JEE MainChemistryToughPhysical Chemistry

Calculate the energy required to excite one litre Hydrogen gas
at 1 atm and 298 K to first excited state of atomic hydrogen. The energy
for the dissociation of H-H bond is 436 kJ mol-1.

Show answer and solution

Determination of number of moles of hydrogen gas

n = \frac{PV}{RT} = \frac{1 \times 1}{0.082 \times 298} = 0.0409

The concerned reaction is

H_{2} \longrightarrow 2H;\Delta H = kJ{mol}^{- 1}

Energy required to bring 0.0409 moles of hydrogen gas to atomic state

= 436 \times 0.0409 = 17.83kJ

1 mole of H2 gas has 6.02 × 1023 molecules
0.0409 mole of H2 gas

= \frac{6.02 \times 10^{23}}{1} \times 0.0409

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

= 2 \times 6.02 \times 10^{23} \times 0.0409\

= 4.92 \times 10^{22}\ atoms

Energy required to excite an electron from the ground state to the next
excited state

= 13.6\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)eV = 13.6 \times \left( \frac{1}{1} - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2eV = 1.632 \times 10^{- 21}kJ

Therefore energy required to excite 4.92 × 1022 electrons

= 1.632 \times 10^{- 21} \times 4.92 \times 10^{22}kJ

= 8.03 \times 10 = 80.3kJ

Therefore total energy required

= 17.83 + 80.3 = 98.17kJ

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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