Question

The reaction, \( A \rightarrow \) Products, follows first-order kinetics. If \([A]\) represents the concentration of reactant at time \( t \), the INCORRECT variation is shown in 

 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

For first-order reactions, the concentration of the reactant decreases exponentially, and the plot of \( \ln[A] \) versus time is linear with a negative slope.

Answer 1

The Correct Option is B

First-order reaction kinetics:

For the reaction A → Products with first-order kinetics:

Rate law: $-\frac{d[A]}{dt} = k[A]$

Integrated form: $[A] = [A]_0 e^{-kt}$ or $\ln[A] = \ln[A]_0 - kt$

Expected variations for first-order kinetics:

(A) [A] vs t:

• Should show exponential decay curve
• Correct representation

(B) -d[A]/dt vs [A]:

• From the rate law: $-\frac{d[A]}{dt} = k[A]$
• This is a direct proportionality
• Should show a straight line through the origin with slope k
• If the graph shows anything other than a straight line (such as a curved line), it is INCORRECT

(C) -d[A]/dt vs t:

• Since rate = k[A] and [A] decreases exponentially with time
• The rate also decreases exponentially
• Should show exponential decay curve
• Correct representation

(D) log[A] vs t:

• From: $\log[A] = \log[A]_0 - \frac{kt}{2.303}$
• Should show a straight line with negative slope
• Correct representation

Identifying the incorrect plot:

For first-order kinetics, the plot of $-\frac{d[A]}{dt}$ versus $[A]$ must be a straight line through the origin, since the rate is directly proportional to concentration.

If graph (B) shows a curved line instead of a straight line, it represents an INCORRECT variation.

Answer: (B)