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

Hess's Law. The experiment conducted was meant to determine the enthalpy of formation of MgO(s) and CaO(s)

Extracts from this document...

Introduction

Hess's Law Data Collection and Processing Table V: Heats of Formation Heat of Formation (kJ/mol) Substance Trial 1 Trial 2 Magnesium oxide - 101.5 � 0.5 - 81.1 � 0.5 Magnesium - 366.1 � 0.4 - 322.2 � 0.4 Calcium oxide N/A - 141.7 � 1.0 Calcium - 192.6 � 0.5 - 173.6 � 0.5 Notes: * The heats of formation were calculated using the formula ?H = Q/nlimiting reactant Ex: Magnesium oxide Trial 1: n(MgO) = (1.08 g � 0.01 g)/40.304 g/mol = 0.026796 mol n(HCl) = (100.0 g � 0.5 g)/36.461 g/mol = 2.7427 mol MgO 1 mol 0.026796 mol ����= ����� = ������������ H2O 1 mol 0.026796 mol HCl 2 mol 2.7427 mol ��� = ����� = ���������� H2O 1 mol 1.3714 mol Therefore MgO is the limiting reactant. Assuming HCl(aq) has the same density as water, the 100.0 mL of HCl(aq) has a mass of 100.0 g. Qsurr = mc?T = (100.0 g)(4.184 J/(g��C)(25.5�C - 19.0�C) = (100.0 g)(4.184 J/g��C)(6.5�C) ...read more.

Middle

+ 2HCl(aq) --> H2(aq) + MgCl2(aq) - 322.2 � 0.4 H2(g) + 1/2O2(g) --> H20(l) - 285.8 Table XI: Enthalpy of Formation of MgO Original Equation Enthalpy of Formation (kJ/mol) Mg(s) + 1/2O2(g) --> MgO(s) - 526.2 � 0.9 Table XII: Enthalpies of Reaction After Manipulating The Given Equations for CaO Manipulated Equation Enthalpy of Reaction (kJ/mol) CaCl2(aq) + H2O(l) --> CaO(s) + 2HCl(aq) 141.7 � 1.0 Ca(s) + 2HCl(aq) --> H2(aq) + CaCl2(aq) - 173.6 � 0.5 H2(g) + 1/2O2(g) --> H20(l) - 285.8 Table XIII: Enthalpy of Formation of CaO Original Equation Enthalpy of Formation (kJ/mol) Ca(s) + 1/2O2(g) --> CaO(s) - 317.7 � 1.5 Notes: * The enthalpy of formation was calculated by adding together the enthalpies of reaction Ex: Combustion of magnesium Trial 1: ?Hf = (101.5 - 366.1 - 285.8)kJ/mol = - 550.4 kJ/mol � 0.9 kJ/mol Error: 0.5 + 0.4 = 0.9 Table XIV: Enthalpy of Formation Averages Original Equation Average Enthalpy of Formation (kJ/mol) Mg(s) ...read more.

Conclusion

was determined experimentally. There were many sources of error that could have accounted for this. When the enthalpy of reaction for magnesium was calculated, the magnesium was first sanded with sandpaper to remove and magnesium oxide that had formed on it. Similarly, calcium oxide forms on calcium, however no attempt was made to remove the calcium oxide. This would mean that more moles of calcium were accounted for than were actually present, and that calcium oxide was also reacting but this was not recorded and the amount of calcium oxide was not measured. As is visible in Table VIII, the enthalpy of reaction of calcium is greater than that of calcium oxide. Since it was not all calcium and some calcium oxide reacting in the trial that was meant to be only calcium, the calcium oxide could be responsible for the enthalpy of reaction being lower than the accepted value. This difference between the magnesium and calcium experiments could be one source of error accounting for the large difference in percent error between the two. The experiment could be improved by sanding the calcium as well. ...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. Experiment - The Empirical Formula of Magnesium Oxide

    Thus, the gradient of the scatter graph ought to be 1; though in the first graph the gradient is 0.6. After the elimination of the anomalous, the gradient became 0.7 which is closer to 1.0. It can be deduced that the off-true-value gradient was caused by some outlier points.

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

    Mg MgO Errors Mass 0.25 g 0.40 g +0.01g Volume acid used 50 cm3 50 cm3 + 1 cm3 Temperature (�C) Trial 1 Initial temperature 21 21 + 0.5�C Final temperature 38 28 + 0.5�C Change in temperature 17 7 + 0.5�C Trial 2 Initial temperature 20 22 + 0.5�C

  1. IB chemistry revision notes

    o Ethene, C2H4 --> Each of the H and H are less than 120� due to the repulsion of the double bond. The molecule is flat (planar) due to the rigidness of the double bond. Intermolecular forces Van der Waal's Weakest Dipole-dipole Hydrogen Bonding Strongest * All the forces are

  2. Lab Experiment : The change in mass when magnesium burns. (Finding the empirical formula ...

    The atomic weight of magnesium is 24.3 g / mole ) 4.Number of moles of oxygen atoms that were used. = 0.0142Mole (the number of moles of oxygen = mass / atomic weight . The atomic weight of magnesium is 16.0 g / mole )

  1. Thermodynamics: Enthalpy of Neutralization and Calorimetry

    With this data the specific heat of NaCl can be collected through the formula (m �?T�s)solution = (m �?T � s)water + (Heat Capacity � ?T)calorimeter . After obtaining the specific heat, we can go back and calculate the

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

    are as follows: 1. Mg (s) + 2HCl (aq.) ? MgCl2 (aq.) + H2 (g) ---------- ?HX 2. MgO (s) + 2HCl (aq.) ? MgCl2 (aq.) + H2O (g) ---------- ?HY 3. H2 (g) + O2 (g) ? H2O (l) ---------- ?HH2O = -285 kJ.mol-1 Adding all the three reactions, we obtained the following reaction: Mg (s)

  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. Analysis of the Standard Enthalpy of Combustion for Alcohols

    Volume of liquid used Measure 100cm3 of distilled water by using 100 cm3 ± 0.08 cm3 graduated pipette for each trial. If the volume was not exactly 100 cm3 it would directly affect the mass of the water which will affect the q=mcâT value and thus the âH value.

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