Question
Observe the following reactions at $T(K)$
I. A → products.
II. $5{Br}^{-}(aq) + {BrO}_{3}^{-}(aq) + 6H^{+}(aq) \rightarrow 3{Br}_{2}(aq) + 3H_{2}O(l)$
Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of $- \frac{\Delta\left\lbrack {Br}^{-} \right\rbrack}{\Delta t}$ at 10.10 am is $2 \times 10^{- 4}\text{ }mol{\text{ }L}^{- 1}{Min}^{- 1}$. The concentration of A at 10.10 am is $10^{- 2}\text{ }mol{\text{ }L}^{- 1}$. What is the first order rate constant (in $\min^{- 1}$ ) of reaction I?
Options
Solution
At $t = 10$ minutes
Rate of reaction $= - \frac{1}{5}\frac{\Delta\left\lbrack {Br}^{-} \right\rbrack}{\Delta t} = \frac{1}{5} \times \left( 2 \times 10^{- 4} \right) = 4 \times 10^{- 5}$
For reaction $A \rightarrow P$
at $t = 10$ minutes
Rate of reaction $= 4 \times 10^{- 5} = k\lbrack A\rbrack$
$k = 4 \times 10^{- 3}{\text{ }\min}^{- 1}$.
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