The relative homology group $H_{1}(\mathbb{R},\mathbb{Q})$

$\require{cancel} \newcommand\R{\mathbb{R}} \newcommand\Q{\mathbb{Q}} \newcommand\Z{\mathbb{Z}}$ Write down the long exact sequence in homology: $$\cancel{H_1(\R)} \to H_1(\R,\Q) \to H_0(\Q) \to H_0(\R)$$ The space $\R$ is contractible so $H_0(\R) = \Z$ and $H_1(\R) = 0$. It's known that $H_0$ counts path components, and each point of $\Q$ is its own path component (the only continuous maps $[0,1] \to \Q$ are constant maps). So you get: $$H_0(\Q) = \bigoplus_{x \in \Q} \mathbb{Z}_x$$ where each $\Z_x$ is a copy of $\Z$. The map $H_0(\Q) \to H_0(\R)$ sends each $1_x \in \Z_x$ to $1 \in H_0(\R) \cong \Z$.

Because of the long exact sequence, $H_1(\R,\Q)$ is the kernel of this map. If you take some chosen base point $x_0 \in \Q$ (for example $x_0 = 0$), then this kernel is the subgroup of $H_0(\Q)$ generated by the $1_x - 1_{x_0}$ for $x \in \Q$. More than that, this set is a basis of $H_1(\R,\Q)$ as an abelian group (so it's free, in general a subgroup of a free abelian group is free abelian).