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

Using Calorimeter Techniques to Indirectly Determine An Enthalpy Change Using Hess Law

Extracts from this document...


Name: Vo Ngoc Kieu Duyen - Cher Date : April 5th 2011 IB Chemistry HL Mr. Brendt Bly Internal Assessment: Title: Using Calorimeter Techniques to Indirectly Determine An Enthalpy Change Using Hess' Law Aim: Determine an enthalpy change using Hess' Law and calorimeter techniques. Procedure: IB Chemistry Investigations - Volume 1 - Core. Data Collection and Processing: Table 1: The volume of 2 mol dm-3 hydrochloric acid and mass of sodium hydrogen carbonate and mass of weighing bottle & sodium hydrogen carbonate when measuring. Volume of Hydrochloric acid (HCL) 100.5 � 0.01 grams Mass of Sodium hydrogen carbonate 14.08 � 0.01 grams Mass of weighing bottle and NaHCO3 41.73 � 0.01 grams Mass of weighing bottle before adding NaHCO3 27.62 � 0.01 grams Mass of weighing bottle after putting NaHCO3 into Hydrochloric acid (HCL) 27.62 � 0.01 grams Table 2: The temperature change during the reaction of Hydrochloric acid and Sodium hydrogen Carbonate Time (s) ...read more.


and Sodium Carbonate (Na2CO3) Calculations: Part 1: HCl + NaHCO3 reaction Change in temperature (?T): ?T = 18.2 �C - 28.0 �C ?T = -9.85 � 1 �C Q = mc(?T) Q = (-9.8 �C)(4.18 J/g�C)(100.5 g) Q = 4116.8 J Q � - 4.12 kJ Sign of Q must be opposite with the sign of delta H. Therefore, Q = 4.12 kJ. Mol NaHCO3 = 14.08 g NaHCO3 ( = 0.167 mol � 0.17 mol NaHCO3 ?H = = 24.23 kJ mol-1 Part 2: HCl + Na2SO3 reaction Change in temperature (?T): ?T = 34.1 �C - 24.2 �C ?T = 9.9 � 1 �C Q = mc(?T) Q = (9.9 �C)(4.18 J/g�C)(100.5 g) Q = 4158.891 J Q � 4.16 kJ Sign (charge) of Q must be opposite with the sign (charge) of delta H. Therefore, Q = -4.16 kJ. Mol Na2SO3 = 8.08 g Na2SO3 ( = 0.076 mol � 0.08 mol Na2SO3 ?H = = - 5.2 kJ mol-1 Enthalpy change calculation: 2NaHCO3 (s) ...read more.


Since I have a high percentage error of 87%, I came up with some random errors that could have occurred throughout our experiment: I have to use an analogue thermometer and I have to open the cup to check the temperature so the reaction might not have the accurate result it is affected by the surroundings. The heat might have spread out to the atmosphere around it when I opened the cup to check the temperature in the second experiment. In the first one, the temperature might be affected by the heat from the surroundings and it couldn't reach the coolest point. Next time, I should use an electric probe (digital thermometer) to minimize the heat loss or heat gain to/from the surroundings. Another thing that has could gone wrong is that throughout the experiment, I put the substance in the acid bit by bit and I moved the cup in circular motions a lot; that could be an error. Because the acid didn't have the enough amount of sodium to react so the results were not very accurate. ...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. Calcium Carbonate and Hydrochloric Acid

    This should though not interfere with the methods used to answer the research question (see methods and explanations below). As mentioned before, to answer the research question, "How does reaction rate change with increasing surface area of calcium carbonate in hydrochloric acid ?", one must either compare the loss of

  2. IB chemistry revision notes

    * Water (Phase Equilibrium) o Rate of vaporization = Rate of condensation. o Any liquid exists in equilibrium with its gas. These equilibria shift their position in exactly the same way as chemical equilibria. o o Heat shifts the equilibrium to the right; pressure, to the left.

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

    The molar enthalpy of one reaction was given and the other molar enthalpies were determined experimentally. A temperature probe was used to calculate the temperature change in both of the reactions. Because of poor isolation, all the graphs were extrapolated according to their cooling rate after the maximum temperature, but

  2. Thermodynamics: Enthalpy of Neutralization and Calorimetry

    The average heat of neutralization was found to be to -2435.453 cal. This enthalpy is negative which means that heat is being released by the system into the surroundings. This also means that the reaction NaOH + HCl --> H2O + NaCl is exothermic.

  1. Can one determine the coefficients of a balanced chemical equation by having the mass ...

    because there were a lot of errors encountered throughout the experiment that changed the value of the coefficients. In addition there is a percentage error of approximately 3% for the coefficient of copper. This is a very large error and this could very well indicate that the mass of the coefficient going on the copper could be far off.

  2. To determine the molecular mass of an unknown alkali metal carbonate, X2CO3.

    In this case, the percentage uncertainty for the readings of the digital balance was 0.67% which is the largest percentage uncertainty in comparison to the other instruments such as the burette or the pipette. If these uncertainties could be avoided a result much closer to the literature would be possible to obtain.

  1. Lab report. Finding the molar enthalpy change of the reaction between Hydrochloric acid and ...

    For we know the mass so in order to fin the number of moles we divide the mass by its molar mass so getting .019 moles when we used 2 grams while when we used 3 grams we got 0.28.

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

    24.0 24.0 0 24.0 24.0 24.0 24.0 15 29.0 30.0 33.5 31.0 30 36.0 37.0 36.0 36.0 45 41.5 42.0 43.5 42.0 60 50.0 51.0 54.0 52.0 75 68.0 69.0 70.0 69.0 90 82.0 83.0 84.0 83.0 105 83.5 84.0 87.0 85.0 120 89.0 90.0 93.0 91.0 135 90.0 89.5

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