10.3: Enthalpy of Neutralization - Data and Report

 

Name: ____________________ Partner(s): ____________________ Date: ____________________

 

DATA AND OBSERVATIONS

Part A: Heat Capacity of the Calorimeter

Table \(\PageIndex{1}\): Part A Data

	Calculation/Result	Units
Temperature of calorimeter (and cold water) before mixing		°C
Temperature of warm water		°C
Final temperature		°C
Temperature change of warm water \(\Delta T_{\text{warm}} = T_{\text{warm}} - T_{\text{final}}\)		°C
Temperature change of calorimeter \(\Delta T_{\text{calorimeter}} = T_{\text{final}} - T_{\text{calorimeter}}\)		°C
Heat lost by the warm water \(q_{\text{warm}} = \Delta T_{\text{warm}} \times 5.0\ \text{g} \times 4.18\ \text{J/g·°C}\)		J
Heat gained by the cold water \(q_{\text{cold}} = \Delta T_{\text{calorimeter}} \times 5.0\ \text{g} \times 4.18\ \text{J/g·°C}\)		J
Heat gained by the calorimeter \(q_{\text{calorimeter}} = q_{\text{warm}} - q_{\text{cold}}\)		J
Heat capacity of the calorimeter \(C_{\text{calorimeter}} = q_{\text{calorimeter}} / \Delta T_{\text{calorimeter}}\)		J/°C

 

Part B: Enthalpy of Neutralization of \(\ce{HCl}\)–\(\ce{NaOH}\)

Table \(\PageIndex{2}\): Part B Data

	Calculation/Result	Units
Temperature of calorimeter (and \(\ce{NaOH}\)) before mixing		°C
Maximum temperature reached		°C
Temperature change \(\Delta T = T_{\text{max}} - T_{\text{calorimeter}}\)		°C
Heat gained by the solution \(q_{\text{solution}} = \Delta T \times 10.0\ \text{g} \times 4.18\ \text{J/g·°C}\)		J
Heat gained by the calorimeter \(q_{\text{calorimeter}} = \Delta T \times C_{\text{calorimeter}}\)		J
Total heat released \(q_{\text{total}} = q_{\text{solution}} + q_{\text{calorimeter}}\)		J

Table \(\PageIndex{3}\): Part B Calculations

	Calculation	Result
Balanced equation for the reaction:
Moles of \(\ce{HCl}\) in 5.0 mL of 1.0 M \(\ce{HCl}\)		mol
Moles of \(\ce{NaOH}\) in 5.0 mL of 1.0 M \(\ce{NaOH}\)		mol
Moles of water produced (\(n_{\text{water}}\))		mol
Enthalpy of neutralization per mole of water \(\Delta H_{\text{neutralization}} = q_{\text{total}} / n_{\text{water}} / (1000\ \text{J/kJ})\)		kJ/mol

 

Part C: Enthalpy of Neutralization of \(\ce{CH3CO2H}\)–\(\ce{NaOH}\)

Table \(\PageIndex{4}\): Part C Data

	Calculation/Result	Units
Temperature of calorimeter (and \(\ce{NaOH}\)) before mixing		°C
Maximum temperature reached		°C
Temperature change \(\Delta T = T_{\text{max}} - T_{\text{calorimeter}}\)		°C
Heat gained by the solution \(q_{\text{solution}} = \Delta T \times 10.0\ \text{g} \times 4.18\ \text{J/g·°C}\)		J
Heat gained by the calorimeter \(q_{\text{calorimeter}} = \Delta T \times C_{\text{calorimeter}}\)		J
Total heat released \(q_{\text{total}} = q_{\text{solution}} + q_{\text{calorimeter}}\)		J

Table \(\PageIndex{5}\): Part C Calculations

	Calculation	Result
Balanced equation for the reaction:
Moles of \(\ce{CH3CO2H}\) in 5.0 mL of 1.0 M \(\ce{CH3CO2H}\)		mol
Moles of \(\ce{NaOH}\) in 5.0 mL of 1.0 M \(\ce{NaOH}\)		mol
Moles of water produced (\(n_{\text{water}}\))		mol
Enthalpy of neutralization per mole of water \(\Delta H_{\text{neutralization}} = q_{\text{total}} / n_{\text{water}} / (1000\ \text{J/kJ})\)		kJ/mol

 

POST-LAB QUESTIONS

1. An unknown metal sample weighing 9.25 g is heated to 97.8 °C and placed in 50.0 g of water at 24.5 °C in a calorimeter. The final temperature is 27.4 °C. What is the specific heat capacity of the metal?

    

    

    

2. A solution is prepared from 2.50 g of \(\ce{NaOH}\) in 75.0 mL of water (both at 25 °C). The temperature rises to 25.5 °C. Is the process exothermic or endothermic? What is the enthalpy of solution of \(\ce{NaOH}\) in kJ/mol?

    

    

    

3. How does the enthalpy in Part B compare to the calculated enthalpy in Pre-Lab Question #3?

    

    

    

4. How does the enthalpy in Part B compare to Part C?