Question

Consider a successive reaction (all first order) \[ A \xrightarrow{k_1} B \xrightarrow{k_2} C \xrightarrow{k_3} D \] The incorrect statement is:

• A. Concentration of \(A\) decreases exponentially with time
• B. Concentration of both \(B\) and \(C\) first increases, reaches maxima, then decreases
• C. If \(k_1
• D. If \(k_1>k_2\) and \(k_2

Hint:

In consecutive first-order reactions:
Slower formation + faster consumption → smaller maximum concentration.
Faster formation + slower consumption → larger accumulation. Always compare rate constants to judge accumulation of intermediates.

Answer 1

The Correct Option is C

Step 1: Behaviour of reactant \(A\). For a first-order reaction, \[ [A]=[A]_0 e^{-k_1 t} \] Thus, concentration of \(A\) decreases exponentially with time. \[ \Rightarrow \text{Statement (A) is correct.} \]
Step 2: Behaviour of intermediates \(B\) and \(C\). In a consecutive reaction: \[ A \rightarrow B \rightarrow C \rightarrow D \] each intermediate is: 
formed from the previous species, 
consumed to form the next species. Hence, both \(B\) and \(C\) first increase, attain a maximum value, and then decrease. \[ \Rightarrow \text{Statement (B) is correct.} \] 
Step 3: Compare \([B]_{\max}\) and \([C]_{\max}\) for k_1 Here:  k_1 \(B\) is formed slowly (small \(k_1\)) but consumed faster (larger \(k_2\)), 
\(C\) is formed relatively faster (from \(B\)) and consumed very fast (\(k_3\)). Under these conditions, accumulation of \(B\) is less compared to \(C\). \[ \Rightarrow [B]_{\max} < [C]_{\max} \] But statement (C) claims: \[ [B]_{\max} > [C]_{\max} \] which is incorrect. 
Step 4: Analyse statement (D). If: \[ k_1 > k_2 \quad \text{and} \quad k_2 < k_3 \] 
\(B\) is formed rapidly (large \(k_1\)), 
\(B\) is consumed slowly (small \(k_2\)), Hence, \(B\) accumulates more than \(C\). \[ \Rightarrow [B]_{\max} > [C]_{\max} \] So statement (D) is correct. 
Hence, the incorrect statement is \[ \boxed{\text{(C)}} \]