The graph which represents the following reaction is:
$$(\text{C}_6\text{H}_5)_3\text{C} - \text{Cl} \xrightarrow[\text{Pyridine}]{\text{OH}^-} (\text{C}_6\text{H}_5)_3\text{C} - \text{OH}$$

The substrate is triphenylmethyl chloride, $$\left(C_6H_5\right)_3CCl$$.

Due to the presence of three bulky phenyl groups around the carbon atom, backside attack by the nucleophile is highly hindered. Therefore, an $$S_N2$$ mechanism is not feasible.

When the carbon-chlorine bond breaks, the triphenylmethyl carbocation,

$$\left(C_6H_5\right)_3C^+$$

is formed.

This carbocation is highly stable because the positive charge is extensively delocalized over the three phenyl rings through resonance.

Hence, the reaction proceeds through an $$S_N1$$ mechanism.

In an $$S_N1$$ reaction, the slow and rate-determining step is the formation of the carbocation.

Therefore, the rate depends only on the concentration of the substrate and is independent of the concentration of the nucleophile.

The rate law is

$$\text{Rate}=k\left[\left(C_6H_5\right)_3CCl\right]$$

Thus, the reaction is first order with respect to the substrate and zero order with respect to $$OH^-$$.

A plot of rate versus substrate concentration must therefore be a straight line passing through the origin.

A plot of rate versus $$OH^-$$ concentration must be a horizontal line because the rate is independent of the nucleophile concentration.

If Option A shows a horizontal line against

$$\left[\left(C_6H_5\right)_3CCl\right]$$

then it cannot be correct, since that would imply zero-order dependence on the substrate.

The graph that correctly represents the rate law

$$\text{Rate}=k\left[\left(C_6H_5\right)_3CCl\right]$$

is the straight-line graph of rate versus substrate concentration.

Hence, the correct graph is Option C.