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Chemical Kinetics and Nuclear Chemistry question

2019 · Shift 1 · Q15
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Chemical Kinetics and Nuclear Chemistry question

2019 · Shift 1 · Q15

JEE AdvancedChemistryChemical Kinetics and Nuclear ChemistryNumerical+3 / −1
Consider the kinetic data given in the following table for the reaction A + B + C →\to→ Product. JEE Advanced 2019 Paper 1 Offline Chemistry - Chemical Kinetics and Nuclear Chemistry Question 17 English The rate of the reaction for [A] = 0.15 mol dm-3, [B] = 0.25 mol dm-3 and [C] = 0.15 mol dm-3 is found to be Y ×\times× 10-5 mol dm-3s-1. The value of Y is .................
Numerical answer
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Correct answer: 6.75

The question refers to kinetic data given in a table, but in the prompt provided, the actual table is missing. Without that table, the reaction orders with respect to AAA, BBB, and CCC, and the rate constant kkk, cannot be determined uniquely.

For such a reaction, the rate law would be of the form:

r=k[A]m[B]n[C]pr = k[A]^m[B]^n[C]^pr=k[A]m[B]n[C]p

To calculate the rate at

[A]=0.15 mol dm−3,[B]=0.25 mol dm−3,[C]=0.15 mol dm−3[A]=0.15\ \text{mol dm}^{-3},\quad [B]=0.25\ \text{mol dm}^{-3},\quad [C]=0.15\ \text{mol dm}^{-3}[A]=0.15 mol dm−3,[B]=0.25 mol dm−3,[C]=0.15 mol dm−3

we must know m,n,pm,n,pm,n,p and kkk, which are normally obtained from the missing table.

Since the table is not available in the prompt, a rigorous derivation is not possible from the given information alone.

However, the stored correct answer is given as:

Y=6.75Y = 6.75Y=6.75

So the rate would be

6.75×10−5 mol dm−3s−16.75\times 10^{-5}\ \text{mol dm}^{-3}\text{s}^{-1}6.75×10−5 mol dm−3s−1

Thus, the required integer/numerical value is:

6.75\boxed{6.75}6.75​

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