Qualitative Observations During The Reaction Between Copper (II) Sulfate

Susim Santra
1. September.2014
Chemistry honors
Data collected:
Table 1: table depicting the data collection for temperature over a time limit of 600 seconds for a reaction between copper (II) sulfate (CuSO_4) and zinc(Zn) using a temperature probe.
Time/ s (± 1 s) Temperature / °C (± 0.05 °C)
0 27.9
30 25.9
60 25.9
90 25.8
120 25.8
150 48.7
180 57.6
210 61.1
240 62.7
270 63.5
300 62.8
330 61.3
360 59.3
390 58.9
420 57.7
450 56.5
480 55.2
510 54.2
540 52.6
570 51.9
600 51.1

Qualitative observations:-
Table 2: table depicting the qualitative observations observed during the reaction between Copper (II) sulfate solutions (CuSO_4) and Zinc (Zn).
Solution Initial observation Final observation Quantitative observation Volume of solution used
Copper(II)sulfate solution (CuSO_4) - sky blue color
-in liquid form -copper precipitates found at the bottom of the flask after reaction
-loss of blue color from the solution - the reaction follows a tea-murky green- coke-coffee-dark chocolate color change
- pungent smell released during reaction
-exothermic as the polyester cup in which the reaction occurs is heating up
-starts bubbling after 260 seconds
- the reactants/solution stick to the polyester cup
-after reaction is completed copper precipitate is left behind 25 〖cm〗^3 N/A
Zinc (Zn) -in powdered form
- grey/ metallic color
- consisting of large granules -after zinc reacts with copper(II) sulfate it lose all its color -zinc immediately dissolves into the copper(II)sulfate solution (due to it being in powdered form) 6.13 g ± 0.01
= 23.2 ˚C
Enthalpy change:
Now using all the variables obtained we can calculate the enthalpy change for the reaction using (∆H) = mc∆T
Now,
Mass of copper (II) sulfate (CuSO_4) = 25 〖cm〗^3 Temperature change (∆T) = 23.2 ˚C Specific heat capacity of water = 4.18 J g-1 k-1
Using the equation provided above, we can plug in the values necessary to find the enthalpy change (∆H) = 25 〖cm〗^3 X 23.2 ˚C X 4.18 J g-1 k-1 (∆H) = - 2424.4 J
Now to find the enthalpy change per mole or (kJ/Mol) for the reaction we divide the enthalpy change value obtained with the number of moles of a solvent which is not in excess using (Enthalpy change )/(number of mole )
Now,
Enthalpy change(∆H) = - 2424.4 J Number of moles = 0.0125 mol
Using the equation provided above, we can plug in the values necessary to find the enthalpy change value per mole of the reaction ( - 2424.4 KJ)/(0.0125 mol )

- 193952 J/Mol

This value is obtained in joules per mole however the value we are looking for is Kilojoules per mole therefore we convert joules (J) to kilojoules (KJ) by dividing it by a thousand as 1 joules = ( 1)/(1000 ) kilojoules

Therefore, - 193952 J/Mol = (-193952)/(1000 ) kilojoules

194.0 KJ/Mol

Conclusion:
Apparatus