Consider the following molecules:

The correct order of rate of hydrolysis is:

The compounds are:

• $$p:\ CH_3CH_2COCl$$ (acid chloride)
• $$q:\ CH_3CH_2CO-O-COCH_3$$ (acid anhydride)
• $$r:\ CH_3CH_2COOCH_2CH_3$$ (ester)
• $$s:\ CH_3CH_2CONH_2$$ (amide)

The rate of hydrolysis of carboxylic acid derivatives depends on the leaving group attached to the carbonyl carbon.

Greater the leaving ability of the group, faster is the hydrolysis.

The leaving groups in the given compounds are:

• In $$p$$, leaving group is $$Cl^-$$
• In $$q$$, leaving group is $$RCOO^-$$
• In $$r$$, leaving group is $$RO^-$$
• In $$s$$, leaving group is $$NH_2^-$$

The leaving group ability follows:

$$Cl^- > RCOO^- > RO^- > NH_2^-$$

Hence, the ease of hydrolysis follows:

$$\text{Acid chloride} > \text{Acid anhydride} > \text{Ester} > \text{Amide}$$

Therefore,

$$p > q > r > s$$

Hence, the correct order of rate of hydrolysis is

$$\boxed{p > q > r > s}$$

Therefore, the correct answer is Option (D).