The problem describes a calorimetry experiment where the dissolution of ammonium nitrate ($\text{NH}_4\text{NO}_3$) leads to a temperature change. Let’s break it down to solve for $\Delta H$ in $\text{kJ/mol}$:

Information provided:

1. Mass of water: $m = 70.00 \, \text{g}$
2. Initial temperature: $T_\text{initial} = 22.0^\circ \text{C}$
3. Final temperature: $T_\text{final} = 17.7^\circ \text{C}$
4. Mass of ammonium nitrate: $m_{\text{NH}_4\text{NO}_3} = 4.15 \, \text{g}$
5. Heat capacity of water: $c = 4.18 \, \frac{\text{J}}{\text{g} \cdot \degree\text{C}}$
6. Assume the solution behaves as water with no heat loss to the calorimeter.

Step 1: Calculate the heat absorbed or released ($q$) by the solution

The heat exchanged is given by: $q = m \cdot c \cdot \Delta T$ where $\Delta T = T_\text{final} - T_\text{initial}$.

$\Delta T = 17.7^\circ \text{C} - 22.0^\circ \text{C} = -4.3^\circ \text{C}$

$q = 70.00 \, \text{g} \cdot 4.18 \, \frac{\text{J}}{\text{g} \cdot \degree\text{C}} \cdot (-4.3^\circ \text{C})$

$q = -1259.66 \, \text{J} \quad \text{(heat released by the system)}$

Step 2: Convert $q$ to $\text{kJ}$

$q = -1.25966 \, \text{kJ}$

Step 3: Determine moles of $\text{NH}_4\text{NO}_3$

The molar mass of $\text{NH}_4\text{NO}_3$ is calculated as: $\text{Molar mass} = 14.01 (\text{N}) + 1.01 \times 4 (\text{H}) + 16.00 \times 3 (\text{O}) = 80.05 \, \text{g/mol}$

$\text{Moles of } \text{NH}_4\text{NO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.15 \, \text{g}}{80.05 \, \text{g/mol}}$

$\text{Moles of } \text{NH}_4\text{NO}_3 = 0.0519 \, \text{mol}$

Step 4: Calculate enthalpy change ($\Delta H$)

The enthalpy change per mole is: $\Delta H = \frac{q}{\text{moles}}$

$\Delta H = \frac{-1.25966 \, \text{kJ}}{0.0519 \, \text{mol}}$

$\Delta H = -24.26 \, \text{kJ/mol}$

Final Answer:

$\Delta H = -24.3 \, \text{kJ/mol}$

Let me know if you’d like a detailed explanation of any step!

Here are 5 related questions to deepen your understanding:

1. How does the assumption of no heat loss affect the calculation?
2. Why is the heat capacity of the solution assumed to be the same as water?
3. What would happen to $\Delta H$ if the mass of water was underestimated?
4. How would the enthalpy change differ for an exothermic process?
5. Can we use this calorimetry setup for reactions with gases? Why or why not?

Tip:

Always ensure the calorimeter is well-insulated to minimize heat exchange with the surroundings!