Explain: Enthalpy as a state function.

(A) The first law of thermodynamics is given by $\Delta U = q + w$.
At constant pressure,the work done is $w = -p \Delta V$,so the equation becomes $\Delta U = q_p - p \Delta V$,where $q_p$ is the heat absorbed by the system at constant pressure.
When the system changes from state $1$ to state $2$,the internal energy changes from $U_1$ to $U_2$ and volume changes from $V_1$ to $V_2$.
Thus,$\Delta U = U_2 - U_1$ and $\Delta V = V_2 - V_1$.
Substituting these into the equation: $U_2 - U_1 = q_p - p(V_2 - V_1)$.
Rearranging gives: $q_p = (U_2 - U_1) + p(V_2 - V_1) = (U_2 + pV_2) - (U_1 + pV_1)$.
We define a new state function,Enthalpy $(H)$,as $H = U + pV$.
Therefore,$q_p = H_2 - H_1 = \Delta H$.
Since $U$,$p$,and $V$ are state functions,$H$ is also a state function.
For a process at constant pressure,$\Delta H = \Delta U + p \Delta V$.
$\Delta H$ is positive for endothermic reactions (heat absorption) and negative for exothermic reactions (heat evolution).