Calculating Enthalpy Change From Experimental Data

Experiment 1 Procedure: The enthalpy change for the reaction CuSO4(s)⟶CuSO4(aq)

CuSO4(s)⟶CuSO4(aq)Using a measuring cylinder, transfer 50.0 mL50.0 mL of distilled water to a polystyrene calorimeter.Weigh accurately (to 2 decimal places) about 4.00 g4.00 g of anhydrous copper(II) sulfate (record the exact mass used).Measure the temperature of the water in the calorimeter to the nearest 0.5 ∘C0.5 ∘C, (record this as 𝑇initialTinitial).Transfer the anhydrous copper(II) sulfate to the calorimeter, cover with a lid, and then stir the contents of the calorimeter carefully with the thermometer. Monitor the temperature closely and note the highest temperature that is reached (record this as 𝑇finalTfinal).Experiment 2 Procedure: The enthalpy change for the reaction CuSO4⋅5H2O(s)⟶CuSO4(aq)CuSO4⋅5H2O(s)⟶CuSO4(aq)

Using a measuring cylinder, transfer 47.75 mL47.75 mL of distilled water to a polystyrene calorimeter (this volume is used to ensure that the total volume of water when considering that the copper sulfate already contains water is 50 mL50 mL). Weigh accurately (to 2 decimal places) about 4.00 g4.00 gof copper(II) sulfate pentahydrate (record the exact mass used).Measure the temperature of the water in the calorimeter to the nearest 0.5 ∘C0.5 ∘C, (record this as 𝑇initialTinitial).Transfer the hydrated copper(II) sulfate to the calorimeter, cover with a lid, and then stir the contents of the calorimeter carefully with the thermometer. Monitor the temperature closely and note the lowest temperature that is reached (record this as 𝑇finalTfinal).(i) Calculate the enthalpy changes for experiments 1 and 2. Show full details of your working.

Experiment 1:

q=-MCT

q=-50g x 4.18 J K-1 g-1 x 7.9K = -1651.1J = -1.65 kJ M=50g

Moles of CuSO4 = 3.94g/159.6 = 0.0247 mol C=4.18JK-1g-1 T=7.9K Enthalpy Change = (-1.65kJ)/(0.0247 mol) = -66.8 kJ mol-1

Experiment 2:

q=-MCT q=-47.75g x 4.18 J K-1 g-1 x -1.6K = 319.4J = 0.32 kJ

M=47.75g Moles of CuSO4 = 7.72g/159.6 = 0.0484 mol C=4.18JK-1g-1 T=-1.6K Enthalpy Change = (-0.32kJ)/(0.0484 mol) = +6.6 kJ mol-1

Is the answer for the enthalpy change I worked out for both experiments correct?

Sample Solution
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