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If "rational rank 1" simply means that the underlying group is isomorphic to $\mathbb{Q}$, then every such lattice-ordered group structure $(\mathbb{Q}, \preceq)$ is totally ordered, in fact isomorphic to $\mathbb{Q}$ with its usual order $\leq$.

By a classical result (see the first sentence here), any abelian lattice-ordered group ($l$-group) $G$ can be embedded (by an $l$-group homomorphism) into a direct product of totally ordered non-trivial abelian groups. By projecting to any one of the factors, there exists a totally ordered $l$-group $G'$ and a non-trivial $l$-group homomorphism $f: G \to G'$. By an epi-mono factorization, we may assume $f$ is surjective.

In the case $G = \mathbb{Q}$, any group quotient $G'$ is divisible, and it is well-known that an $l$-group $G'$ is torsionfree (see e.g. here, corollary 1.2.6 page 16). But divisible torsionfree abelian groups are rational vector spaces, so in this situation the quotient $G'$ is isomorphic to $\mathbb{Q}$ and $f: G \to G'$ is a group isomorphism. But $f$ is also a lattice homomorphism. Since bijective lattice homomorphisms are lattice isomorphisms, $f: G \to G'$ is a lattice isomorphism, and hence $G$ is totally ordered, as desired.

Finally, a totally ordered $l$-group structure $(\mathbb{Q}, \preceq)$ is given by two submonoids $P, N = -P$ whose intersection is $\{0\}$ and whose union is $\mathbb{Q}$. To show $(\mathbb{Q}, \preceq)$ is isomorphic to $(\mathbb{Q}, \leq)$, we need only show that $P$ contains every multiple $\frac{m}{n}p$ where $p \in P$ is any chosen non-zero element and $0 < m, n$. But clearly the submonoid generated by $\frac{m}{n}p$ contains $0 \neq mp \in P$. Thus $\frac{m}{n}p \in N$ is impossible. This completes the proof.

Edit: Following on Andreas's first interpretation of "rational rank 1" (subgroup $B \subseteq \mathbb{Q}$), the same results hold. For again, we have a surjective $l$-group homomorphism $f: B \to B'$ to a totally ordered $l$-group $B'$, where $B'$ is nontrivial and torsionfree. If $A = \ker(f)$, we have $\mathrm{rank}(B) = \mathrm{rank}(A) + \mathrm{rank}(B')$, where $\mathrm{rank}(B') = 1$ since $B'$ is torsionfree. Thus $\mathrm{rank}(A) = 0$, meaning $A$ is a torsion subgroup of a torsionfree group, meaning $A = 0$. Thus $f: B \to B'$ is a group isomorphism, and we deduce as before it is a lattice isomorphism, so $B$ is totally ordered. We deduce that $B$ is isomorphic to $B$ with its standard lattice structure, by essentially the same argument as in the last paragraph before this edit.

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If "rational rank 1" simply means that the underlying group is isomorphic to $\mathbb{Q}$, then every such lattice-ordered group structure $(\mathbb{Q}, \preceq)$ is totally ordered, in fact isomorphic to $\mathbb{Q}$ with its usual order $\leq$.

By a classical result (see the first sentence here), any abelian lattice-ordered group ($l$-group) $G$ can be embedded (by an $l$-group homomorphism) into a direct product of totally ordered non-trivial abelian groups. By projecting to any one of the factors, there exists a totally ordered $l$-group $G'$ and a non-trivial $l$-group homomorphism $f: G \to G'$. By an epi-mono factorization, we may assume $f$ is surjective.

In the case $G = \mathbb{Q}$, any group quotient $G'$ is divisible, and it is well-known that an $l$-group $G'$ is torsionfree (see e.g. here, corollary 1.2.6 page 16). But divisible torsionfree abelian groups are rational vector spaces, so in this situation the quotient $G'$ is isomorphic to $\mathbb{Q}$ and $f: G \to G'$ is a group isomorphism. But $f$ is also a lattice homomorphism. Since bijective lattice homomorphisms are lattice isomorphisms, $f: G \to G'$ is a lattice isomorphism, and hence $G$ is totally ordered, as desired.

Finally, a totally ordered $l$-group structure $(\mathbb{Q}, \preceq)$ is given by two submonoids $P, N = -P$ whose intersection is $\{0\}$ and whose union is $\mathbb{Q}$. To show $(\mathbb{Q}, \preceq)$ is isomorphic to $(\mathbb{Q}, \leq)$, we need only show that $P$ contains every multiple $\frac{m}{n}p$ where $p \in P$ is any chosen non-zero element and $0 < m, n$. But clearly the submonoid generated by $\frac{m}{n}p$ contains $0 \neq mp \in P$. Thus $\frac{m}{n}p \in N$ is impossible. This completes the proof.