Reactivity of chlorobenzene and benzene in electrophilic substitutions
considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?


Let's expand the question to include all of the halobenzenes.

All of the halogens ($\ce{F, Cl, Br, I}$) are more electronegative than hydrogen.  Therefore they will inductively remove electron density from a benzene ring. However, you can also draw resonance structures for the various halobenzenes where the halogen substituent is donating electron density to the ortho and para positions of the benzene ring through resonance.

Therefore, in the case of the halobenzenes these two effects are working in opposite directions.  In order to determine which effect controls the reaction, we need some data.

The following table compares the relative rates of electrophilic aromatic nitration of the halobenzenes to the rate for benzene itself. If we assign benzene a relative rate of 1, we see that all of the halobenzenes react slower than benzene itself. This tells us that the inductive effect is stronger than the resonance effect in the halobenzene series.
\begin{array} \hline
 \text{Relative rates of aromatic electrophilic nitration} \\ \hline
\end{array}
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} &amp; \text{Relative rate}  \\ 
\hline
\ce{Ar-H} &amp; 1.0 \\ \hline
\ce{Ar-F} &amp; 0.11 \\ \hline
\ce{Ar-Cl} &amp; 0.02 \\ \hline
\ce{Ar-Br} &amp; 0.06  \\ \hline
\ce{Ar-I} &amp; 0.13 \\ \hline
\end{array}

(data source: see p. 26 here)

However, when the halobenzenes do undergo electrophilic aromatic substitution, they react preferentially at the ortho and para positions, because, as explained above, the resonance effect donates electron density to the ortho and para positions.  Therefore, these two positions are the least deactivated and react preferentially.