Question

which process is shown in the following enthalpy diagram:
enthalpy diagram with h, enthalpy on y - axis, rxn on x - axis, h₂o(g) at higher enthalpy, h₂o(l) at lower enthalpy, δh = - 40.7 kj/mol
options:

• 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 $\Delta H = -40.7\ \text{kJ/mol}$ for the process from $\ce{H2O(g)}$ to $\ce{H2O(l)}$. A negative $\Delta H$ means the system releases heat (exothermic).

Step2: Match Reaction and Energy

For an exothermic process, the energy is a product (or added as a release). The reaction $\ce{H2O(g) -> H2O(l)}$ has $\Delta H = -40.7\ \text{kJ/mol}$, which can be written as $\ce{H2O(g) -> H2O(l) + 40.7\ \text{kJ}}$ (since energy is released).

Step3: Eliminate Incorrect Options

• Endothermic processes have positive $\Delta H$ (absorb heat), so eliminate endothermic options.
• The exothermic reaction with the correct energy direction (release) is $\ce{H2O(g) -> H2O(l) + 40.7\ \text{kJ}}$.

Answer:

A. exothermic; $\ce{H2O(g) -> H2O(l) + 40.7\ \text{kJ}}$ (assuming the first option is this, as per the description)