Does MOPAC allows calculation of Standard Enthalpy of reaction in this way?
The free energy change of a reaction occurring in solution can be computed using a thermodynamic cycle:
$$\Delta G^\circ_{r,soln} = \Delta G^\circ_{r,gas} + \Delta\Delta G^\circ_{r,solv} $$
$\Delta G^\circ_{r,soln}$ can also be written as
$$\Delta G^\circ_{r,soln} = G^\circ_{soln}(B)- G^\circ_{soln}(A)$$
where 
$$G^\circ_{soln}(X) = G^\circ_{gas}(X) + \Delta G^\circ_{solv} $$
and 
$$G^\circ_{gas}(X) = E_{gas}(X) + G^\circ_{gas,RRHO}(X)$$
The solvation energy is usually computed using a continuum as the difference in electronic energy with and without the continuum. 
$\Delta G^\circ_{solv}(X) = E^\prime_{soln}(X)-E^\prime_{gas}(X)$
The parameters in the continuum model are usually optimized for a particular level of theory, which may not be appropriate for the computation of $\Delta G^\circ_{r,gas}$, so $E_{gas}(X)$ is not necessarily the same as $E^\prime_{gas}(X)$

However, if $E_{gas}(X) = E^\prime_{gas}(X)$ then 
$$G^\circ_{soln}(X) = E_{soln}(X) + G^\circ_{gas,RRHO}(X) $$

The parameters in the continuum model are usually determined using the gas phase geometry $(X)$.  But if $E_{gas}(X) = E^\prime_{gas}(X)$ then many people often choose to use the geometry optimized in solution
$G^\circ_{soln}(X_{soln}) = E_{soln}(X_{soln}) + G^\circ_{soln,RRHO}(X_{soln})$.

For semiempirical calculations, the electronic energy corresponds to the enthalpy of formation at 298K, so if one simply makes the substitution
$$E(X) \rightarrow \Delta H^\circ_{f,298}(X)$$
then (at 298K)
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) -TS^\circ_{gas,RRHO}(X)$$ $G^\circ_{soln}(X)$ doesn't correspond to a measurable free energy because $S^\circ_{gas,RRHO} \ne S^\circ_{f}$ but it can still be used to compute $\Delta G^\circ_{r,soln}$ since the missing terms cancel.

For temperatures other than 298K
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) + (H^\circ_{gas,RRHO}(X,T)-H^\circ_{gas,RRHO}(X,T=298K)) -TS^\circ_{gas,RRHO}(X)$$