Q13 / 75 · JEE Main 2026 (23 January Shift 2) · Chemistry
Observe the following reactions at $\mathrm{T}(\mathrm{K})$.
I. $\mathrm{A} \rightarrow$ products.
II. $5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_{3}{ }^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of $-\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}$ at 10.10 am is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$. The concentration of A at 10.10 am is $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the first order rate constant (in $\mathrm{min}^{-1}$) of reaction $I$ ?
I. $\mathrm{A} \rightarrow$ products.
II. $5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_{3}{ }^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of $-\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}$ at 10.10 am is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$. The concentration of A at 10.10 am is $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the first order rate constant (in $\mathrm{min}^{-1}$) of reaction $I$ ?
- A$10^{-2}$
- B$10^{-3}$
- C$2 \times 10^{-3}$
- Correct answerD$4 \times 10^{-3}$
Solution
For reaction II:
5Br$^-$ + BrO$_3^-$ + 6H$^+$ $\rightarrow$ 3Br$_2$ + 3H$_2$O,
The overall reaction rate is:
$r = -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{1}{3}\frac{\Delta[Br_2]}{\Delta t}$.
Given $\frac{\Delta[Br^-]}{\Delta t} = 2 \times 10^{-4}$ mol L$^{-1}$ min$^{-1}$
The reaction rate is $r = \frac{2 \times 10^{-4}}{5} = 4 \times 10^{-5}$ mol L$^{-1}$ min$^{-1}$.
Since the rates of both reactions are equal at 10.10 am and reaction I is first-order:
$r = k[A]$, so $4 \times 10^{-5} = k \times 10^{-2}$, giving $k = 4 \times 10^{-3}$ min$^{-1}$.
5Br$^-$ + BrO$_3^-$ + 6H$^+$ $\rightarrow$ 3Br$_2$ + 3H$_2$O,
The overall reaction rate is:
$r = -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{1}{3}\frac{\Delta[Br_2]}{\Delta t}$.
Given $\frac{\Delta[Br^-]}{\Delta t} = 2 \times 10^{-4}$ mol L$^{-1}$ min$^{-1}$
The reaction rate is $r = \frac{2 \times 10^{-4}}{5} = 4 \times 10^{-5}$ mol L$^{-1}$ min$^{-1}$.
Since the rates of both reactions are equal at 10.10 am and reaction I is first-order:
$r = k[A]$, so $4 \times 10^{-5} = k \times 10^{-2}$, giving $k = 4 \times 10^{-3}$ min$^{-1}$.
Hence, the correct answer is option: 'D'.