Part A: the heat required for a change of state

Firstly, I needed to boil the water in the kettle, so that it can be used in the experiment.

Also I weighted the ice I was going to put in the water.

m= mass of ice- mass of ice in the paper= 25g – 7.1 g = 17.9 g of ice

1 mole — 18g

X — 17.9g

X= 0.99 moles

Than, I started the experiment.

Later on, the graph was plotted.

Graph 1: Graph showing the course of the measurement

Using informations from the table and the extrapolations from graph, I am able to calculate the enthalpy change.

∆T= temp.max – temp.min

∆T= 59.8 – 30.5

∆T= 29.3ºC

Q= m·c·∆T where m- mass of solution (g)

c- specific heat capacity of H2O (J g–1 K–1)

∆T- change of temperature (K)

Mass of water=100 cm3 * 1 g cm-3= 100 g

Qwater= m·c·∆T

Qwater=100g* 4.18 J g–1 K–1* 29.3 K=12247.4 J

From the first law of thermodynamics we get the equation:

Qwater= -Qice

So the Q of ice is equal to -12247.4 J, so -12.2474 kJ.

What is more, we know that the heat gained by ice was used to melt the ice first and then to raise the temperature of the whole solution:

Qice=Qfus + (∆Tice ·c ·mice)

Qfus=Qice - (∆Tice ·c ·mice)

Qfus=-12247.4 – 30.5 K* 4.18 J g–1 K–1 * 17.9g

Qfus= - 14529.71 J

1 mole H2O ice — x

0.99 moles of H2O ice — - 14529.71 J

x= -14676.47 J mol-1

The enthalpy change for this reaction is -14676.47 J mol-1.

Part B: enthalpy change for displacement reaction

First step in the experiment was to create 1M copper sulphate. To do this solution, some calculations were needed:

mCuSO4 · 5H2O = =250 g

to create 50 cm3 1M solution I need:

x=12.5g of copper sulphate

So I dissolve 12.5 g copper sulphate in 50 cm3 of water.

Also, I calculated how many moles of zinc I am going to add to the solution.

MZn=65g

1 mole — 65 g

x— 5 g

x=0.077 moles of Zn

Since I don't know, I decided to calculate how many moles of zinc are actually going to react with copper sulphate.

The equation for the reaction looks as follow:

CuSO4 (aq) + Zn (s) → ZnSO4 (aq) + Cu (s)

or

Cu2+(aq) + Zn (s) → Zn2+ (aq) + Cu(s)

1 mole — 250 g

x— 12.5 g

x=0.05 moles of CuSO4

As the proportion between Zn and CuSO4 is 1:1,

1 mole CuSO4 — 1 mole Zn

0.05 moles — x

x=0.05 moles of Zn will actually react

Which means that zinc is in excess.

Next step was to record-firstly-the temperature of the solution, then add zinc.

First two minutes of the temperature was measured for the solution of CuSO4 only.

Then, the powdered zinc was added and the solution was stired rapidly until the maximum temperature has been reached ( which happened in 3.5 minute).

Later on, the graph was plotted.

Graph2: Graph showing the course of the measurement

Using information from the table and extrapolations, I am able to calculate the enthalpy change.

∆T= temp.max – temp.of CuSO4

∆T= 70.0-18.8

∆T= 51.2ºC

Q= m·c·∆T where m- mass of solution (g)

c- specific heat capacity of H2O (J g–1 K–1)

∆T- change of temperature (K)

m of solution=50 cm3 * 1 g cm-3= 50 g

∆H=50 g* 4.18 J g–1 K–1 * 51.2 K = - 10700 J

1 mole CuSO4 — x

0.05 moles of CuSO4 — -10.7 kJ

x= -214.0 kJ mol-1

The enthalpy change for this reaction is -214.0 kJ mol-1.

CE: