Do we neglect the temperature changes during a reaction while calculating delta(H)?
Yes. By saying that $\Delta (pV) = \Delta (nRT) = RT\Delta n$ you have made the assumption that $T$ is constant. When you take anything out of a $\Delta$ sign, you are saying that the initial and final values are the same. Otherwise you would have to write $\Delta (nRT) = R \Delta (nT) = R(n_2T_2 - n_1T_1)$ - this only simplifies to $RT(n_2 - n_1)$ if you assume $T_1 = T_2 = T$.



By the way, I want to be picky about the formulae that you are using. Look at $$\mathrm{d}H = \mathrm{d}U + (c + d - a - b)RT$$

The problem with this is that $\mathrm{d}H$ and $\mathrm{d}U$ are infinitesimals, but $(c + d - a - b)RT$ is a finite quantity. Do not ever add or subtract these two in the same equation. It is like trying to add a mass to a length - what is $30 \mathrm{~kg} + 25 \mathrm{~cm}$? It doesn't make sense.

Same thing applies for $\mathrm{d}H = \mathrm{d}U + (pV)$.