Question

which process is shown in the following enthalpy diagram: h, enthalpy h₂o(g) δh = - 40.7 kj/mol h₂o(l) rxn → exothermic: h₂o(g) → h₂o(l) + 40.7 kj exothermic: h₂o(g) + 40.7 kj → h₂o(l) endothermic: h₂o(g) → h₂o(l) + 40.7 kj endothermic: h₂o(g) + 40.7 kj → h₂o(l)

Explanation:

Step1: Analyze Enthalpy Change

The enthalpy diagram shows $\Delta H = -40.7\ \text{kJ/mol}$, meaning the system releases energy (exothermic, since $\Delta H < 0$). The reaction is $\ce{H2O(g) -> H2O(l)}$, and energy is a product (released), so the equation can be written as $\ce{H2O(g) -> H2O(l) + 40.7\ \text{kJ}}$ (or with energy on the product side, indicating exothermic).

Step2: Evaluate Options

• Exothermic reactions have $\Delta H < 0$ (energy released). The reaction is $\ce{H2O(g) -> H2O(l)}$, and the energy is released (40.7 kJ). Check the options: the first option (exothermic: $\ce{H2O(g) -> H2O(l) + 40.7\ \text{kJ}}$) matches the exothermic nature (negative $\Delta H$) and the reaction direction (gas to liquid, condensation, which releases heat).

Answer:

exothermic: $\ce{H2O(g) -> H2O(l) + 40.7\ \text{kJ}}$ (the first option in the list, likely labeled as the first radio button option with that text)