Consider the kinetic data given in the following table for the reaction $\mathrm{A}+\mathrm{B}+\mathrm{C} \rightarrow$ Product. | Experiment No. | [A] $\left(\mathrm{mol} \mathrm{dm}^{-3}\right)$ | [B] $\left(\mathrm{mol} \mathrm{dm}^{-3}\right)$ | [C] $\left(\mathrm{mol} \mathrm{dm}^{-3}\right)$ | Rate of reaction $\left(\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\right)$ | | --- | --- | --- | --- | --- | | 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 $[\mathrm{A}]=0.15 \mathrm{~mol} \mathrm{dm}^{-3},[\mathrm{~B}]=0.25 \mathrm{~mol} \mathrm{dm}^{-3}$ and $[\mathrm{C}]=0.15 \mathrm{~mol} \mathrm{dm}^{-3}$ is found to be $\mathbf{Y} \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. The value of $\mathbf{Y}$ is ____

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Chemical Kinetics

Consider the kinetic data given in the following table for the reaction

A + B + C \to

Product.

Experiment No.	[A] $(mol dm^{- 3})$	[B] $(mol dm^{- 3})$	[C] $(mol dm^{- 3})$	Rate of reaction $(mol dm^{- 3} s^{- 1})$
1	0.2	0.1	0.1	$6.0 \times 1 0^{- 5}$
2	0.2	0.2	0.1	$6.0 \times 1 0^{- 5}$
3	0.2	0.1	0.2	$1.2 \times 1 0^{- 4}$
4	0.3	0.1	0.1	$9.0 \times 1 0^{- 5}$

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 1 0^{- 5} mol dm^{- 3} s^{- 1}

. The value of

Y

is ____

Answer: 6.75

Rate law:

r = k [A]^{n_{1}} [B]^{n_{2}} [C]^{n_{3}}

.

Comparing experiments 1 and 2 (B doubles, rate unchanged):

n_{2} = 0

.

Comparing experiments 1 and 3 (C doubles, rate doubles):

n_{3} = 1

.

Comparing experiments 1 and 4 (A increases 1.5×, rate increases 1.5×):

n_{1} = 1

.

So

r = k [A] [C]

.

From experiment 1:

k = \frac{6.0 \times 1 0 ^{- 5}}{0.2 \times 0.1} = 3 \times 1 0^{- 3} mol^{- 1} dm^{3} s^{- 1}

For the given concentrations:

r = (3 \times 1 0^{- 3}) \times 0.15 \times 0.15 = 6.75 \times 1 0^{- 5} mol dm^{- 3} s^{- 1}

Y = 6.75

. The source key accepts 6.70 to 6.80.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer