Is it possible to have a finitely generated lattice - ordered group has rational rank bigger than the number of generators of the group?
This answer suppose, given the tag, that we are dealing with an Abelian ordered group $G$. (Actually the fact that $G$ is ordered does not come into play once supposed the abelianity)
Then we can suppose, without loss of generality, $G$ to be torsion free. Indeed if $T(G)$ is the torsion group of $G$ then $r(G)=r(G/T(G))$. So we can suppose $G=G/T(G)$ which has equal rank and maybe a smaller set of generators.
But then $G$ is abelian, finitely generated and torsion free, so it is an abelian free group with $n$ generators and it has rank $n$.
In conclusion it is impossible to have such a group.