21Energy = (273 * 436 * 3)/(298 * 22.4 * 4) Kcal
energy = (273 * 436 * 3 * 4.2)/(298 * 22.4 * 4) KJ
Calculate the energy required to excite one litre of hydrogen gas at 1 atm and 298K to the first excited state of atomic hydrogen. the energy of dissociation of H-H bond is 436 kJmol-1
-
UP 0 DOWN 0 0 5
5 Answers
1moles of H2 =PV/RT =1/0.082x298= 4.092x10-2
energy to break H-H bond =4.092x10-2 x436kj
=17.84kj
for H atom E1=-2.179x10-18J/atom
E2=-2.179x10-18/4
=-0.5448x10-18J
delta E = 1.6355x10-18Jper atom
deta Etotal=1.6355x10-18x2x4.092x10-2x6.023x1023
=80.62x103
toatl energy req.=17.84+80.62=98.46kJ
21
1n = .04 moles using gas equation
436*.04=17.82 kJ
.04 *6.023*10^23 *(13.6/12 - 13.6/22)* 1.6 * 10-19 = 39318 J = 39.3 kJ
answer = 17.82 kJ + 39.3 kJ
1061*1 = n*0.0821*298
==> n = 0.041 mol
So, total energy for BD = 0.041*436 = 17.876 kJ
no. of H atoms = 2*0.041*6.02*1023
Now E1 = -1312 kJmol-1
So, E1 for the atoms = -1312*2*0.041 kJ
E2 = -1312/4 kJ/mol
So, E2 for the atoms = -1312/4*2*0.041 kJ
So, ΔE = 3/4*1312*2*0.041 = 20.172 kJ
So, total energy supplied = 20.172 + 17.876 = 38.048 kJ