For the reaction \( R \to P \), half life is independent of initial concentration of the reactant, R. Which one of the following graphs is not correct for this reaction?
Show Hint
Key graphical signatures:
- Zero Order: \( [R] \) vs \( t \) is linear.
- First Order: \( \ln[R] \) vs \( t \) is linear.
- Second Order: \( 1/[R] \) vs \( t \) is linear.
Step 1: Identify Reaction Order:
The statement "half life is independent of initial concentration" defines a First Order Reaction.
For 1st order: \( t_{1/2} = \frac{0.693}{k} \) (Constant).
Step 2: Analyze Validity of Graphs for First Order:
Option A: The integrated rate equation is \( \ln[R] = \ln[R]_0 - kt \). Plotting \( \ln[R] \) vs \( t \) gives a straight line with slope \( -k \). Correct.
Option B: Plotting \( [R] \) vs \( t \). Since \( [R] = [R]_0 e^{-kt} \), this graph should be an exponential decay curve. A straight line with negative slope \( -k \) corresponds to a Zero Order reaction (\( [R] = [R]_0 - kt \)). Thus, this graph is Incorrect for a first-order reaction.
Option D: From the rate law, \( \log \frac{[R]_0}{[R]} = \frac{k}{2.303} t \). Plotting \( \log \frac{[R]_0}{[R]} \) vs \( t \) gives a straight line passing through origin with slope \( k/2.303 \). Correct.
Step 3: Conclusion:
Graph (B) represents a zero-order reaction, not first-order.
Step 4: Final Answer:
Graph (B) is not correct.
