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.
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:
Qualitative Data – Observations:
- Adding zinc to copper sulfate resulted in a displacement reaction that created copper metal to precipitate as a solid. The color of the solid was dark greyish/black.
CuSO4 (aq) + Zn(s) ---> ZnSO4 (aq) + Cu(s)
- There was a temperature increase, hence the reaction was exothermic.
Data Processing
I have used the results for Trial 1 in order to determine the enthalpy change of the reaction.
Calculations:
- Mass of Zinc powder = (Mass of zinc + filter paper) – (Mass of filter paper)
= (7.52 ± 0.01 g) – (1.50 ± 0.01 g)
= (7.52 – 1.50) ± (0.01 ...
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- Adding zinc to copper sulfate resulted in a displacement reaction that created copper metal to precipitate as a solid. The color of the solid was dark greyish/black.
CuSO4 (aq) + Zn(s) ---> ZnSO4 (aq) + Cu(s)
- There was a temperature increase, hence the reaction was exothermic.
Data Processing
I have used the results for Trial 1 in order to determine the enthalpy change of the reaction.
Calculations:
- Mass of Zinc powder = (Mass of zinc + filter paper) – (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
- 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) can be obtained from the graph, as shown below:
Note: The horizontal error bars for time, were too small to be visible on the graph.
- Substituting these values into the equation we obtain the Enthalpy Change of the reaction:
Q = mc ∆T
= (31.02 ± 0.02g) × 4.18 × (52.0 °C ± 0.5) – (23.5 °C ± 0.5)
= (31.02 ± 0.02g) × 4.18 × (29 °C ± 1)
Measurement: 31.02 × 4.18 × 29 = 3760.2444 J
Relative error of ∆T: 1 ÷ 29 = 0.03448275862
Relative error of m: 0.02 ÷ 31.02 = 0.00064474532
Sum of Relative errors: 0.03448275862 + 0.00064474532 = 0.03512750395
Absolute Uncertainty = 0.03512750395 × 3760.2444 = 132.088
J
Final Answer = 3760 J ± 132
- Convert the enthalpy into Kilojoules:
∴ Enthalpy Change = (3760 ± 132) ÷ 1000
= 3.8K J ± 0.1
- To obtain the enthalpy change for 1 mole of Zn and CuSO4 solution, we calculate the moles of the limiting reagent. 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
- 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
- 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