Consider the following reaction, the rate expression of which is given below: \( \text{A} + \text{B} \to \text{C} \) \(\text{rate} = k [\text{A}]^{1/2} [\text{B}]^{1/2}\) The reaction is initiated by taking 1M concentration of A and B each. If the rate constant (\(k\)) is \(4.6 \times 10^{-2} \, \text{s}^{-1}\), then the time taken for A to become 0.1 M is ______ sec. (nearest integer)
Step 1: Given data
Reaction: A + B → C
Rate law: rate = k [A]¹ᐟ² [B]¹ᐟ²
Initial concentrations: [A]₀ = [B]₀ = 1 M
Rate constant: k = 4.6 × 10⁻² s⁻¹
We need the time when [A] = 0.1 M.
Step 2: Simplify the rate law using stoichiometry
Since A and B react in a 1:1 ratio and both start at the same concentration, their concentrations will remain equal throughout the reaction.
Therefore, [A] = [B] = x at any time t.
So, the rate law becomes:
\[
-\frac{d[A]}{dt} = k [A]^{1/2} [B]^{1/2} = k [A]
\]
Hence, the reaction follows an effective **first-order** rate law with respect to [A].
Step 3: Write and integrate the rate equation
For a first-order reaction:
\[
-\frac{d[A]}{dt} = k [A]
\]
\[
\Rightarrow \frac{d[A]}{[A]} = -k\,dt
\]
Integrating from [A]₀ to [A]:
\[
\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k \int_{0}^{t} dt
\]
\[
\ln\frac{[A]_0}{[A]} = k t
\]
\[
t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right)
\]
Substitute values:
\[
t = \frac{1}{4.6 \times 10^{-2}} \ln\left(\frac{1}{0.1}\right)
\]
\[
t = \frac{1}{4.6 \times 10^{-2}} \ln(10)
\]
\[
\ln(10) = 2.303
\]
\[
t = \frac{2.303}{4.6 \times 10^{-2}} = 50.07 \, \text{s}
\]
So, \( t \approx 50 \, \text{s} \).
