• 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. hess's law

    * ?Ha = (- 78.4 � 10.2 KJ mol-1) Calculating ?Hp: No. of moles of Hydrated copper (II) sulphate = 0.020 moles Mass of solution = 50 g Specific Heat Capacity of water = 4.18 J g-1 K-1 Change in Temperature = (4.0 � 1.0)

  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. The Enthalpy of Neutralization

    x 100 = Percentage error of recorded temperature] (1�12.25)x100= 8.2% [Percentage value of volume error + percentage value of temperature error = relative error] 2%+8.2% = 10.2% relative error Enthalpy change in joules with relative error percentage Q= -2560J � 10.2% Conversion of Relative Error to Absolute Error (2560�100)

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

    (28 + 0.5)�C (27 + 0.5)�C (28 + 0.5)�C EXPERIMENT 1 A time/s(�1sec) Temp.0C (�0.50C ) HCl 0 21 30 21 60 21 90 21 120 21 150 21 180 21 HCl+Mg 190 22 200 24 210 26 220 28 230 30 240 31 250 32 260 33 270 34

  1. IB chemistry revision notes

    * Lewis structures o Lewis states the idea that atoms tend to bond in order to have eight electrons in the outer shell. This idea became known as the octet rule. E.g. Fluorine: or * Covalent Bond Lengths o The minimum distance between the two nuclei of the atoms is 1 covalent bond length.

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

    * c (H2O) * ΔT (H2O) Q – heat energy M – mas of water C- heat capacity of water , the value is 4.20 J g-1 K-1 ΔT – change in temperature In the method it has been said that the volume of water is 40 cm ³.

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

    was -5160 ± 20kJ/mol % error = |1-Experimental value/literature value|x100% = |1-(-4982.66)/(-5160)|x 100% = 3.44% Discussions This experiment was a success as each objective of the experiment was completed and the experimentally determined enthalpy of combustion of solid naphthalene was very close to its literature value.

  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)

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