Consider the kinetic data given in the following table for the reaction A + B + C $\to$Product.

Experiment No.	[A] (mol $d{m}^{-3}$ )	[B] (mol $d{m}^{-3}$ )	[C] (mol $d{m}^{-3}$ )	Rate of reaction (mol $d{m}^{-3} s^{-1}$ )
1	0.2	0.1	0.1	$6.0\times {10}^{-5}$
2	0.2	0.2	0.1	$6.0\times {10}^{-5}$
3	0.2	0.1	0.2	$1.2\times {10}^{-4}$
4	0.3	0.1	0.1	$9.0\times {10}^{-5}$

The rate of the reaction for [A] = 0.15 mol $d{m}^{-3},$ [B] = 0.25 mol $d{m}^{-3}$ and [C] = 0.15 mol $d{m}^{-3}$ is found to be $Y\times {10}^{-5}$ mol $d{m}^{-3}{s}^{-1}.$ The value of Y is______

Kinetics Problem: Rate Law Determination

Consider the reaction rate expression:

rate = k [A]^x [B]^y [C]^z

Using experimental data analysis:

• From experiments 1 and 2: With [A] and [C] constant and [B] doubled, rate does not change → y = 0
• From experiments 1 and 3: With [B] and [C] constant and [A] halved, rate halves → x = 1
• From experiments 1 and 3: Doubling [C] doubles rate → z = 1

Therefore, rate law is:

rate = k [A][C]

Calculating rate constant k using experiment 1:

6.0 × 10^-5 = k × (0.2) × (0.1) ⇒ k = 3.0 × 10^-3

Calculating rate for [A] = 0.15, [B] = 0.25, and [C] = 0.15:

rate = 3.0 × 10^-3 × 0.15 × 0.15 = 6.75 × 10^-5 mol dm⁻³ s⁻¹

Final answer: The value of Y is 6.75.