• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Experimental Molar Enthalpy of Neutralization for Sodium Hydroxide Solution

Extracts from this document...

Introduction

Yundi Wang October 22, 2012 Molar Enthalpy of Neutralization ________________ 1. For information regarding the problem, prediction, materials and procedure, please see attached Measurements Table for Molar Enthalpy of Neutralization for Sodium Hydroxide Solution Substance Instrument Used Measurement Sodium hydroxide 100mL graduated cylinder (±0.2mL) 50.0mL Sulfuric acid 100mL graduated cylinder (±0.2mL) 30.0mL Temperature of sodium hydroxide solution Thermometer (±0.2ËC) 26.0ËC Temperature of the sulfuric acid Thermometer (±0.2ËC) 24.0ËC Final temperature reached by solution Thermometer (±0.2ËC) 34.5ËC Initial and Final Temperatures of Solutions Temperature of sodium hydroxide solution (±0.2ËC) 26.0ËC Temperature of the sulfuric acid (±0.2ËC) 24.0ËC Final temperature reached by solution (±0.2ËC) 34.5ËC Neutralization Reaction Taking Place Pre-Lab Calculations – Volume of Sulfuric Acid Needed Average Initial Temperature of Solutions Calculation Experimental Molar Enthalpy of Neutralization for Sodium Hydroxide Solution Calculation Solution 1. The experimental molar enthalpy of neutralization for sodium hydroxide solution was found to be -64±3.3KJ/mol. Calculation of Uncertainties 34.5±0.2ËC – 25.0±0.2ËC =9.5±0.2ËC 50±0.2mL + 30±0.2mL =80±0.2mL 9.5 ± 0.4ËC = 4.210…% 80 ± 0.4mL = 0.5% 50 ± 0.2mL = 0.4% =5.11…% =5.1% Percent Difference Conclusion Through a pre-lab calculation the amount of sulfuric acid solution needed was found to be 30.0m±0.2mL. ...read more.

Middle

In saying that, it is possible within this lab the reactants were concentrated in one area causing the experimental change in enthalpy to be quite large. Because it is impossible to see into the calorimeter to see if the reaction is concentrated or when the reaction is complete the reactants could easily have been concentrated in one area. Furthermore, by not knowing when the reaction is complete, the temperature might be measured too soon or too late causing inaccurate results. In general, because the calorimeter is an isolated environment it results in the experiment having many errors because how the reaction is occurring and when the reaction is finished is unknown. A way to eliminate this error is by inserting an electronic stirring rod to stir the reactants so they do not become concentrated in one area. Furthermore, another reason contributing to the large enthalpy change is the impurity of the substances used. As a result, because the substances are impure, they could have had a higher concentration of reactants. With a higher concentration of reactants, the reaction rate will increase and there will be a greater reaction than wanted. With a larger reaction at an increased rate, the final temperature of the solutions will spike higher than wanted generating a larger enthalpy change. ...read more.

Conclusion

A hole is needed to be made to insert the thermometer. And there were many holes between the lid of the calorimeter and the calorimeter itself. Due to this ineffectiveness of the Styrofoam calorimeter, some of the heat from the reaction would have escaped through the many holes causing a lower final temperature of the reaction and the experimental enthalpy change to be lower than the theoretical (actual) value. As a result, the experimental value is usually lower than the theoretical (actual) value. Another reason includes the fact that some of the heat released during the reaction would have been transferred to the calorimeter itself instead of transferring to the thermometer. As a result, when the calorimeter and/or glass of the thermometer absorb the heat, it causes the thermometer to absorb less heat than it should. The final temperature will then be lower than it should be causing a lower enthalpy change. Even though this is not a main reason why the experimental molar enthalpy should be lower than the actual molar enthalpy it still contributes to it. As a result, with the combination of these factors the experimental enthalpy change should be lower than the theoretical value because a lot of heat is able to escape into the calorimeter and into the air due to there being holes in the calorimeter. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Investigate the rate of reaction of luminol in various factors. The objective was to ...

    Store In a dark bottle in a cool place. Take care when removing the cap as pressure may have built up. 100'vol' hydrogen peroxide should be kept with other corrosive liquids, non-acid and should only be diluted immediately before use (as the inhibitor, which slows down decomposition, is diluted as well).

  2. Enthalpy Change Design Lab (6/6)How does changing the initial temperature (19C, 25C, 35C, and ...

    or HCl(aq) is negligible itself - even more so considering the dilute concentration being used in the investigation. Another component of measuring the molar enthalpy change of a neutralization reaction is the change from the initial temperature of the reactants, to the final temperature of the resulting solution.

  1. IB IA: Determination of Heat of Neutralization

    Therefore, 0.05 moles of HCl reacts with 0.05 moles of NaOH and gives 0.05 moles of water. Therefore, heat of neutralization = 2.508 � 0.05 = -50.16kJmol-1 = -50.2kJmol-1 (3 s.f.)

  2. Hess's Law. The experiment conducted was meant to determine the enthalpy of formation of ...

    Evaluation Clearly, while determining the enthalpy of formation of CaO(s), there were many more errors present than when the enthalpy of formation of MgO(s)

  1. Hesss Law Lab, use Hesss law to find the enthalpy change of combustion of ...

    �change in temperature= 22 kJ/mol Q (energy)= m(mass of HCl solution)* c(specific heat capacity of water) *?T (temperature change) Q = 0.05kg x 4200 J/kg�C(from data booklet) x 17 �C = 4620 J = 4.62 kJ I will carry out the following calculations to find the limiting reactant The ratio

  2. The aim of this experiment is to examine the enthalpy of combustion of the ...

    * c (H2O) * ΔT (H2O) Q – heat energy M – mas of water C- heat capacity of water , the value is 4.20 J g-1 K-1 ΔT – change in temperature In the method it has been said that the volume of water is 40 cm ³.

  1. To determine the standard enthalpy of formation of Magnesium Oxide using Hess Law.

    Once 120 seconds were up, the calorimeter was emptied and cleaned so that another trial of the same experiment could be performed. Part Y ? With Magnesium Oxide Powder 1. The calorimeter was cleaned thoroughly ensuring no chemicals were left behind which would hinder in the reaction in part Y.

  2. Bomb calorimetry. The goal of this experiment was to use temperature data over ...

    0 - 0 = 8.0 oC = 281K For naphthalene, â T= 23.2 – 15.5 – 0 - 0 = 7.7 oC = 280.7K The eq. (7) rearranges to CV = (âcUbenzoic acid+ âcUcotton)/(- âTbenzoic acid) CV = (-26.434 kJ/g x 0.7408 g– 0.059 kJ)/[- 281 K] = 0.0699 kJ

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work