For the equilibrium $$A \rightleftharpoons B$$, the variation of the rate of the forward (a) and reverse (b) reaction with time is given by:

We are dealing with the reversible chemical change $$A \rightleftharpoons B$$. The two simultaneous reactions involved are

$$A \xrightarrow{k_f} B \quad\text{(forward)}$$

$$B \xrightarrow{k_r} A \quad\text{(reverse)}$$

At any instant the rate of the forward reaction is given, by definition, as

$$r_f = k_f\,[A]$$

and the rate of the reverse reaction is

$$r_r = k_r\,[B]$$

Initially, that is at time $$t = 0$$, only reactant $$A$$ is present while product $$B$$ has not yet been formed. Hence we have

$$[A] = [A]_0 \quad\text{and}\quad [B] = 0$$

Substituting these concentrations in the two rate expressions gives

$$r_f(t=0) = k_f\,[A]_0 \;>\; 0$$

$$r_r(t=0) = k_r\,[B]_{t=0} = k_r\times 0 = 0$$

So the forward rate starts from a finite positive value, while the reverse rate starts from zero.

As time passes, $$A$$ is consumed and its concentration $$[A]$$ continuously falls. According to the equation $$r_f = k_f\,[A]$$, the forward rate therefore keeps decreasing smoothly from its initial value. Simultaneously, $$B$$ is being produced, so $$[B]$$ keeps rising from zero. From $$r_r = k_r\,[B]$$ we see that the reverse rate increases smoothly from zero.

This opposite monotonic behaviour continues until a particular moment when

$$r_f = r_r$$

This equality of the two opposing rates is the very definition of chemical equilibrium. Once equilibrium is reached, the concentrations $$[A]$$ and $$[B]$$ no longer change with time, so both $$r_f$$ and $$r_r$$ remain constant and equal thereafter. Consequently each rate curve attains a horizontal plateau, the two plateaux being super-imposed because the numerical values of the rates are identical at equilibrium.

Putting all these deductions together, the required plot of rate versus time must possess the following characteristic features:

1. Forward curve (labelled a): starts at a non-zero value on the ordinate and falls asymptotically to a constant value.
2. Reverse curve (labelled b): starts exactly from the origin on the ordinate, rises smoothly and approaches the same constant value.
3. The two curves intersect exactly once, at the equilibrium point, and then run parallel (actually coincident) afterwards because both rates have become equal and time-independent.

Among the four sketches provided in the question, graph (2) alone shows a curve a that descends from a positive initial value, a curve b that ascends from zero, the two meeting once and then remaining flat and equal. Hence graph (2) uniquely satisfies every theoretical requirement we have just derived.

Hence, the correct answer is Option 2.