Explain: $\Delta H = q_{p}$

(A) The first law of thermodynamics is given by $\Delta U = q + w$.
At constant pressure,the work done by the system is $w = -p \Delta V$,so we can write $\Delta U = q_{p} - p \Delta V$.
Here,$q_{p}$ is the heat absorbed by the system at constant pressure.
When the system undergoes a change 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 $q_{p} = (U_{2} - U_{1}) + p(V_{2} - V_{1}) = (U_{2} + pV_{2}) - (U_{1} + pV_{1})$.
Defining enthalpy as $H = U + pV$,we get $q_{p} = H_{2} - H_{1} = \Delta H$.
Since $U, p,$ and $V$ are state functions,$H$ is also a state function.
For processes 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).