It is a little harder than sodium metal. When it reacts with water, it would produce calcium hydroxide , hydrogen gas and a lot of heat. That is because when calcium element during losing two electrons from the atom and form lattice this process, its enthalpy change is very high.
But why the reaction is not exciting as sodium metal that is because the high melting point of calcium metal. If the calcium would melt during the reaction, the surface of calcium to contact with water would become very large, so that the reaction would become exciting.
Ca(OH)2 deposit.
However, compared with sodium,calcium has much stronger covalent bonds between each atoms, so it has higher melting point.
Let's use thermodynamics to check how calcium metal emit heat into the surrounding.
First, we have to write out the reaction formula and figure out the enthalpy change in the reaction.
ΔfHm 0.0 2 x (-285.8) -985.2 0.0 (kJ/mloe)
=> ΔH =
(-985.2+0) - (0.0+ 2 x (-285.8))
=
413.6 kJ/mole
Second, we are going to do an experiment to measure the heat transformation from the reaction.
|
Calcium
metal weight
|
0.12 g
|
|
Molar
of calcium
|
3.00 x 10-3
mole
|
|
Water
weight
|
20.22 g
|
|
Origin
temperature of water
|
30.59 ℃
|
Theoretical
ΔT of water:
ΔH = m x s x ΔT
=>ΔT = ΔH / ( m x s)
= (( - (-413.6 x 103 x
(0.12/40.078))) / (4.186)) / (20.22 x 1)
= 14.63 ℃
=> T = 30.59 + 14.63
= 45.22 ℃
|
Theoretical
temperature
|
45.22 ℃
|
|
Measured
temperature
|
40.32 ℃
|
|
Surrounding
temperature
|
29.44 ℃
|
|
Lose
heat
|
98.00 cal
|
|
Heat
capacity of the beaker
|
2.87 cal/℃
|
Heat
lose in the beaker:
(40.32 – 30.59)
x (2.87)
=
27.93 cal
Energy
lose in other forms(gas kinetic energy, radiation…):
98.00 – 27.93
=
70.07 cal
=
293.02 J
|
Total
energy released
|
1238.32 J
|
|
Energy
in heat form
|
945.30 J
|
|
Heat
transform efficiency
|
76.3%
|
We can know not all the enthalpy change would turn into heat that is why the water is only 40.32℃ in the experiment.
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