The internal energy U can be calculated (in principle!) by the state parameters $P,V, T$ and $n_i$. Therefore $U$ $+$ $P\cdot V$ can be calculated (in principle!) also by the state parameters. Let's call it enthalpy and give it the symbol $H$:

Definition of enthalpy: $H$ $=$ $U$ $+$ $P\cdot V$

$H$ is a state function of the system, but why did we define this function ?

The aim of this definition arises from the fact that most usual chemical transformations happen at constant pressure $P$ (often the atmospheric pressure). The heat of such a transformation is measured in an open calorimeter, in contact with the outer air. Considering a transformation from a state 1 to a state 2, we have: $\Delta H $ $=$ $H_f$ $-$ $H_i$ (consider that H is a state function, f: final state, i: initial state) $=$ $(U_f$ $+$ $P\cdot V_f)$ $-$ $(U_i$ $+$ $P\cdot V_i)$ (see the definition of $H$, $P$ constant pressure) $=$ $(U_f$ $-$ $U_i)$ $+$ $P(V_f$ $-$ $V_i)$ $=$ $\Delta U$ $+$ $P\cdot \Delta V$ $= $ $(Q$ $+$ $W)$ $+$ $P\cdot \Delta V$ (see the → Variation of internal energy!) $= $ $(Q$ $-$ $P\Delta V)$ $+$ $P\cdot \Delta V$ (see the → Received work!) $=$ $Q$

At constant pressure: $\Delta H$ $=$ $Q$ The variation of enthalpy of a system is equal to the heat received by the system at constant pressure (measured in an open calorimeter)!

In case of a chemical reaction involving one or more gaseous reagents or products, only the gas species are considered for calculating volume variations and the (comparatively) small volume variations of non gaseous species are neglected: Let be $n_f$ the numbers of moles of gases in the products, $n_i$ the numbers of moles of gases in the reagents: $\Delta H $ $= $ $\Delta U $ $+$ $P\Delta V$ $= $ $\Delta U$ $ +$ $PV_f$ $-$ $PV_i$ $= $ $\Delta U$ $ +$ $n_fRT$ $-$ $n_iRT$ (see the → Ideal gas law!) $= $ $\Delta U$ $ +$ $\Delta n RT$

In the case of gases: $\Delta H$ $=$ $\Delta U$ $+$ $\Delta n RT$

The variation of enthalpy of the following reaction can be determined easily in an open calorimeter (at $298\;K$, $1 \;atm$), because it is simple combustion $2CO$ $+$ $O_2$ $\longrightarrow$ $2CO_2$ Details:

Initial state | Final state | $\Delta H$ |

2mol $CO$ + 1 mol $O_2$ | 2 mol $CO_2$ | $\Delta H$ $=$ $-563.2kJ$ |

Let's calculate the variation of internal energy of this reaction: $\Delta U$ $=$ $\Delta H$ $-$ $\Delta n RT$ $=$ $-563.2\cdot 10^3$ $-$ $(2$ $-$ $(2$ $+$ $1))\cdot 8,3 \cdot 298$ $ =$ $ -560726\; J$ $ =$ $\approx $ $-560.7\;kJ$ Let's analyse the result: - At constant volume, the heat transmitted by this reaction to the surroundings equals $-\Delta U$ $ = $ $ 560.7 kJ$. There is no work transmitted. - At constant pressure, the heat transmitted by this reaction to the surroundings equals $-\Delta H$ $=$ $563.2 kJ$. But there is a compression work of $-2.5 kJ$ transmitted to the surroundings. Indeed as the number of moles of gases decreases, the volume of the system decreases too at constant pressure and the surroundings perform therefore a compression work on the system. In either of the two cases, the energy produced by the reaction occuring in the system is the same!