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Using Hess's law to calculate enthalpy change

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Introduction

Chemistry lab report (5) Using Hess law to calculate enthalpy change Aim: To calculate the enthalpy change for the reaction of: (MgSO4(s) + 7H2O MgSO4.7H2O(s) ) Hypothesis: The literature value according to research is -104 Kj/mol Variables: Dependant variable: The final temperature of the solution Independent variable: --------------------------------- Controlled variables: - Mass of water - Mass of magnesium sulphate anhydrous - Mass of MgSO4.7H20 Materials: 1) Safety spectacles 2) 2 filter papers 3) Spatula 4) Digital balance 5) 2 polystyrene cups 6) Thermometer 7) Magnesium sulphate anhydrous 8) Magnesium sulphate-7- water 9) Distilled water 10) Measuring cylinder Procedure: Part A 1) Weigh 3.01g of MgSO4 anhydrous to the nearest 0.01g on a filter paper, using the digital balance. 2) Weigh 45.00g of water to the nearest 0.01g into a polystyrene cup using the balance. 3) Measure the initial temperature of the measured amount of water using the thermometer and record this value. 4) Add the measured amount of MgSO4 anhydrous to the water and stir. Measure and record the maximum temperature obtained using the thermometer. Part B 1) ...read more.

Middle

Number of moles of water = mass / molar mass = 45.1 / 18.0 = 2.50 moles *Number of moles of MgSO4(s) = Mass/ Molar mass = 3.02 / (24.31 + 32.06 + (4 x 16.00)) = 3.02 / 120 = 0.0251 moles * Ratio of moles = 1 : 100 Thus both reactants are limiting * Enthalpy change = Q / number of moles = 603 / 0.0251 = - 24023.90 (rounded to 3 significant figures) = -24000 J/mol (Exothermic reaction) Enthalpy change for part B: (MgSO4.7H20(s) + 93H2O(l) MgSO4(aq,100H2O)) * Q = m x c delta T * M = 6.19 + 41.9 = 48.09 (rounded to 3 significant figures) = 48.1 g * C = 4.18 J/g.K * Delta T = 16 - 18 = - 2 K (place only the absolute value) = 2.00 Q = 402.116 (rounded to 3 significant figures) = 402 J * Number of moles of water = mass / molar mass = 41.9 / 18.0 = 2.33 moles *number of moles of MgSO4.7H2O = mass / molar mass = 6.19 / (24.31 + 32.06 + (4 x 16.00) ...read more.

Conclusion

x 100 = 0.313 % Total percentage uncertainties = 0.166 + 0.0808 + 0.0111 + 0.0119 + 0.278 + 0.238 + 0.278 + 0.313 = 1.3768 = 1.38 % The enthalpy change = - 40 � ((1.38/100) x 40) = - 0 � 0.552 Kj/ mole Conclusion: Percentage error: difference between calculated value and literature value divided by literature value times 100 Percentage error = ((40 - 104) / - 104) x 100 = (- 64 / 104) x 100 = 61.5 % (take absolute value) Sources of error: 1) Heat lost to the surrounding 2) The specific heat capacity used in calculations was that of water, not the mixture itself 3) Parallax error 4) Position of thermometer 5) Stirring was not constant Improvements: 1) Use a well insulated vessel with lid 2) Find the specific heat capacity of the mixture itself 3) Use digital thermometer connected to a computer 4) Ensure that the line of sight is perpendicular to the reading of the thermometer and measuring cylinder 5) Use magnetic stirrer for uniform stirring 6) Repeat the experiment several times, taking an average value for the heat emitted or gained by the mixture ...read more.

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