Question:
Rate constant of decomposition of hydrogen peroxide is \(0.0204 \text{minutes}^{-1}\), calculate half life of reaction.
Rate constant of decomposition of hydrogen peroxide is \(0.0204 \text{minutes}^{-1}\), calculate half life of reaction.
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For a first-order reaction t half = 0.693 divided by k.
Updated On: Oct 1, 2026
- \(17.56\) minutes
- \(33.97\) minutes
- \(51.50\) minutes
- \(68.70\) minutes
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The decomposition of hydrogen peroxide is first order, and the half-life of a first-order reaction does not depend on concentration.
Step 2: Key Formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Step 3: Calculation:
\[ t_{1/2} = \frac{0.693}{0.0204\ \text{min}^{-1}} = 33.97\ \text{min} \]
Step 4: Check the Other Options:
17.56 min would need \(k = 0.0395\), 51.50 min would need \(k = 0.01346\), and 68.70 min would need \(k=0.0101\). Since the given \(k\) is 0.0204, only 33.97 min is consistent. Option (B) is correct.
Final Answer:
The half-life is 33.97 min, option (B). \[ \boxed{\text{(B) } 33.97\ \text{min}} \]
The decomposition of hydrogen peroxide is first order, and the half-life of a first-order reaction does not depend on concentration.
Step 2: Key Formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Step 3: Calculation:
\[ t_{1/2} = \frac{0.693}{0.0204\ \text{min}^{-1}} = 33.97\ \text{min} \]
Step 4: Check the Other Options:
17.56 min would need \(k = 0.0395\), 51.50 min would need \(k = 0.01346\), and 68.70 min would need \(k=0.0101\). Since the given \(k\) is 0.0204, only 33.97 min is consistent. Option (B) is correct.
Final Answer:
The half-life is 33.97 min, option (B). \[ \boxed{\text{(B) } 33.97\ \text{min}} \]
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