Let me attempt an answer for the case of split groups over $\mathbb{C}$ showing that rational, algebraic and homological equivalence coincide. A similar argument can be done for other characteristic $0$ fields using cycle class maps to algebraic de Rham cohomology instead of singular cohomology below.

First, the presentation of the Chow ring of a connected linear group can be found in the following paper of Grothendieck (see remark on p. 21)

- A. Grothendieck. Torsion homologique et sections rationelles. Sem. C. Chevalley, tome 3, exp 5 (1958), 1-29.

Consider a split semisimple group $G$ with Borel subgroup $B$. The Chow ring of $G$ (with integral coefficients) can be computed as quotient of ${\rm CH}^\ast(G/B)$ modulo the ideal generated by the image of the characteristic morphism $X(B)\to {\rm CH}^\ast(G/B)$ which maps a character $\chi$ of $B$ to the class of the associated line bundle $\mathcal{L}(\chi)$ in ${\rm CH}^1(G/B)$.

Furthermore, the remark on p. 29 of Grothendieck's article states that the composition $X(B)\to {\rm CH}^\ast(G/B)\to {\rm H}^{2\ast}(G/B(\mathbb{C}))$ of the algebraic characteristic morphism with the cycle class map is the topological characteristic map. Note that the cycle class map is an isomorphism in this case.

It remains to check that the Serre spectral sequence for the fibration $B(\mathbb{C})\to G(\mathbb{C})\to G/B(\mathbb{C})$ identifies the image of ${\rm H}^\ast(G/B(\mathbb{C}),{\rm H}^0(B(\mathbb{C}),\mathbb{Z}))$ in ${\rm }H^\ast(G(\mathbb{C}))$ with the quotient modulo the ideal generated by the image of the characteristic morphism. Bascially, this requires that the differential $$d_2:X(B)\cong {\rm H}^0(G/B(\mathbb{C}),{\rm H}^1(B(\mathbb{C}),\mathbb{Z}))\to {\rm H}^2(G/B(\mathbb{C}),{\rm H}^0(B(\mathbb{C}),\mathbb{Z}))$$ is the topological characteristic map. I don't know a proper reference for that, likely it's in Mimura-Toda "Topology of Lie groups I,II".

Combining these statements implies that the cycle class map ${\rm CH}^\ast(G)\to {\rm H}^{\ast}(G(\mathbb{C}))$ is injective, so that rational and homological equivalence coincide in this case. But then also rational and algebraic equivalence coincide and ${\rm CH}_{rat}(G)\cong {\rm CH}_{alg}(G)$.