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Class 12 Chemistry Experiment 6 - To Determine the Enthalpy of Dissolution of Copper Sulphate

Class 12 · Chemistry · Thermochemistry · CBSE · ENGLISH · 0 views

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Experiment 6Class XII · CBSEChemistry 043

To Determine the Enthalpy of Dissolution of Copper Sulphate

Blue crystals stirred into water in a polystyrene cup, and the thermometer goes down, not up. Dissolving is not always a warming business.

SectionThermochemistry
In the examContent based experiment · 6

01Aim

To determine the enthalpy of dissolution of copper sulphate pentahydrate, CuSO4·5H2O, in water using a simple polystyrene cup calorimeter.

02Requirements

No.MaterialSpecification / purpose
1Polystyrene cup with a lidThe calorimeter; it loses very little heat
2Beaker (250 ml)To hold the cup steady
3ThermometerReading to 0.1 °C
4StirrerA glass or plastic rod
5Measuring cylinders50 ml and 25 ml
6Chemical balanceReading to 0.01 g
7Copper sulphate pentahydrateAbout 5 g, finely powdered
8Distilled water100 ml

03Principle

The enthalpy of dissolution is the heat change when one mole of a substance dissolves in so much solvent that further dilution produces no further heat change.

Two opposing energies decide the sign. Breaking the crystal lattice absorbs energy, the lattice enthalpy; hydrating the separated ions releases energy, the hydration enthalpy. The sum is what the thermometer shows:

ΔHsol=ΔHlattice+ΔHhydration(1)

For the anhydrous salt the hydration enthalpy wins and the process is exothermic. For the pentahydrate the ions are already hydrated, so there is little hydration energy left to release, the lattice term wins, and dissolving is endothermic.

The heat taken from the solution is found from its mass, its specific heat and the fall in temperature:

q=mcΔT(2)
m = mass of the solution in gramsc = specific heat capacity, taken as 4.18 J g⁻¹ K⁻¹ΔT = change in temperature in kelvin

Dividing by the number of moles gives the molar value:

ΔHsol=qnin J mol−1(3)
The sign convention: the system is the dissolving salt. If the temperature of the surroundings falls, the system has absorbed heat and ΔH is positive, that is endothermic.

04The calorimeter

Plate IA polystyrene cup calorimeter with a thermometer and a stirrer; the temperature before and after mixing gives the enthalpy change.ENTHALPY OF DISSOLUTION OF COPPER SULPHATEstirrerthermometerpolystyrene cup inside a beaker — a simple calorimeterstart 28.0 °Cend 26.4 °CΔTWHAT IS MIXEDA100 ml distilled waterB5 g powdered CuSO₄·5H₂Ostirred and the steady temperature readWHAT IT MEANSthe temperature fallsENDOTHERMICΔH is positive

A polystyrene cup inside a beaker, with a thermometer and a stirrer through the lid. The whole temperature change is read off in one or two minutes.

05Procedure

  1. Weigh the empty polystyrene cup and record its mass.
  2. Measure 100 ml of distilled water into the cup and stand it in the beaker.
  3. Fit the lid, put in the thermometer and the stirrer, and leave it for two or three minutes.
  4. Stir gently and record the steady initial temperature to 0.1 °C.
  5. Weigh out about 5 g of finely powdered CuSO4·5H2O accurately.
  6. Add it to the water all at once, replace the lid and stir continuously.
  7. Watch the thermometer and record the lowest steady temperature reached.
  8. Repeat the whole determination and take the mean.

06Observations

QuantitySymbolReading
Mass of CuSO4·5H2O takenw5.00 g
Volume of waterV100 ml
Mass of the solutionm105.0 g
Initial temperatureT128.0 °C
Final temperatureT226.4 °C
Fall in temperatureΔT1.6 °C
Molar mass of CuSO4·5H2OM249.7 g mol−1
  • The crystals dissolved to give a clear pale blue solution.
  • The thermometer fell steadily for about a minute and then held.
  • The outside of the cup felt slightly cold to the touch.
  • The anhydrous salt, tried separately, gave a rise in temperature instead.

07Calculation

Heat absorbed by the solution
q=mcΔT=105.0×4.18×1.6=702.2J
Moles of copper sulphate dissolved
n=5.00249.7=0.0200mol
Enthalpy of dissolution
ΔHsol=+702.20.0200=+35110J mol−1=+35.1kJ mol−1
The sign is positive because the temperature fell: heat flowed from the water into the dissolving salt. Quoting it without the sign, or with the wrong one, loses the mark.

08Result

The enthalpy of dissolution of CuSO4·5H2O in water is +35.1 kJ mol−1. The positive value shows that dissolving the pentahydrate is an endothermic process, because the ions are already hydrated and the lattice energy is not repaid by further hydration.

09Precautions

  • Use a polystyrene cup, not glass. Glass absorbs a good deal of heat itself and the answer comes out low.
  • Keep the cup covered and stir gently and continuously.
  • Read the thermometer to 0.1 °C, with the bulb fully immersed and not touching the wall.
  • Note the highest or lowest steady temperature reached, not the first reading after mixing.
  • Work quickly; the longer the mixture stands the more heat is exchanged with the room.
  • Both solutions must be at the same starting temperature.
  • Powder the crystals finely so they dissolve quickly, and add them all at once.

10Viva voce

Q1Define the enthalpy of solution.

ANSThe heat change when one mole of a substance dissolves in so much solvent that further dilution causes no further heat change.

Q2Why is dissolving the pentahydrate endothermic while the anhydrous salt is exothermic?

ANSIn the pentahydrate the ions are already hydrated, so very little hydration energy is released and the lattice energy absorbed dominates. The anhydrous salt still has its full hydration energy to give out.

Q3Why is a polystyrene cup used instead of a glass beaker?

ANSPolystyrene is a poor conductor and has a very small heat capacity, so almost no heat is lost to the calorimeter itself.

Q4What is the specific heat capacity of water?

ANS4.18 J per gram per kelvin.

Q5What is lattice enthalpy?

ANSThe energy released when one mole of an ionic solid is formed from its gaseous ions; the same magnitude of energy is needed to break the lattice apart.

Q6Why must the solution be stirred?

ANSTo keep the temperature uniform, so that the thermometer reads the temperature of the whole solution and not of one layer.

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