# What is the difference between enthalpy (H) and heat (q)?

It seems that in some cases it's the same (that is, $\Delta Q_p=\Delta H$) yet in others they differ. For example, \begin{align} \ce{2H2(g) + O2(g) &->2H2O(l)} & \Delta H &= -572~\mathrm{kJ} \end{align}

The answer for one mole of $\ce{H2O}$ produced is $\mathrm{-572\times 0.5~kJ}$. Is this under the assumption of constant pressure? Therefore it's $\Delta Q_p=\Delta H$?

Enthalpy and heat are entirely different things.

Enthalpy is a function of state. If you know the state of a system, you know its enthalpy. If you know the starting and ending states of a process, you can find the enthalpy change.

Heat, on the other hand, is an inexact differential. Knowing the initial and final states of a process is not enough information to tell you the heat transfer. Instead, the heat transfer depends on the particular path taken between the states.

For a simple example of why this is important, consider a heat engine. The process is a cycle, meaning the initial and final states are the same, so a cycle of a heat engine has zero enthalpy change. However, the entire point is that the heat engine converts heat into work, so the heat exchange during a cycle is not zero.

That is why the heat exchanged is equal to the enthalpy change only under constant pressure. We need to give extra information (such as that the pressure is constant) to know the path taken during the process in order to say anything meaningful about heat exchange.

Enthalpy is defined as $$H= E + PV$$

Using First Law, we get $$\mathrm d H =\delta Q - \left(\underbrace{\sum Y\,\mathrm dX + W'}_{\delta W}\right) + \mathrm d(PV)$$

where \begin{align} \sum Y\,\mathrm dX &= \,\,\textrm{configuration work}\\ W'& = \,\,\textrm{dissipative work}\end{align}

When the process is reversible, then the dissipative work $W'$ is zero.

We can further write the expression for constant pressure as \begin{align}\mathrm dH & = \delta Q - \underbrace{A'}_\textrm{non-compression work} - P\,\mathrm dV + P\,\mathrm dV\\ &= \delta Q- A'\end{align}

Thus, basically $$\mathrm dH = \delta Q - A'$$

When $A' = 0,$ then only $\mathrm dH = \delta Q$ at constant pressure.

# Further look:

$\bullet$ Thermodynamics, Kinetic Theory, and Statistical Thermodynamics by Sears, Salingar.

Strictly speaking, that $$\Delta H$$ assumes that the initial state consists of 2 moles of pure $$\ce{H2}$$ and 1 mole of pure $$\ce{O2}$$, each at 298 K and 1 bar, and the final state consists of 2 moles of pure liquid water at the same temperature and pressure. Since enthalpy is a state function independent of path, this enthalpy change is independent of any and all processes used for transitioning the system from state 1 to state 2.

Since, at this pressure, $$\ce{H2}$$ and $$\ce{O2}$$ behave essentially as ideal gases (and their mixing is ideal), the gases can be mixed first and then allowed to react at constant pressure. In this case, the heat added to the system with be equal to the stated $$\Delta H$$ (in this case negative).