Question

Observe the following reactions at T(K).
I. $A \to$ products.
II. $5Br^- + BrO_3^- + 6H^+ \to 3Br_2 + 3H_2O$.
Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of $-\frac{d[Br^-]{dt}$ 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 $10^{-2} \text{ mol L}^{-1}$. What is the first order rate constant (in $\text{min}^{-1}$) of reaction I?}

• A. $2 \times 10^{-3}$
• B. $4 \times 10^{-3}$
• C. $10^{-2}$
• D. $10^{-3}$

Hint:

Rate of reaction relates to rate of disappearance by stoichiometric coefficients: Rate $= \frac{1}{n_i} \frac{d[X_i]}{dt}$.

Answer 1

The Correct Option is B

Rate of reaction II is defined as $R_{II} = -\frac{1}{5} \frac{d[Br^-]}{dt}$.
Given $-\frac{d[Br^-]}{dt} = 2 \times 10^{-4}$.
$R_{II} = \frac{1}{5} (2 \times 10^{-4}) = 4 \times 10^{-5}$.
Given Rate I = Rate II at 10.10 am.
$R_I = k [A] = 4 \times 10^{-5}$.
Given $[A] = 10^{-2}$.
$k (10^{-2}) = 4 \times 10^{-5} \implies k = 4 \times 10^{-3} \text{ min}^{-1}$.