Questions: ZnO(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l) ΔH° = +14.4 kJ/mol + Spontaneous at low temp. Co(s) + Cl₂(g) → CoCl₂(s) ΔH° = +10.4 kJ/mol + Spontaneous at low temp. NiAl(s) → Ni(s) + Al(s) ΔH° = +85.1 kJ/mol + Spontaneous at high temp. NaN + O₂(g) → Na₂O(s) ΔH° = -180.7 kJ/mol - Spontaneous at all temps. 3O(s) → O₃(g) exothermic - Spontaneous at low temps.

Solution Steps

We are given five reactions with their respective enthalpy changes (\(\Delta H^\circ\)):

1. \( \text{ZnO(s) + 2HCl(aq) → ZnCl}_2\text{(aq) + H}_2\text{O(l)} \) with \( \Delta H^\circ = +14.4 \text{ kJ/mol} \)
2. \( \text{Co(s) + Cl}_2\text{(g) → CoCl}_2\text{(s)} \) with \( \Delta H^\circ = +10.4 \text{ kJ/mol} \)
3. \( \text{NiAl(s) → Ni(s) + Al(s)} \) with \( \Delta H^\circ = +85.1 \text{ kJ/mol} \)
4. \( \text{NaN + O}_2\text{(g) → Na}_2\text{O(s)} \) with \( \Delta H^\circ = -180.7 \text{ kJ/mol} \)
5. \( \text{3O(s) → O}_3\text{(g)} \) is exothermic

We are given the entropy change (\(\Delta S^\circ\)) for each reaction:

1. \( \Delta S^\circ = + \)
2. \( \Delta S^\circ = + \)
3. \( \Delta S^\circ = + \)
4. \( \Delta S^\circ = - \)
5. \( \Delta S^\circ = - \)

To determine the spontaneity of each reaction, we use the Gibbs free energy equation:

\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]

A reaction is spontaneous if \(\Delta G^\circ < 0\).

1. For \( \text{ZnO(s) + 2HCl(aq) → ZnCl}_2\text{(aq) + H}_2\text{O(l)} \):

   • \( \Delta H^\circ = +14.4 \text{ kJ/mol} \)
   • \( \Delta S^\circ = + \)
   • Spontaneous at low temperatures because \(\Delta H^\circ > 0\) and \(\Delta S^\circ > 0\).

2. For \( \text{Co(s) + Cl}_2\text{(g) → CoCl}_2\text{(s)} \):

   • \( \Delta H^\circ = +10.4 \text{ kJ/mol} \)
   • \( \Delta S^\circ = + \)
   • Spontaneous at low temperatures because \(\Delta H^\circ > 0\) and \(\Delta S^\circ > 0\).

3. For \( \text{NiAl(s) → Ni(s) + Al(s)} \):

   • \( \Delta H^\circ = +85.1 \text{ kJ/mol} \)
   • \( \Delta S^\circ = + \)
   • Spontaneous at high temperatures because \(\Delta H^\circ > 0\) and \(\Delta S^\circ > 0\).

Final Answer