Question

substance | equation for dissociation of compound | δh value (negative for exothermic, positive for endothermic) | sketch of energy diagram | sketch of summation of heat diagram
sodium hydroxide | | | |
potassium chloride | | | |

analysis questions:

1. which salt is the best choice for a hot pack? why?
2. if the hot pack is made from 12.4 g of the best choice, what is the quantity of heat released in the process?

Explanation:

To solve this, we first recall the dissociation reactions and enthalpy changes:

1. Sodium Hydroxide (NaOH) Dissociation:

The dissociation equation is $\ce{NaOH(s) -> Na^{+}(aq) + OH^{-}(aq)}$. This process is exothermic, so $\Delta H^{\circ}$ is negative (heat is released, suitable for a hot pack).

1. Potassium Chloride (KCl) Dissociation:

The dissociation equation is $\ce{KCl(s) -> K^{+}(aq) + Cl^{-}(aq)}$. This process is endothermic (heat is absorbed), so $\Delta H^{\circ}$ is positive (not suitable for a hot pack).

Analysis Question 1: Best Salt for a Hot Pack

A hot pack requires a reaction that releases heat (exothermic). Sodium hydroxide’s dissociation is exothermic (negative $\Delta H$), while potassium chloride’s is endothermic (positive $\Delta H$). Thus, sodium hydroxide is the best choice because its dissociation releases heat, warming the pack.

Analysis Question 2: Heat Released from 12.4 g of NaOH

First, find the molar mass of $\ce{NaOH}$:
Molar mass = $22.99\ (\ce{Na}) + 16.00\ (\ce{O}) + 1.008\ (\ce{H}) = 40.00\ \text{g/mol}$ (approx).

Moles of $\ce{NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{12.4\ \text{g}}{40.00\ \text{g/mol}} = 0.31\ \text{mol}$.

The enthalpy of dissociation for $\ce{NaOH}$ (exothermic) is typically $\Delta H = -44.5\ \text{kJ/mol}$ (experimental value).

Heat released ($q$) = moles $\times |\Delta H|$ = $0.31\ \text{mol} \times 44.5\ \text{kJ/mol} \approx 13.8\ \text{kJ}$ (or $13800\ \text{J}$).

Energy Diagrams

• NaOH (Exothermic): The energy of products is lower than reactants. The diagram shows a downward arrow (energy released) with reactants at higher energy, products at lower energy, and activation energy (Ea) for the dissociation.
• KCl (Endothermic): Products have higher energy than reactants. The diagram shows an upward arrow (energy absorbed) with reactants at lower energy, products at higher energy, and Ea.

Summation of Heat Diagrams

• NaOH: Heat is released (system loses heat, surroundings gain heat). The diagram shows heat flowing from the system (dissociation) to the surroundings (hot pack).
• KCl: Heat is absorbed (system gains heat, surroundings lose heat). The diagram shows heat flowing into the system from the surroundings.

Answer:

1. Best Salt for Hot Pack

Sodium hydroxide is the best choice because its dissociation in water is exothermic (releases heat), which warms the hot pack. Potassium chloride’s dissociation is endothermic (absorbs heat), so it would cool the pack.

2. Heat Released from 12.4 g NaOH

Using $\Delta H_{\text{dissociation}}^{\circ}(\ce{NaOH}) \approx -44.5\ \text{kJ/mol}$:
Moles of $\ce{NaOH} = \frac{12.4\ \text{g}}{40.00\ \text{g/mol}} = 0.31\ \text{mol}$.
Heat released = $0.31\ \text{mol} \times 44.5\ \text{kJ/mol} \approx 13.8\ \text{kJ}$ (or $13800\ \text{J}$).

(Note: For energy/heat sketches, label reactants, products, $\Delta H$ direction, and Ea. For NaOH, $\Delta H$ is negative (exothermic); for KCl, $\Delta H$ is positive (endothermic).)