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# Chemistry Lab- Determining enthalpy change of a reaction. Adding zinc to copper sulfate resulted in a displacement reaction that created copper metal to precipitate as a solid.

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Introduction

Name: Laraib Hussain Class: IB 1 Chemistry Lab: Experiment 22 - Determining an enthalpy change of reaction Exp.22 Determining the Enthalpy Change of a reaction Data Collection Quantitative Data - Table of Results: Trial 1 Mass of filter paper: 1.50 � 0.01 g Mass of zinc + filter paper: 7.52 � 0.01 g Trial 2 Mass of filter paper: 1.49 � 0.01 g Mass of zinc + filter paper: 7.58 � 0.01 g Time (� 0.01s) Temperature (� 0.5�C) Time (� 0.01s) Temperature (� 0.5�C) 0.00 24.0 0.00 25.0 30.00 24.0 30.00 25.0 60.00 24.0 60.00 25.0 90.00 23.5 90.00 25.0 120.00 23.5 120.00 25.0 150.00 23.5 150.00 25.0 Addition of Zinc Powder at 3 minutes, then temp. recorded for additional 6 minutes Addition of Zinc Powder at 3 minutes, then temp. ...read more.

Middle

- (Mass of filter paper) = (7.52 � 0.01 g) - (1.50 � 0.01 g) = (7.52 - 1.50) � (0.01 + 0.01) = 6.02 � 0.02 g 2. To Calculate the Enthalpy Change (? H) of the reaction we will use the formula: Where m = mass of solution (g) c = specific heat capacity of water: 4.18 ?T = rise in temperature (� C) Note: We assumed that the products in the reaction have the same specific heat capacity as water. Also we assume that the reactants have a density of 1g hence the mass of the reactants will be: Mass = density � volume = 1g � 25 = 25 g Total mass of solution = (6.02 � 0.02 g) + 25 g = 31.02 g �0.02 The rise in temperature (?T) ...read more.

Conclusion

The limiting reagent is CuSO4. No. of moles of CuSO4 = concentration � volume = 1 mol dm-3 � (25cm3 � 1000cm3) = 1 � 0.025 dm3 = 0.025 moles 6. Hence, the Enthalpy Change of 1 mole of Zn and CuSO4: ?H = = Measurement: -3.8 � 0.025 = -152 KJ mol-1 Relative error: 0.1 � -3.8 = -0.02631578947 Absolute Uncertainty: -0.02631578947 � -152 = � 4 Final Answer: -152KJ mol-1 � 4 ?H of this reaction is -152 KJmol-1 � 4 7. Percentage error= � 100% Literature value of Enthalpy Change for this reaction is -217KJmol-1 = � 100% = [-217 - (- 152 �4)] �100% -217 = (-65�4) �100% -217 Measurement: -65� -217= - 0.2995391705 Relative error: (4� -65) �100= - 6.153846154 % Absolute uncertainty: - 0.2995391705 � - 6.153846154 % = � 0.018 � �0.02 Final answer: (- 0.30 � 0.02) � 100 = - 30 % � 2 ...read more.

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