# 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.

It seems that in some cases it's the same (that is, $$\Delta Q_p=\Delta H$$) yet in others they differ.

Heat is an energy transfer, whereas enthalpy is a kind of energy content of a sample. When we want to know the change in enthalpy (e.g. before and after a reaction), we use the "$$\Delta$$" prefix to indicate that. For heat, which already refers to a transfer of energy, the "$$\Delta$$" prefix is not used. So it should say $$Q_p=\Delta H$$

For example, \begin{align} \ce{2H2(g) + O2(g) &->2H2O(l)} & \Delta H &= -572~\mathrm{kJ} \end{align}

If you want to describe the enthalpy of reaction without knowing the amount that reacts, it is best to use the molar enthalpy with units $$\pu{kJ/mol}$$. This depends on how you write the reaction, e.g.

$$\ce{2H2(g) + O2(g) ->2H2O(l)}\ \ \ \Delta_r H = \pu{-572 kJ/mol}$$

or

$$\ce{H2(g) + 1/2 O2(g) -> H2O(l)}\ \ \ \Delta H_f(\ce{H2O}) = \pu{-286 kJ/mol}$$

Unfortunately, there is no generally accepted and followed convention for distinguishing enthalpy change and molar enthalpy change, but the unit tells us which one is meant.

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 $$Q_p=\Delta H$$?

Here, $$\Delta H$$ symbolizes the extensive quantity enthalpy (i.e. not molar enthalpy) so the equation is dimensionally correct. There is one more assumption we need - no work other than pressure-volume work. So if you run this reaction in a fuel cell to do electrical work, less heat will be exchanged.

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).