Explanation

For reaction II, the given rate is expressed in terms of bromide ion disappearance.

Reaction II is:

$$ \mathrm{5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O} $$

From stoichiometry, rate of reaction is related to bromide ion as:

$$ \mathrm{Rate = -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t}} $$

Given:

$$ \mathrm{-\frac{\Delta[Br^-]}{\Delta t} = 2 \times 10^{-4}\ mol\ L^{-1}\ min^{-1}} $$

Therefore, rate of reaction II is:

$$ \mathrm{Rate = \frac{1}{5} \times 2 \times 10^{-4}} $$

$$ \mathrm{Rate = 4 \times 10^{-5}\ mol\ L^{-1}\ min^{-1}} $$

It is given that the rates of reaction I and reaction II at 10.10 am are same.

So, rate of reaction I at 10.10 am is:

$$ \mathrm{Rate = 4 \times 10^{-5}\ mol\ L^{-1}\ min^{-1}} $$

Reaction I is first order:

$$ \mathrm{Rate = k[A]} $$

Given concentration of A at 10.10 am:

$$ \mathrm{[A] = 10^{-2}\ mol\ L^{-1}} $$

Substituting values:

$$ \mathrm{k = \frac{4 \times 10^{-5}}{10^{-2}}} $$

$$ \mathrm{k = 4 \times 10^{-3}\ min^{-1}} $$

Hence, the first order rate constant of reaction I is 4 × 10⁻³ min⁻¹.

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