Free or projective resolutions for definition of Ext In Hatcher, Lemma 3.3.1 the author shows that two free resolutions $F,F'$ of an abelian group $H$ are homotopy equivalent. He later notes that this is still true for $R$-modules $H$ over any principal ideal ring $R$ and this allows him to define the functor $\operatorname{Ext}_R(\,\_\,,H)$.
Diving a bit further into the theory, I noticed that in general people use projective resolutions in order to define $\operatorname{Ext}^n_R(\,\_\,,H)$ since there is an analogous statement to Lemma 3.3.1 for projective resolutions of modules over arbitrary rings.
However, it seems to me that for the definition of $\operatorname{Ext}^n_R(\,\_\,,H)$ we actually don't have to use the greater generality of the statement since we know that there is always a free resolution and that every free resolution is projective. My question is: would it make any difference to define $\operatorname{Ext}^n_R(\,\_\,,H)$ via free instead of projective resolutions in the general case? If not, why do people define it via projectives when they compute it via free resolutions anyway (most of the time, at least that's my impression).
EDIT: Taking into account Pedro's comment I would like to rephrase the question as follows:
Can we define a model structure on $\operatorname{Ch}_{\bullet \geq 0}(\operatorname{Mod}_R)$ by taking free resolutions to be the cofibrant objects so that the corresponding homotopy category is the same as for the projective model structure?
 A: Indeed, it doesn't make a difference whether you use free resolutions or projective resolutions in the definition, since free resolutions are a special case of projective resolutions and they always exist.  The notion of projective resolution generalizes better to broader situations, though.  For instance, it makes sense to talk about projective resolutions in an arbitrary abelian category and use them to define Ext (and other derived functors), but an abstract abelian category does not have any notion of a "free object".
A: To address another part of the question than Eric did ("Why do people define it via projectives when they compute it via free resolutions anyway (most of the time, at least that's my impression)"):
In fact, when (nonfree) projective modules are well understood, they are often used for computations, and often make them more transparent than using free modules.
For example, this is often true for finite dimensional algebras over fields. Here's a very simple example to illustrate how insisting on sticking with free modules obscures what is going on:
Let $R$ be the ring of $2\times 2$ upper triangular matrices over a field. Then (up to isomorphism) $R$ has two indecomposable projective modules $P_1$ and $P_2$, and two simple modules $S_1$ and $S_2$, and the smallest projective resolutions of the simple modules have the forms:
$$\dots\to0\to0\to P_2\to P_1\to S_1\to0$$
and $$\dots\to0\to0\to0\to P_2\to S_2\to0,$$
from which it is evident that $S_2$ is projective and $S_1$ has projective dimension $1$ (i.e., $\text{Ext}^i(S_1,-)=0$ iff $i>1$).
The smallest free resolutions have the forms:
$$\dots\to R^2\to R^2\to R^2\to R\to S_1\to0$$
and
$$\dots\to R\to R\to R\to R\to S_2\to0,$$
from which the elementary homological properties are much less evident without carefully examining the differentials.
