Can the RDS be different in different direction in a multi-step reversible reaction? The following mechanism is given in Peter Sykes when they are &quot;talking about de-esterification&quot; of an ester where the alcohol part is bulky-

Here the second step is given as RDS. According to me, in the reversed reaction(esterification), this step will remain RDS and so the reaction is bimolecular. Someone argues that RDS will be the formation of carbocation then and so the reaction is unimolecular.
After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put:

$\ce{A&lt;=&gt;B&lt;=&gt;[slow]C &lt;=&gt; D}$ doesn't imply that $\ce{B &lt;=&gt; C}$ is the RDS in both $\ce{A -&gt; D}$ and $\ce{D -&gt; A}$.

Rather than this, we need to consider $\ce{A-&gt;D}$ and $\ce{D-&gt; A}$ as two separate reactions. Doing so for the two reactions given in the question (acid catalysed hydrolysis ($\ce{A -&gt;D}$) and  esterification ($\ce{D -&gt; A}$)), we see the mechanisms are as follows:
Acid catalyzed hydrolysis

Here the RDS is the formation of the carbocation $\ce{^+CMe3}$.
Esterification of an acid and a bulky alcohol

Here the RDS is the formation of carbocation but it happens at a different step and not the same step.
Therefore the two reactions have different RDS steps as they are different reactions