Total change in enthalpy (in kJ) going from State 1 to State 3 is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta H$ from State 1 to 2 is $27\text{ J} = 0.027\text{ kJ}$. For heating the liquid from 353 K to 373 K, $\Delta H \approx \Delta U = 75.6\text{ kJ}$. For vaporizing 25 moles, $\Delta H_{\text{vap}} = 25 \times 40\text{ kJ} = 1000\text{ kJ}$. Total enthalpy change $\Delta H = 0.027 + 75.6 + 1000 \approx 1075.6\text{ kJ}$.