Match List I with List II List I List II A. Linear plot with +ve slop…

The correct answer is option 3. A-III, B-IV, C-II, D-I.

List I	List II
A. Linear plot with +ve slope and an intercept	III. \(\frac{1}{[A]}\) vs time for the 2nd order
B. Linear plot with -ve slope and an intercept	IV. Conc. \([A]\) vs time for zero order
C. Linear horizontal plot	II. \(t_{1/2}\) vs \([A]_0\) for 1st order
D. Linear plot passing through the origin	I. \(t_{1/2}\) vs \(\frac{1}{[A]_0}\) for 2nd order

Let us delve into the details of each match in the context of chemical kinetics.

A. Linear plot with +ve slope and an intercept: III. \(\frac{1}{[A]}\) vs time for the 2nd order

For a second-order reaction, the integrated rate law can be expressed as:

\(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\)

Here:

\([A]\) is the concentration of the reactant at time \(t\),

\(k\) is the rate constant,

\([A]_0\) is the initial concentration of the reactant.

This equation is of the form \(y = mx + c\), where:

\(y = \frac{1}{[A]}\),

\(x = t\),

the slope \(m = k\) (which is positive for a second-order reaction),

the intercept \(c = \frac{1}{[A]_0}\).

A plot of \(\frac{1}{[A]}\) vs. \(t\) will give a straight line with a positive slope \(k\) and an intercept equal to \(\frac{1}{[A]_0}\).

B. Linear plot with -ve slope and an intercept: IV. Conc. \([A]\) vs time for zero order

In a zero-order reaction, the concentration of the reactant \([A]\) decreases linearly with time. The rate of the reaction is independent of the concentration of the reactant, and the equation for a zero-order reaction can be expressed as:

\([A] = [A]_0 - kt\)

Where:

\([A]\) is the concentration of the reactant at time \(t\).

\([A]_0\) is the initial concentration of the reactant.

\(k\) is the rate constant (with units of concentration/time).

\(t\) is the time.

This equation represents a straight line with:

A negative slope equal to \(-k\).

A y-intercept equal to \([A]_0\).

The linear plot of \([A]\) vs. time will have a downward (negative) slope, indicating the decrease in concentration over time.

C. Linear horizontal plot: II. \(t_{1/2}\) vs \([A]_0\) for 1st order

For a first-order reaction, the half-life \(\left(t_{1/2}\right)\) is independent of the initial concentration \([A]_0\). The formula for the half-life of a first-order reaction is:

\(t_{1/2} = \frac{0.693}{k}\)

Here, \(t_{1/2}\) is the half-life, and \(k\) is the rate constant. Since \(t_{1/2}\) does not depend on \([A]_0\), a plot of \(t_{1/2}\) versus \([A]_0\) for a first-order reaction will be a horizontal line, indicating that \(t_{1/2}\) remains constant regardless of the initial concentration.

D. Linear plot passing through the origin: I. \(t_{1/2}\) vs \(\frac{1}{[A]_0}\) for 2nd order

For a second-order reaction, the half-life \(\left(t_{1/2}\right)\) is inversely proportional to the initial concentration \(\left([A]_0\right)\). The formula for the half-life of a second-order reaction is:

\(t_{1/2} = \frac{1}{k[A]_0}\)

Here:

\(t_{1/2}\) is the half-life,

\(k\) is the rate constant,

\([A]_0\) is the initial concentration of the reactant.

This equation can be rewritten as:

\(t_{1/2} = \frac{1}{k} \times \frac{1}{[A]_0}\)

This is of the form \(y = mx\), where:

\(y = t_{1/2}\),

\(x = \frac{1}{[A]_0}\),

and the slope \(m = \frac{1}{k}\).

A plot of \(t_{1/2}\) versus \(\frac{1}{[A]_0}\) for a second-order reaction will be a straight line that passes through the origin, with the slope equal to \(\frac{1}{k}\).

Thus, the correct matching is option 3: A-III, B-IV, C-II, D-I.