# Enthalpy of Dissolution of Copper Sulphate or Potassium Nitrate

A substance’s molar solution heat is the heat absorbed or released when one of the substance’s molecules is dissolved in water. It is denoted by ΔH. The overall heat of the solution can be either endothermic or exothermic, depending on the relative amount of energy needed to break bonds initially, as well as how much is released upon the formation of solute-solvent bonds.

## Aim:

To determine the enthalpy of dissolution of given Copper Sulphate or Potassium Nitrate at room temperature using water as reaction medium.

## Theory:

Aqueous solutions are generally mixed in thermochemical measurements, Water in the medium of reaction changes in temperature result from chemical reactions in solution.

The sum of enthalpy changes occurring in the calorimeter either loss or energy gain must be zero, according to the energy conservation law. We can, therefore, write the following equation,

ΔH1 + ΔH2 + ΔH3 + ΔH4 = 0

Solution formation often goes hand in hand with heat changes. Solution enthalpy is the amount of heat released or absorbed when one mole of a solvent (solid/liquid) is dissolved in such a large amount of solvent (usually water) that further dilution does not change heat.

## Materials Required:

1. Beakers
2. Thermometer
3. Glass rod
4. Weight box
5. Physical balance
6. Stirrer
7. Measuring cylinder
8. Hydrated copper sulfate
9. Potassium nitrate
10. Distilled water
11. Small wooden block
12. Cotton wool
13. Small piece of cardboard
14. Filler

## Apparatus Setup:

Determining the Enthalpy of Dissolution of Given Solid Copper Sulphate in Water at Room Temperature

## Procedure:

### 1. Enthalpy of Dissolution of Copper Sulphate:

1. Take 100 mL of distilled water in the beaker with a constant calorimeter and place it on a wooden block in a larger 500 mL beaker.
2. Pack the empty space with cotton wool between the bigger and the smaller beaker and cover with cardboard.
3. Record water temperature. Let it be T1°C.
4. Transfer copper sulphate powder which is already weighed. Remove the solution until the entire sulfate of copper dissolves.
5. Keep intact the thermometer and stirrer on the foam case mouth.
6. Note down the solution’s maximum temperature after copper sulphate is added. Let it be T2°C.

### 2. Enthalpy of Dissolution of Potassium Nitrate:

1. Take 100 mL of distilled water in the beaker with a constant calorimeter and place it on a wooden block in a larger 500 mL beaker.
2. Pack the empty space with cotton wool between the bigger and the smaller beaker and cover with cardboard.
3. Record water temperature. Let it be T1°C.
4. Transfer potassium nitrate powder which is already weighed. Remove the solution until the entire potassium nitrate dissolves.
5. Keep intact the thermometer and stirrer on the foam case mouth.
6. Note down the solution’s maximum temperature after potassium nitrate is added. Let it be T2°C.

## Observation and Inference:

 Enthalpy of Dissolution of Copper Sulphate Enthalpy of Dissolution of Potassium Nitrate Weight of hydrated copper sulfate dissolved. w g Weight of hydrated potassium nitrate dissolved. w g Volume of water taken into the bottle. 200 ml 200 gms (assuming sp density =1) Volume of water taken into the bottle. 200 ml 200 gms (assuming sp density =1) Temperature of water. t1oC Temperature of water. t1oC Temperature of water after dissolving hydrated copper sulfate. t2oC Temperature of water after dissolving potassium nitrate. t2oC Water equivalent of the polythene bottle. W g (given) Water equivalent of the polythene bottle. W g (given)

Assuming density and specific heat of the solution to be the same as that of water heat evolved or absorbed for dissolution of wg of the solute.

$Q = (W+200)(t_{1}-t_{2})cals$ $Q = (W+200)(t_{1}-t_{2}) ## Calculations: \times 4.2 joules$ $Q =(W+200)\times \Delta t \times 4.2 joules$

For w/M moles of the solute dissolved heat change = $(W+200)(t_{1}-t_{2})\times 4.2$ Joules

For 1 mole of the solute dissolved heat change = $\frac{(W+200)(t_{1}-t_{2})\times 4.2\times M}{W}$ Joules

 Enthalpy of dissolution = $\frac{(W+200)(t_{1}-t_{2})\times 4.2\times M}{W}$Joules

Where

M = formula mass of the solute

w = mass of the solute

W = water equivalent of the calorimeter

Δt = change in temperature

Enthalpy of dissolution ΔH is positive if heat is absorbed and negative if heat is evolved.

Similar way we can find the enthalpy of dissolution of potassium nitrate. For that dissolve 5.5g of KNO3 in 200ml of water. Here the mole ratio of solute and solvent is 1:200

## Results and Discussions:

1. The heat of solution of potassium nitrate KNO3 is __________J/mole.
2. The heat of solution of copper sulfate solution CuSO4.5H2O is ________J/mole.
3. % Error for KNO3 = $\frac{Heat\ of\ solution\ determined}{Experimental\ value}\times 100$
4. % Error for CuSO4.5H2O = $\frac{Heat\ of\ solution\ determined}{Experimental\ value}\times 100$

## Precautions:

1. Record the temperature of hot water just before mixing when calorimeter constant is determined.
2. To dissolve the solid and record the temperature, stir the solution well. Avoid too much stirring, because of friction it can produce heat.
3. Carefully weigh copper sulphate as it is naturally hygroscopic.
4. The potassium nitrate should be taken in small amount that is 2.27g to be dissolved in 100ml of water.
5. In order to create isolation between the two beakers, use cotton wool.
6. Use a 0.1°C thermometer to record the water temperature.

## Viva Questions:

1. Define calorie.

Ans: A calorie is an energy unit. There are actually two units that have been widely used with that name. The small calorie or gram calorie (usually referred to as calorie) is the amount of heat energy required to raise one gram of water temperature by one degree Celsius (or one kelvin).

2. Identify the types of reaction if the value of ΔH is negative.

Ans: A system that releases heat to the environment, an exothermic reaction, has a conventional negative H because the product enthalpy is lower than the system’s reactant enthalpy.

3. Define specific heat.

Ans: Specific heat is the thermodynamic property that specifies the amount of heat required by one degree of temperature for a single unit of mass of a substance.

4. What is the enthalpy of neutralization?

Ans: Neutralization enthalpy is the change in enthalpy that occurs when one acid equivalent and one base equivalent undergoes a water and salt neutralization reaction. It’s a special case of reaction enthalpy.

5. What is the difference between endothermic and exothermic reaction?

Ans: Endothermic reactions attract and store energy in the form of chemical reaction bonds. As the reaction progresses, an exothermic reaction sheds heat energy, meaning it radiates heat while it is going on.