Calculating Enthalpy of Neutralization in Calorimetry Experiment

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SUMMARY

The discussion centers on calculating the enthalpy of neutralization for a calorimetry experiment involving 100 mL of 0.500 M acetic acid and 100 mL of 0.500 M sodium hydroxide. The temperature change observed was from 25.00 °C to 27.55 °C, with a calorimeter heat capacity of 150.48 J/°C and a solution density of 1.034 g/mL. The calculated enthalpy of neutralization is definitively -50.2 kJ/mol. Participants emphasize the importance of demonstrating prior effort before seeking assistance.

PREREQUISITES
  • Understanding of calorimetry principles
  • Knowledge of molarity and solution density
  • Familiarity with specific heat capacity calculations
  • Basic thermodynamics concepts related to enthalpy
NEXT STEPS
  • Study the calculation of enthalpy changes in chemical reactions
  • Learn about the principles of calorimetry and heat transfer
  • Explore the concept of specific heat capacity in detail
  • Investigate the effects of concentration on enthalpy of neutralization
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Chemistry students, laboratory technicians, and educators involved in thermodynamics and calorimetry experiments will benefit from this discussion.

boredooom
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Could I get a solution to this question?

In a laboratory experiment in calorimetry, 100 mL of 0.500 M of acetic acid is mixed with 100 mL of 0.500 M sodium hydroxide in a calorimeter. The temperature rises from 25.00 C to 27.55 C. The heat capacity of the calorimeter is 150.48 J/C, and the density of the resulting solution is 1.034g/mL. Given that the specific heat capacity of 0.250 M of sodium acetate is 4.034 Jg-1C-1, calculate the enthalpy of neutralization of acetic acid.

Answer: -50.2kJ/mol
 
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boredooom said:
Could I get a solution to this question?

In a laboratory experiment in calorimetry, 100 mL of 0.500 M of acetic acid is mixed with 100 mL of 0.500 M sodium hydroxide in a calorimeter. The temperature rises from 25.00 C to 27.55 C. The heat capacity of the calorimeter is 150.48 J/C, and the density of the resulting solution is 1.034g/mL. Given that the specific heat capacity of 0.250 M of sodium acetate is 4.034 Jg-1C-1, calculate the enthalpy of neutralization of acetic acid.

Answer: -50.2kJ/mol

perhaps you should explain what you've done so far. We don't help you with your homework unless you demonstrate you have at least tried to do it yourself.
 

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