Question

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)

Answer 1

Correct Answer: 50

For this reaction:
\[K = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\]
Given: \( k = 4.6 \times 10^{-2} \, \text{s}^{-1}, \, [A]_0 = 1 \, \text{M}, \, [A] = 0.1 \, \text{M} \)
\[4.6 \times 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}\]
\[4.6 \times 10^{-2} = \frac{2.303}{t} \times 1\]
\[t = \frac{2.303}{4.6 \times 10^{-2}} \approx 50 \, \text{sec.}\]

Answer 2

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, \( t \approx 50 \, \text{s} \).

Step 4: Final answer

50