A student performed a calorimetry experiment. The neutralization reaction that took
place during the experiment is represented by the equation below.
HNO3(aq) + LiOH(aq) → LiNO3(aq) + HOH(I
Numeric Response
2. 61.1 mL of 1.00 mol/L HNO3(aq) was combined with 78.0 mL of 1.00 mol/L
LiOH(aq). The initial temperature of both solutions was 22.90°C and the highest
temperature recorded was 29.00°C. Using this data, the molar enthalpy of
neutralization of the nitric acid is +/..

To determine the molar enthalpy of neutralization of nitric acid (\(\Delta H_{\text{neut}}\)), we can use the data from the calorimetry experiment. The following steps outline the process:

1. **Determine the heat absorbed or released by the solution (\(q\))**:
   The heat change \(q\) can be calculated using the formula:
   \[
   q = m \cdot c \cdot \Delta T
   \]
   where:
   - \(m\) is the mass of the solution (in grams)
   - \(c\) is the specific heat capacity of the solution (assumed to be the same as water, \(4.18 \, \text{J/g}^\circ\text{C}\))
   - \(\Delta T\) is the change in temperature (\(T_{\text{final}} - T_{\text{initial}}\))

2. **Calculate the total volume and mass of the solution**:
   Assuming the density of the solutions is approximately \(1.00 \, \text{g/mL}\):
   \[
   \text{Total volume} = 61.1 \, \text{mL} + 78.0 \, \text{mL} = 139.1 \, \text{mL}
   \]
   \[
   \text{Total mass} = 139.1 \, \text{g}
   \]

3. **Calculate the change in temperature**:
   \[
   \Delta T = 29.00^\circ\text{C} - 22.90^\circ\text{C} = 6.10^\circ\text{C}
   \]

4. **Calculate the heat change (\(q\))**:
   \[
   q = 139.1 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 6.10^\circ\text{C}
   \]
   \[
   q \approx 3544.89 \, \text{J}
   \]
   Since \(1 \, \text{kJ} = 1000 \, \text{J}\), we convert \(q\) to kJ:
   \[
   q \approx 3.54489 \, \text{kJ}
   \]

5. **Determine the number of moles of \( \text{HNO}_3 \) reacted**:
   \[
   \text{Number of moles of } \text{HNO}_3 = 61.1 \, \text{mL} \times \frac{1.00 \, \text{mol}}{1000 \, \text{mL}} = 0.0611 \, \text{mol}
   \]

6. **Calculate the molar enthalpy of neutralization (\(\Delta H_{\text{neut}}\))**:
   \[
   \Delta H_{\text{neut}} = \frac{q}{\text{moles of } \text{HNO}_3}
   \]
   \[
   \Delta H_{\text{neut}} = \frac{3.54489 \, \text{kJ}}{0.0611 \, \text{mol}} \approx 58.03 \, \text{kJ/mol}
   \]

So, the molar enthalpy of neutralization of nitric acid is approximately \( \boxed{58.03} \, \text{kJ/mol} \).

Would you like further details or have any questions?

Here are 5 questions you might want to ask:
1. How do you convert temperature changes to heat energy in calorimetry?
2. What assumptions are made in calorimetry calculations?
3. How does specific heat capacity affect the heat change in a solution?
4. Can you explain the steps to balance a neutralization reaction equation?
5. What are some common sources of error in calorimetry experiments?

**Tip:** When performing calorimetry experiments, ensure that the calorimeter is well-insulated to minimize heat exchange with the surroundings for more accurate measurements.

Learn how to calculate the molar enthalpy of neutralization of nitric acid using calorimetry data. This problem involves determining heat changes, moles of reactants, and applying specific heat capacity principles. Suitable for high school students studying thermochemistry.

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