# Do we neglect the temperature changes during a reaction while calculating delta(H)?

Suppose for a reaction: $$\ce{aA + bB -> cC + dD }$$

we know the $\Delta U$ as $\mathrm dU$ and all the components of the reaction are gas.

Then to calculate the $\Delta H$ at a temperature $T$ we use the expression $\mathrm dH=\mathrm dU+\left(c+d-a-b\right)RT$.

Then while using this expression did we assume the temperature to remain constant as it wouldn't really be constant as the actual expression for $\mathrm dH$ is: $\mathrm dH=\mathrm dU+\left(pV\right)$ and so we can write $\mathrm dH=\mathrm dU+\mathrm d(nRT)$ and for $\mathrm dH=\mathrm dU+\left(c+d-a-b\right)RT$, $T$ has to be constant?

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 $\mathrm{d}H = \mathrm{d}U + (pV)$.
• In addition, the equation $dH = dU + d(pV)$ already assumes that temperature is constant before anything involving $\Delta$ terms. The fundamental equation is $dU = T dS - p dV$ or if you prefer $dH = T dS + V dP$ . The differential entropy $dS$ of an ideal gas depends on both temperature and volume. To neglect that term requires assuming $T$ is constant. – Curt F. Oct 2 '15 at 15:46