Question:

For the following graphs, 

Choose from the options given below, the correct one regarding order of reaction : 
 

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A quick way to distinguish: If \(t_{1/2}\) is proportional to initial concentration, it's zero order. If \(t_{1/2}\) is independent of concentration, it's first order.
Updated On: Feb 3, 2026
  • (a) and (b) Zero order, (c) and (e) First order
  • (a) and (b) Zero order, (e) First order
  • (b) Zero order, (c) and (e) First order
  • (b) and (d) Zero order, (e) First order
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
Chemical kinetics describes how reaction rates and half-lives depend on reactant concentrations.
For Zero Order: Rate = \(k\), and \(t_{1/2} = \frac{[A]_0}{2k}\).
For First Order: Rate = \(k[A]\), and \(t_{1/2} = \frac{\ln 2}{k}\).
Step 2: Key Formula or Approach:
Analyze the functional dependence in each graph:
- Zero order: Rate is constant, \(t_{1/2} \propto [A]_0\).
- First order: Rate \(\propto [A]\), \([A]\) decreases exponentially with time.
Step 3: Detailed Explanation:
- Graph (a): Rate vs Time is horizontal. Rate is independent of time and concentration. \(\implies\) Zero Order.
- Graph (b): \(t_{1/2}\) vs Initial concentration is linear through the origin. \(\implies\) Zero Order.
- Graph (c): Concentration vs Time is a curve (exponential decay). \(\implies\) First Order.
- Graph (e): Rate vs Concentration is linear through the origin (\(Rate = k[A]\)). \(\implies\) First Order.
Combining these, (a) and (b) are zero order; (c) and (e) are first order.
Step 4: Final Answer:
The correct option is (A).
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