• 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. The Enthalpy of Neutralization

    x 10.2= 261 Absolute Uncertainty = �261J --> Q= -2560J � 261J (�1000) [To change to kilojoules] Q= -2.56 kJ � 0.261kJ (�0.050mol H2O) [To find kilojoules per mole of water] Q = -51.2 kJ�mol-1 H2O (� 5.22 kJ�mol-1 H2O)

  2. 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.

  1. Thermodynamics: Enthalpy of Neutralization and Calorimetry

    cal 285.5615 �?.286 cal (x cal/gC�) (1268.394 �.00991 gC�) = 285.5615 �?.286 cal x cal/gC� = .225 �?.001?cal/gC� Specific Heat of NaCl Mass of NaCl Solution 101.862 �.001 g -7.921 �.001g 93.941 �.0014

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

    (21 +0.5)�C (20 +0.5)�C (21 +0.5)�C Highest temperature recorded of solution (Tf) (38 + 0.5)�C (39 + 0.5)�C (39 + 0.5)�C Table 2: Data observed for Reaction #2 between hydrochloric acid solution and magnesium oxide Trial 1 Trial 2 Trial 3 Mass of magnesium oxide (mMgO)

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

    + 0.0001 =27.9850g Mass of MgO inside crucible with lid in grams after �heating to constant mass`=28.2125g Processed Data: Percentage uncertainty= 0.04% 1.Mass of magnesium (in grams) + 0.0001= 0.3450g (The mass of magnesium ribbon = The mass of the crucible, lid and magnesium- mass of the crucible and lid)

  2. Confirming Hess's Law Experiment

    since the solution was excessively diluted. Therefore, the molar enthalpies for both reactions were calculated. The reaction involving the magnesium oxide and hydrochloric acid was determined twice, one with a rubber stopper and one without. The trial with the rubber stopper, Trial #1, proved to be more accurate, with a

  1. Analysis of the Standard Enthalpy of Combustion for Alcohols

    Pressure of surroundings For standard enthalpy of combustion the pressure must be 1 atm, however in a classroom this is hard to obtain, so all experiments will be done in a room with the same pressure. Might influence the vapour pressure point, which will affect the q=mcâT value, and thus the âH.

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

    Qualitative data Mass of methanol Mass 1 Mass 2 Mass 3 Average mass Initial mass g ± 0.01 133.47 136.52 138.42 136.13 Final mass g ± 0.01 132.15 135.31 137.39 134.95 Mass of methanol used g ± 0.02 1.32 1.21 1.03 1.18 Ethanol ( C2H5OH ) Time ( s )

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