• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15

Research question: how to convert NaOH to NaCl by two different routes , and measure the enthalpy changes for each one in order to test Hesss law ?

Extracts from this document...


Name: Rand Ishaq School: katedralskolan chemistry IA 2011-12-17 converting NaOH to NaCl by two different routes , and measuring the enthalpy changes for each one in order to test Hess’s law . aim: the aim is to measure and compare the quantity of heat involved in three reactions and to provide experimental verification of Hess's Law research question: how to convert NaOH to NaCl by two different routes , and measure the enthalpy changes for each one in order to test Hess’s law ? general background: The foundation of the study of thermochemistry was laid by the chemist Germain Hess, who investigated heat in chemical reactions during the last century. One statement of the law that bears Hess's name says: The enthalpy change for any reaction depends on the products and reactants and is independent of the pathway or the number of steps between the reactant and product In this experiment, the enthalpy of several reactions will be determined using the method of calorimetry. The reactions that will be considered are the following: 1: NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) All of the reactions in this experiment are exothermic. As heat is given off for each reaction, the temperature of the mixture will increase. The amount of heat given off by the reaction can be calculated according to the following equation: q= mcΔT Where: m = mass (g) c = specific heat capacity (The specific heat capacity of each mixture can be approximated as that of water, 4.18 J/g ºC.) ...read more.


From graph (2) highest temperature could be determined to 32.2ËC. Calculations: uncertainties Percentage uncertainty Mass of H2O and HCl (Assuming that the solution has the same density of water) 100.0g H2O: 1.0% HCl : Temperature 0.1 º C 1.6% Number of moles (NaOHaq) Since n = Un. for the mass of NaOH = 0.1g Un. for the volume of water = 5x10-4 dm3 Un. for the volume of HCl = 3x10-5dm3 = -m (H2O + HCl) × c (H2O) × âT = -100.0× 4.18 × = -53.5 kJ mol-1 Total percentage uncertainty of =1.0%+ 0.06% + 1.6% + 2.5% + 0.06%+1% ≈ 6.22% to count the absolute uncertaity of the = = -54kJmol-1 (± 3.4) So : NaOH (s) + 2M (l) 2M NaOH (aq) = 31 kJmol-1(±1.5 kJmol-1) + 2M NaOH (aq) + 2M HCl (aq) 1M NaCl (aq) = 54 kJmol-1 (±3.4 kJmol-1) NaOH (s) + 2M (l) + 2M HCl (aq) 1M NaCl (aq) âH = -85 (±4.9 kJmol-1) Table 3 : the table below shows the raw data collected when adding 4.0 g of sodium hydroxide NaOH to 50.00 cm3 of HCl in the thermos and measuring the temperature change by applying route (B) : NaOH (s) + 2M (l) 2M NaCl (aq) 1. The bolded and underlined data presents the original temperature of the HCl before adding the sodium hydroxide. time (t) /s ±1 temperature (T) / ºc ±0.1 0 23.8 30 23.8 60 59.6 120 58.8 150 55.8 180 54.4 210 53.8 240 53.1 270 52.2 300 51.3 330 50.3 360 49.3 390 48.5 420 47.3 450 46.3 Graph (3) ...read more.


by swirling the solutions in both route (A) (B) that created a hole in the solution, in a result temperature readings could have been inaccurate because the thermometer would not encounter that much of the solution molecules and may be take the reading of the air temperature in that hole, and therefore the final calculations are incorrect. In addition: the NaOH was not grind well, so there were some pieces that were relatively large, so they took longer time to dissolve which might have changed the reaction results , an improvement of this error could be to use an electrical mortar or a coffee grinder so that all NaOH is grind evently One last point is that the thermos was washed immediately with water after each reaction , and then the next reaction was performed the thermos was not dried carefully. It is highly possible that some areas of the thermos were not washed down properly and some drops of water were left. This would have changed the outcome of the reaction as well as the enthalpy calculation at the end because it would give different temperature changes. So the error that encountered in this experiment is systematic errors more than random errors. The experiment was successful to the degree that the procedure was carried out correctly. Equipment problems and other unavoidable sources of error served to cause a large percent error for each part of the experiment. Assuming that the solution has the same density and specific heat capacity as water According to "Chemistry" by Raymond Chang (9th ed), it is -56.2 kj/mol p.243. http://www2.ucdsb.on.ca/tiss/stretton/chem2/enthlab2.htm 50.0 cm3 of Water added 50.0 cm3 of Water added âT % error = ( i found it to be -56kjmol-1) ...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. IB chemistry revision notes

    Given in * Ways of measurement: o Change in mass as gas escapes. o Collect a gas given off and look at volume changes. o The formation of a p.p.t. o Use of an indicator to show the end of reaction (different [A] and T)

  2. IB IA: Determination of Heat of Neutralization

    Experiment 2: Equation for the experiment: HNO3(aq) + KOH(aq) � KNO3(aq) + H2O(l) Heat evolved = mc? = 0.1 x 4.18 x 5.75 = 2.4035 kJ Number of moles of acids used = 1 x 0.05 = 0.05 moles From the equation, 1 mole of HNO3 reacts with 1 mole of KOH and gives 1 mole of water.

  1. flame test lab

    Energy can be added to atoms many different ways. It can be in the form of light, an electric discharge or heat. This added or extra energy is emitted when the excited electrons in the atoms give off light and fall back to lower shells. The light emitted has wavelengths and colors that depend on the amount of energy originally absorbed by the atoms.

  2. The Enthalpy of Neutralization

    x 100 = Percentage error of volume] (1�50) x100= 2% [(Nearest unit of temperature measurement observed � recorded temperature) x 100 = Percentage error of recorded temperature] (1�13) x 100= 7.7% [Percentage value of volume error + percentage value of temperature error = relative error] 2%+7.7% = 9.7% relative error Enthalpy change in joules with relative error percentage

  1. Research question - How many molecules are there in a liquid drop?

    the average mass of one drop is the final value needed). I have rounded off those averages to three decimal places (instead of one) as the values are very small. The average mass of one drop has been rounded off to the same number of places as the standard deviation, that is two significant figures.

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

    If one of the solutions of KOH(aq) or HCl(aq) in the 150 cm3 beaker reaches the desired temperature of 19�C quicker, it can be briefly removed from the ice bath and re-introduced as necessary until the other solution catches up. Throughout the cooling process, the temperatures of both 40.0 cm3 samples of 1.00 mol dm-3 KOH(aq)

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

    = Tmin – Tmax = 24.0 °C – 74 °C = 50°C ΔT (H2O) = 50K Knowing this , we can calculate the total amount of energy that has been transformed to water : Q = 40.0 g * 4.20 J g-1 K-1 * 50 K Q = 8400J =

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

    The same ruler was used for all trials and 3cm of the Magnesium strip was cut each time using strong wire cutters. Volume of HCl No. of moles of HCl remains constant. The 15cm3 was measured using a burette to reduce the percentage uncertainty.

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