HW#8
1 (a) 1 mole of water equals 18 grams, so 1 kg of water is
1000 g
= 55.5 mol
18 g/mol
55.5 mol ¥ 285.8 kJ/mol = 15877.78 kJ ª 15.9 MJ = 1.59 ¥ 10 7 J
(b) On the left hand side, there are 2 protons. 2 H has 2-1=1 neutron, 3 H has 3-1=2
neutrons. So altogether 3 neutrons 2 protons on the left hand side.
On the right hand side, 4 He has 2 protons, 4-2=2 neutrons. So there has to be an
extra neutron released. We get 2 protons and 3 neutrons on the right hand side
too.
Every 2 H atom reacted will lead to energy release of 17.6MeV. 1 mol of 2 H
equals 2 grams, so .5 kg of 2 H equals
500 g
= 250 mol
2 g/mol
250 mol ¥ 6.02 ¥ 10 23 atom/mol ¥ 17.6 MeV = 2.65 ¥ 10 27 MeV
1 MeV=1.60e-13 J,
Thus E = 2.65 ¥ 10 27 MeV ¥ 1.60 ¥ 10 -13 J/MeV = 4.24 ¥ 1014 J
This is 7 orders of magnitude larger than the energy released from chemical
reaction.
2. Left hand side, total energy equals
LFS = 6m p c 2 + 6m n c 2 - 92.16 MeV + 8m p c 2 + 8m n c 2 - 127.62MeV
Total energy on right hand side is
RFS = 14m p c 2 + 14m n c 2 - 236.54 MeV
Energy available=LFS-RFS
= 6m p c 2 + 6m n c 2 - 92.16 MeV + 8m p c 2 + 8m n c 2 - 127.62MeV
- 14m p c 2 + 14m n c 2 - 236.54 MeV
=236.54 MeV -92.16 MeV -127.62 MeV
=16.76 MeV
3.
Uranium has 92 protons, 238-92=146 neutrons
Cesium has 55 protons, 140-55=85 neutrons
Rubidium has 37 protons, 92-37=55 neutrons
The number of protons are equal on both sides of the reaction.
146-85-55=6.
So there should be 6 neutrons released from the reaction.
By looking at the figure, at A=238, EA/A=7.6, at A=140, EA/A=8.5, at A=92,
EA/A=8.8
E= 92m p c 2 + 146m n c 2 - 7.6 ¥ 238 MeV
- (55m p c 2 + 85m n c 2 - 8.5 ¥ 140 MeV + 37 m p c 2 + 55m n c 2 - 8.8 ¥ 92MeV)
=-1808.8 MeV+1190 MeV+809.6 MeV=190.8 MeV