JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Observe the following reactions at T(K):
I. \( A \rightarrow \text{products} \)
II. \( 5Br^{-}(aq) + BrO_{3}^{-}(aq) + 6H^{+}(aq) \rightarrow 3Br_{2}(aq) + 3H_{2}O(l) \)
Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of \( -\frac{\Delta[Br^{-}]}{\Delta t} \) at 10.10 am is \( 2 \times 10^{-4} \text{ mol L}^{-1} \text{Min}^{-1} \). The concentration of A at 10.10 am is \( 10^{-1} \text{ mol L}^{-1} \). What is the first order rate constant (in \( min^{-1} \)) of reaction I?
- A \( 2 \times 10^{-3} \)
- B \( 10^{-3} \)
- C \( 10^{-2} \)
- D \( 4 \times 10^{-4} \)
Answer & Solution
Correct Answer
(D) \( 4 \times 10^{-4} \)
Step-by-step Solution
Detailed explanation
Rate of reaction II = \( -\frac{1}{5}\frac{\Delta[Br^{-}]}{\Delta t} \) Rate of reaction II = \( \frac{1}{5} \times (2 \times 10^{-4} \text{ mol L}^{-1} \text{Min}^{-1}) = 4 \times 10^{-5} \text{ mol L}^{-1} \text{Min}^{-1} \) Rate of reaction I = Rate of reaction II =…
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