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

Thermodynamics: Enthalpy of Neutralization and Calorimetry

Extracts from this document...


Thermodynamics: Enthalpy of Neutralization/ Calorimetry Introduction The First Law of Thermodynamics states that energy can be neither created nor destroyed. This means that energy, instead of 'disappearing', is either transformed, transferred, dispersed, or dissipated. When energy is lost by a system, it will be acquired by the surroundings. Heat can be described as the amount of energy needed to cause the temperature of a substance to rise and it is transferred from warmer areas to cooler ones. In order to be able to measure the change in heat or enthalpy of a reaction, a colorimeter can be used. The calorimeter was first introduced in the 18th century and can be used with any procedure that involves the flow heat between a system and it's surroundings (CACT). It is capable of measuring the heat created or exchanged after a reaction has occurred in a system with a constant pressure. A calorimeter can be used to find the specific heat of a substance or even the heat of neutralization between a base and an acid. A basic calorimeter is composed of two Styrofoam cups (to provide insulation and prevent heat from entering or exiting the system), a lid covered with aluminum, a thermometer, and a stirrer. Since the calorimeter will still release heat, the first step is to find the heat capacity of it which is the heat absorbed by the calorimeter. In order to find the heat capacity of a substance, the item must be weighed, heated, dropped into a calorimeter filled with cool water, and allowed to cool for a minute. During this whole process data will be collected including the weight of the item and the water, the temperature of the water and of the item; this information will form the base for the formula through which the specific heat of the item can be measured. In this experiment the specific heat for a unknown metal will be calculated. ...read more.


g Mass of Hot Water 148.68 �.01 g -101.862 �.001 g 46.818 �.01 g Specific Heat of NaCl Solution (m �?T � s)water = (m �?T�?s)solution+ (Heat Capacity � ?T)calorimeter (1.00 cal/gC�) (46.818 �.001 g) (65.6 �.5 C�- 39.3 �.5 C�) = (x cal/gC�) (93.941 �.0014 g)(39.3 �.5 C�-34.5 �.5 C�) + (39.3 �.5 C�-34.5 �.5 C�) (12.456 �.0351cal/C� ) 65.6 �.5 C�- 39.3 �.5 C� (39.3 C�-34.5 C�(�.5)) 26.3 �.71 C� 4.8 �.71 C� (1.00 cal/gC�) (46.818 �.01 g) (26.3 �.71 C�) = (x cal/gC�) (93.941 �.0014 g)(4.8 �.71 C�) + (4.8 �.71 C�) (12.456 �.0351cal/C� ) 46.818 �.01 g � 26.3 �.71 C� 93.941 �.0014 g �?4.8 �.71 C� 1231.3134 �?.027 gC� 450.9168 gC� �.148 4.8 �.71 C� � 12.456 �.0351cal/C� ???????� .148 cal (1.00 cal/gC�) (1231.3134 �?.027 gC�) = (x cal/gC�) (450.9168 gC��.148) + (??????� .148 cal) 1231.3134 �?.027 cal = (x cal/gC�) (450.9168 gC��.148) + (??????� .148 cal) (1231.3134 �?.027 cal) - (??????� .148 cal)= (x cal/gC�) (450.9168 gC��.148) (1231.3134 �?.027 cal) - (??????� .148 cal) 1171.5244 �??????cal 1171.5244 �??????cal = (x cal/gC�) (450.9168 gC��.148) x cal/gC� = 2.598 �.00035 cal/gC� Neutralization of NaOH with HCl Mass of NaCl 101.862 �.001 g - 7.921 �.001 g 93.941 g �.0014 Heat Evolved from Neutralization q neutralization = (m �?T � s)Solution + (Heat Capacity � ?T)calorimeter q neutralization = (m (Final Temp - � s)Solution + (Heat Capacity � ?T)calorimeter q neutralization = (93.941 �.0014 g) ((35.5 �.5 C�)-) (2.598 �.00035 cal/gC�) + (12.456 �.0351cal/C� )((35.5 �.5 C�)- ) 23.55 �.71 C� q neutralization = (93.941 �.0014 g) (35.5 �.5 C�- 23.55 �.71 C�) (2.598 �.00035 cal/gC�)+ (12.456 �.0351cal/C� )(35.5 �.5 C� - 23.55 �.71 C�) 35.5 �.5 C� - 23.55 �.71 C� 11.95 �.868 C� q neutralization = (93.941 �.0014 g) (11.95 �.868 C�) (2.598 �.00035 cal/gC�)+(12.456 �.0351cal/C� )(11.95 �.868 C�) 93.941 �.0014 g � 11.95 �.868 C� 12.456 �.0351cal/C� �11.95 �.868 C� 1122.59 �.0726 gC� 148.849 �.0727 cal q neutralization cal = (1122.59 �.0726 gC�)(2.598 �.00035 cal/gC�) ...read more.


In the data this can be seen through the way that the difference between the temperature of the hot water and the final temperature decreases significantly in the third and final trial by which time the procedures were followed promptly. The use of the stirring rod, even though useful to helping keep the same temperature throughout all the content of the calorimeter could also have added to a source of error because it was inside the system while the neutralization reaction was taking place, and glass like all substances absorbs heat which wasn't accounted for in the calculations. These are all small sources of error but they impact the results non-the-less. Additionally, one last source of error is the fact that we changed the calorimeter used in the first two parts of the experiment (heat capacity of the calorimeter and the specific heat of the metal) to the other last two parts (the heat of neutralization and the specific heat of the NaCl solution). Since we used a completely different calorimeter in part two, using the heat capacity of the calorimeter of part one will have given us inaccurate results because the right values weren't used to complete the calculations. Were the experiment to be performed again, several steps could be taken in order to improve the process. One is to reinforce the calorimeter in order to improve it's efficiency by adding more insulation. Another is to become familiar with the procedures beforehand. Instead of using a glass rod for the experiment perhaps an automatic stirrer could be used and it's heat capacity could be calculated and then added to the equation in order to get a more accurate value. Calorimetry overall is a very useful process. With this technology we could test several chemical reactions and test to see which ones have higher enthalpies. With this alternative fuels that are equally or more efficient than the gas and oil we use now could be found. It could also be used by dieticians to investigate what foods react with the body to give it more energy. ...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. Peer reviewed

    Calculating the specific heat of a metal

    4 star(s)

    used for calculation: HLM = HGW Note: - ?t1 is the change in the temperature of lead - ?t2 is the change in temperature of the water in the cup V. Conclusion: In this experiment, the specific heat was determined through calculation using the measurements found in Table 1 and throughout data collection.

  2. IB IA: Determination of Heat of Neutralization

    Therefore, 0.05 moles of HNO3 reacts with 0.05 moles of NaOH and gives 0.05 moles of water. Therefore, heat of neutralization = 2.299 � 0.05 = -45.98kJmol-1 = -46.0kJmol-1 (3 s.f.) Experiment 2: Equation for the experiment: HNO3(aq) + KOH(aq)

  1. The Enthalpy of Neutralization

    Notably however, was that the mixture of any two reagents gave off heat; meaning that the reaction in all cases was definitely exothermic. But again this was to be expected. Any change in temperature of the two aqueous reagents whatsoever was to be expected, as that is what we were looking to measure and observe in the first place.

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

    Therefore, I will use the mathematical procedure to find the heat of combustion of magnesium. We have determined the following values for the two reaction and one is given to us. ?H 1 = Mg(s) + 2HCl (aq) � MgCl2(aq)

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

    The two 150 cm3 beakers containing 40.0 cm3 of 1.00 mol dm-3 KOH(aq) and HCl(aq) separately, will then be placed in the ice bath at the same time until they reach the desired temperature of 19�C. If one of the solutions of KOH(aq)

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

    The bomb, used in bomb calorimetry, is a completely sealed and oxygen filled metal container. This is placed in an insulated jacket containing a pail of water and a thermometer, all combined to form the calorimeter (http://jr.stryker.tripod.com/physchem/calorimetry.html). The bomb allows for both a constant volume of the container and for

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

    Average mass Initial mass g ± 0.01 139.42 144.21 135.15 139.59 Final mass g ± 0.01 137.01 141.82 133.78 137.5 Mass of methanol used g ± 0.02 2.41 2.39 1.37 2.09 Propanol ( C3H7OH ) Time ( s ) ±1 s Temperature in °C ± 1 °C Average temperature in

  2. Measuring the fatty acid percentage of the reused sunflower oil after numerous times of ...

    5.096 ± 0.020% = 0.7% ± 0.8% ml/g oleic acid 26 ASLAN Özge Cemre D129077 Means of the trials of the samples: Mean of sample 1 = (3.6% ± 0.1 + 3.6% ± 0.1 + 3.7% ± 0.1)

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