Prove that the change in enthalpy of a system in which a chemical reaction occurs is equal to the heat gained by the system at constant pressure.

(N/A) According to the first law of thermodynamics,$\Delta U = q + w$.
At constant pressure,the work done by the system is $w = -p \Delta V$.
Substituting this into the equation,we get $\Delta U = q_p - p \Delta V$.
Here,$q_p$ is the heat absorbed by the system at constant pressure.
When the system undergoes a change from state $1$ to state $2$,we have $\Delta U = U_2 - U_1$ and $\Delta V = V_2 - V_1$.
Substituting these values: $U_2 - U_1 = q_p - p(V_2 - V_1)$.
Rearranging the terms: $q_p = (U_2 - U_1) + p(V_2 - V_1)$.
$q_p = (U_2 + pV_2) - (U_1 + pV_1)$.
Defining enthalpy as $H = U + pV$,we get $q_p = H_2 - H_1 = \Delta H$.
Thus,the change in enthalpy $\Delta H$ is equal to the heat absorbed at constant pressure $q_p$.