Question:
What is the concentration (in mol L\(^{-1}\)) of the product after 20 s in the following reaction:
\[
\text{A} \rightarrow 3 \text{B}, \quad \text{rate} = k[\text{A}]^0
\]
What is the concentration (in mol L\(^{-1}\)) of the product after 20 s in the following reaction:
\[ \text{A} \rightarrow 3 \text{B}, \quad \text{rate} = k[\text{A}]^0 \]
\[ \text{A} \rightarrow 3 \text{B}, \quad \text{rate} = k[\text{A}]^0 \]
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For zero-order reactions, use \([A] = [A]_0 - k t\) and account for stoichiometry when calculating product concentration.
Updated On: Jun 26, 2026
- 6.6 × 10\(^{-2}\)
- 1.32 × 10\(^{-1}\)
- 1.98 × 10\(^{-1}\)
- 2.2 × 10\(^{-2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Identify reaction order.
Rate law given: \( \text{rate} = k[\text{A}]^0 \) → zero-order reaction with respect to A.
Step 2: Zero-order rate equation.
\[ [A] = [A]_0 - k t \]
Step 3: Determine k using data at t = 15 s.
\[ [A] = 0.05 = 0.1 - k (15) \implies k = \frac{0.1 - 0.05}{15} = 3.33 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} \]
Step 4: Find [A] at t = 20 s.
\[ [A] = 0.1 - 3.33 \times 10^{-3} \times 20 = 0.0334 \text{ mol L}^{-1} \]
Step 5: Calculate concentration of product B.
From stoichiometry, 1 mol A produces 3 mol B: \[ [B] = 3 ([A]_0 - [A]) = 3 (0.1 - 0.0334) = 0.198 \approx 1.98 \times 10^{-1} \text{ mol L}^{-1} \]
Step 6: Conclusion.
The concentration of the product after 20 s is \[ \boxed{1.98 \times 10^{-1} \text{ mol L}^{-1}} \]
Rate law given: \( \text{rate} = k[\text{A}]^0 \) → zero-order reaction with respect to A.
Step 2: Zero-order rate equation.
\[ [A] = [A]_0 - k t \]
Step 3: Determine k using data at t = 15 s.
\[ [A] = 0.05 = 0.1 - k (15) \implies k = \frac{0.1 - 0.05}{15} = 3.33 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} \]
Step 4: Find [A] at t = 20 s.
\[ [A] = 0.1 - 3.33 \times 10^{-3} \times 20 = 0.0334 \text{ mol L}^{-1} \]
Step 5: Calculate concentration of product B.
From stoichiometry, 1 mol A produces 3 mol B: \[ [B] = 3 ([A]_0 - [A]) = 3 (0.1 - 0.0334) = 0.198 \approx 1.98 \times 10^{-1} \text{ mol L}^{-1} \]
Step 6: Conclusion.
The concentration of the product after 20 s is \[ \boxed{1.98 \times 10^{-1} \text{ mol L}^{-1}} \]
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