rate depends only on [X][X][X].

  1. Use Experiment I to find kkk

From Experiment I:

[X]=0.1,r=2×10−3[X] = 0.1, \qquad r = 2\times 10^{-3}[X]=0.1,r=2×10−3

Using r=k[X]r = k[X]r=k[X],

2×10−3=k(0.1)2\times 10^{-3} = k(0.1)2×10−3=k(0.1)

k=2×10−30.1=2×10−2k = \frac{2\times 10^{-3}}{0.1} = 2\times 10^{-2}k=0.12×10−3​=2×10−2

  1. Find LLL using Experiment II

In Experiment II:

[Y]=0.2,r=4×10−3[Y]=0.2, \qquad r = 4\times 10^{-3}[Y]=0.2,r=4×10−3

Since rate is independent of YYY,

r=k[X]=(2×10−2)Xr = k[X] = (2\times 10^{-2})Xr=k[X]=(2×10−2)X

So,

4×10−3=(2×10−2)L4\times 10^{-3} = (2\times 10^{-2})L4×10−3=(2×10−2)L

L=4×10−32×10−2=0.2L = \frac{4\times 10^{-3}}{2\times 10^{-2}} = 0.2L=2×10−24×10−3​=0.2

  1. Find MMM using Experiment III

In Experiment III:

[X]=0.4,[Y]=0.4,r=M×10−3[X]=0.4, \qquad [Y]=0.4, \qquad r = M\times 10^{-3}[X]=0.4,[Y]=0.4,r=M×10−3

Again,

r=k[X]=(2×10−2)(0.4)=8×10−3r = k[X] = (2\times 10^{-2})(0.4) = 8\times 10^{-3}r=k[X]=(2×10−2)(0.4)=8×10−3

Thus,

M=8M = 8M=8

  1. Calculate the ratio of numerical values of MMM and LLL

We have:

M=8,L=0.2M = 8, \qquad L = 0.2M=8,L=0.2

Therefore,

ML=80.2=40\frac{M}{L} = \frac{8}{0.2} = 40LM​=0.28​=40

  1. Answer

The required nearest integer is

40\boxed{40}40​

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