The Cayley parametrization of $O(n),$ as in my answer to this question makes one wonder: which algebraic groups are actually rational? I am sure this is very well understood, just not by me...

A reductive group $G$ over a ground field $k$ need not be a rational variety over $k$ (although the group $G_{/\ell}$ obtained by scalar extension is a rational $\ell$ variety, where $\ell$ is an algebraic closure of $k$). Maybe a good place to look concerning these matters is Merkurjev's 1996 Publ. Math. IHES paper "Requivalence and rationality problem for semisimple adjoint classical algebraic groups". In the intro to that paper, Merkurjev makes some interesting observations:
In Merkurjev's paper, you will find explicit examples of groups $G$ which are not stably rational  a $k$variety $X$ is stably rational if $X \times \operatorname{Aff}^d$ is a rational $k$variety for some $d \ge 0$. One way to show that $G$ is not stably rational (and hence not rational) is to show that the "group of $R$equivalence classes" for $G$ is nontrivial; this group of $R$equivalence classes was introduced by Manin. Merkurjev's paper is devoted to the computation of the group of $R$equivalence classes for semisimple, adjoint, classical groups. 


The following answer applies to affine algebraic groups over algebraically closed fields, or more generally for quasisplit groups, as Scott Carnahan suggests. I don't really know about these rationality (in the numbertheoretic sense) questions. Under these conditions, every algebraic group is an extension of a reductive group by a unipotent group (since a reductive group is one whose unipotent radical, i.e. maximal normal unipotent connected subgroup, is trivial). Every unipotent group is rational, since it is isomorphic as a variety to $\mathbb{A}^n$. Every reductive group is rational by the Bruhat decomposition. Every Zariskilocally trivial bundle over a rational variety whose fibers are rational, is rational. 


Suppose G is a connected affine algebraic group over a field k. If G is reductive or k is perfect, then G is unirational. A reference is Borel, Linear Algebraic Groups, Theorem 18.2. This doesn't quite give rationality, but is the best result in general that I know of, and is sufficient for some applications. (Ryan's answer of course gives more precise information under stronger hypotheses). An affine counterexample over an imperfect field of characteristic p is given by y^p=x+tx^p where t is not a pth power in k. (This is a subgroup of G_{a}^{2}). Now if G is connected and not affine, then it surjects onto an abelian variety. So if G were rational, then we could get a nonconstant morphism from P^1 to an abelian variety. Now let me just quote http://math.stackexchange.com/a/114611 so no nonaffine conntected G can be rational, or even unirational. 

