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

To determine the first-order rate constant for reaction I, we need to use the information given about both reactions.

The reactions are:

1. Reaction I: A \to \text{products}
2. Reaction II: 5Br^- + BrO_3^- + 6H^+ \to 3Br_2 + 3H_2O

The rate of disappearance of Br^− in reaction II is given as -\frac{d[Br^-]}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1} \text{min}^{-1}.

The stoichiometry of reaction II shows that for every 5 moles of Br^{-} consumed, 1 mole of BrO_3^{-} reacts. Therefore, the rate of the entire balanced reaction can be based on the stoichiometry and the rate of change of Br^−:

Rate of reaction II = \frac{-1}{5}\frac{d[Br^-]}{dt}

\frac{-1}{5}(2 \times 10^{-4}) = 4 \times 10^{-5} \text{ mol L}^{-1} \text{min}^{-1}

Given that the rates of both reactions are the same at 10.10 am:

Rate_{\text{reaction I}} = Rate_{\text{reaction II}}

k[A] = 4 \times 10^{-5}

where k is the rate constant for reaction I and [A] = 10^{-2} \text{ mol L}^{-1} at 10.10 am.

Therefore, solve for k:

k \cdot 10^{-2} = 4 \times 10^{-5}

k = \frac{4 \times 10^{-5}}{10^{-2}} = 4 \times 10^{-3} \text{ min}^{-1}

Thus, the first-order rate constant for reaction I is 4 \times 10^{-3} \text{ min}^{-1}.