Question

Calculate the energy required to excite one litre of hydrogen gas at 1 atm and 298 k to the first excited state of atomic hydrogen

Answer 1

Determination of number of moles of hydrogen gas,

N = PV/RT = 1 *1/0.082 *298 = 0.0409

The concerned reaction is H2 → 2H ; ∆H = 436 kJ mol-1

Energy required to bring 0.0409 moles of hydrogen gas to atomic state = 436 *0.0409 = 17.83 kJ

Calculate of total number of hydrogen atom in 0.0409 mole of H2 gas

1 mole of H2 gas has 6.02 * 1023 molecules

0.0409 mole of H2 gas = 6.02 *1023/1 * 0.0409 molecules

Since 1 molecule of H2 gas has 2 hydrogen atoms 6.02 * 1023 * 0.0409 molecules of H2 gas

=2 * 6.02 * 1023 * 0.0409 = 4.92 * 1022 atoms of hydrogen since energy required to excite an electron from the ground state to the next excited state is given by

E = 13.6 \left ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right )eV = 13.6 * (1/1 – 1/4) = 13.6 * 3/4 = 10.2eV

= 1.632 * 10-21 kJ

Therefore energy required to excite 4.92 * 1022 electrons

= 1.632 * 10-21 * 4.92 * 1022 kJ = 8.03 * 10 = 80.3 kJ

Therefore total energy required = 17.83 + 80.3 = 98.17 kJ

hope it helpss uuhh

Answer 2

Answer:

Amount of hydorgen gas is

n1=PVRT=(1atm)(1L)(0.082LatmK−1mol−1)(298K)

=0.04092mol

Energy consumed in dissociating hydorgen molecule

E1=(0.04092mol)(436kJmol−1)=17.84kJ

Since each hydrogen molecule given two hydrogen atoms, the amount of hydorgen atoms will be

n2=2n1=2×0.04092mol=0.08184mol

Energy required to excite one mole of hydorgen atoms from the ground electronic state to the first excited state is

ΔE=Na.Rh. c(112−122)=34NARHhc

=(34)(6.023×1023mol−1)(109676cm−1)×(6.626×10−34J)×(3×108cms−1)

=9848.5Jmol−1

For exciting 0.08184mol of hydrogen atoms, we will have

E2=(0.08184mol)(9.8485kJmol−1)=0.806kJ

Hence, the total energy required will be

E=E1+E2=(17.84+0.806)kJ=18.65kJ

Think it helps