Why does the sign of Delta H indicate whether the reaction is exothermic or endothermic? If $\Delta H = Q + W$ (assuming constant conditions), then there are two terms involved in calculating $\Delta H$, only one of which measures heat gained/released. It is possible then for $\Delta H$ to be negative (if $W$ were very negative), even with $Q$ being positive? (And vice versa too.)
So why do we say that the sign of $\Delta H$ indicates whether a reaction is exo- or endo- thermic?
 A: Your original equation is incorrect.  For a process occurring in a closed system, the equation should read$$\Delta U=Q+W$$where U is the internal energy of the system, Q is the heat added to the system, and W is the work done by the surroundings on the system, $W=-\int{PdV}$.  If the process takes place at a constant pressure, then $W=-P\Delta V$, and the change in internal energy becomes:$$\Delta U=Q-P\Delta V$$.  But, from the definition of enthalpy, we have  $\Delta H=\Delta U+\Delta (PV)$.  So, finally, $$\Delta H=Q$$
So, for a process carried out at constant pressure, if the heat added to the system is positive (endothermic), $\Delta H$ is positive and if the heat added to the system is negative (exothermic, heat removed from system), $\Delta H$ is negative.
A: The symbol $\Delta$ (here) simply means difference, e.g.:
$$\Delta H=H_2-H_1$$
So, enthalpy of the end state minus enthalpy of the initial state.
In the case of an exothermic reaction the system has lost enthalpy, so:
$$H_2<H_1$$
Thus, for an exothermic reaction:
$$\boxed{\Delta H=H_2-H_1<0}$$

It is possible then for $ΔH$ to be negative (if $W$ were very negative), even with $Q$ being positive? (And vice versa too.)

Yes.
