Question

identify the exothermic processes. check all that apply.

n₂o₅(g) + h₂o(l) → 2hno₃(l), δh = -18.1 kcal
2nh₃(g) → 3h₂(g) + n₂(g), δh = +21.9 kcal
caco₃(s) → cao(s) + co₂(g), δh = +179.2 kj
n₂(g) + 2o₂(g) → 2no₂(g), δh = +15.9 kcal
co(g) + h₂(g) + o₂(g) → co₂(g) + h₂o(g), δh = -525. kj
none of the above

Explanation:

An exothermic process is one where the enthalpy change ($\Delta H$) is negative, meaning the system releases heat. Let's analyze each reaction:

1. For $\ce{N2O5(g) + H2O(l) -> 2HNO3(l)}$, $\Delta H = -18.1$ kcal. A negative $\Delta H$ indicates heat is released, so this is exothermic.
2. For $\ce{2NH3(g) -> 3H2(g) + N2(g)}$, $\Delta H = +21.9$ kcal. Positive $\Delta H$ means heat is absorbed (endothermic).
3. For $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$, $\Delta H = +179.2$ kJ. Positive $\Delta H$ (endothermic).
4. For $\ce{N2(g) + 2O2(g) -> 2NO2(g)}$, $\Delta H = +15.9$ kcal. Positive $\Delta H$ (endothermic).
5. For $\ce{CO(g) + H2(g) + O2(g) -> CO2(g) + H2O(g)}$, $\Delta H = -525.$ kJ. Negative $\Delta H$ (exothermic).

So the exothermic processes are the first and fifth reactions.

Answer:

• $\ce{N2O5(g) + H2O(l) -> 2HNO3(l)}, \Delta H = -18.1$ kcal
• $\ce{CO(g) + H2(g) + O2(g) -> CO2(g) + H2O(g)}, \Delta H = -525.$ kJ