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Shouldn't enthalpy be measured at constant volume? Then the work done on and by the gas is 0 meaning that the temperature change is caused solely by heat released or absorbed by the system. You'd be able to calculate $$\mathrm dQ = \mathrm dT \cdot C_\mathrm s$$ (specific heat).
However, since enthalpy is at constant pressure the substance will do or have work done on it and thus there is no easy way to relate the enthalpy change to the change in temperature. Am I mistaken somehow?

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    $\begingroup$ If you want to define a new function then go ahead , but it wouldn't be enthalpy. $\endgroup$ – Avnish Kabaj Feb 10 '18 at 9:35
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Enthalpy is heat flow associated with a constant pressure process. Here is how:

Start with the familiar first law of thermodynamics for energy change:

$\Delta E=q+w$

$E$=energy

$q$=heat flow into the system

$w$=work done on the system

The work done is rendered thusly:

$w=-\int_\text{start}^\text{end}P\,\mathrm dV$, $P$=pressure, $V$=volume

Pressure is deemed constant so:

$w=-\int_\text{start}^\text{end}P\,\mathrm dV=-P\Delta V=-\Delta(PV)$

And then our first law balance equation becomes:

$\Delta E=q-\Delta(PV)$

$q=\Delta(E+PV)$

So heat flow is the change in a combination of state variables involving energy, pressure and volume. This combination is what we define as enthalpy and usually label $H$. Thereby, this enthalpy variable is designed to automatically include (mechanical) work inherent to a constant pressure process.

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  • $\begingroup$ Trying to figure out the answer to my question I've read and tried to understand this equation a million times. I see how the terms cancel but if we picture the situation things start falling apart. React a gas exothermically so that it expands and it will do work. The enthalpy change is no longer equal to the change in heat. $\endgroup$ – Edward Garemo Feb 10 '18 at 13:41
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    $\begingroup$ Why not, if the reaction was run at constant pressure? We see above that enthalpy change must equal heat flow in a constant pressure process. Your example just has a big work term included in the enthalpy change. $\endgroup$ – Oscar Lanzi Feb 10 '18 at 14:48
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Enthalpy was just defined for constant pressure.

It's not like that there was a need of new quantity, and it was taken at constant pressure, but there was a need of a quantity defined for constant pressure. This is because in real life most reactions are not carried out at constant volume, but in test tubes or flasks under constant pressure, that is, atmospheric pressure.

Also, there is already a quantity defined for constant volume, Qv. It's not necessary that a new symbol should be given to a quantity, as you said, it will be equal to heat absorbed or released, therefore it's symbol is related directly to heat.

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