Chemical spontaneity is characterized by $\Delta G<0$. Why is nuclear spontaneity characterized by $Q>0$ and not $\Delta G<0$ as well? Chemical reactions that occur at constant temperature and pressure are spontaneous if and only if the reaction reduces the systems Gibbs free energy ($\Delta G=\Delta H -T\Delta S<0$). Clearly this criterion permits spontaneous endothermic chemical reactions provided that the entropy of the system increases by enough (provided $T\Delta S >\Delta H$).
In my nuclear physics textbook (Krane) however, the author characterizes the emission of nuclear matter from a large nucleus (as in alpha decay) as being spontaneous only if the decay satisfies $Q=\Delta K >0$ where $\Delta K$ is the net kinetic energy released. Thus the $Q$ value of a nuclear decay seems to correspond to the change in enthalpy of the decay(so we should expect that $Q=\Delta H$). If this is the case then why do we not have an additional term of the form $-T\Delta S$ in the criterion for nuclear spontaneity? After all, alpha decay is occurring at constant temperature and pressure and in these conditions, $\Delta G <0$ is equivalent to $\Delta S_{universe}>0$. The fact that nuclear spontaneity is characterized by $Q>0$ seems to indicate that for some reason (?) endothermic nuclear reactions are impossible yet endothermic chemical reactions are. Why is this the case? Why do we not use $\Delta G<0)$ to characterize spontaneity of alpha emission?
Any help on this would be most appreciated!
 A: 
Chemical reactions that occur at constant temperature and pressure are
spontaneous if and only if the reaction reduces the systems Gibbs free
energy ($ΔG=ΔH−TΔS<0$). Clearly this criterion permits spontaneous
endothermic chemical reactions provided that the entropy of the system
increases by enough (provided $TΔS>ΔH$).

That's a very simplistic view of things.
Let's say we have a simple chemical equilibrium:
$$\mathrm{A}\leftarrow\rightarrow \mathrm{B}\tag{1}$$
The 'tendency' of this equilibrium to lean to the left or to the right depends on the equilibrium constant $K$:
$$K=\frac{[B]}{[A]}$$
where the bracketed quantities are (somewhat simplified) concentrations. Obviously if $K\gg 1$ then the reaction leans very much to the left (because $[B]\gg[A]$) but $K \ll 1$ then it leans much to the left.
$K$ relates to $\Delta G$ through:
$$\Delta G=-RT\ln K$$
So, very negative $\Delta G$ lead to large $K$ and the equilibrium leaning to the right of $(1)$.
Note also that thermodymamics says nothing about kinetics: a reaction may have a very high $K$, yet proceed at imperceptibly small rate.
