JEE Advance - Chemistry (2000 - No. 6)
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. The energy for the dissociation of H-H bond 436 kJ mol-1.
436 kJ
17.83 kJ
80.49 kJ
98.32 kJ
1312 kJ
Explanation
No. of moles of H2 gas,
$$n = {{PV} \over {RT}} = {{1 \times 1} \over {0.0821 \times 298}} = 0.0409$$
Energy required to dissociate 0.0409 mol H2 into atoms $$ = 0.0409 \times 436 = 17.83$$ kJ
No. of moles of H atoms
$$ = 2 \times 0.0409 = 0.0818$$
$${E_n} = - {{1312} \over {{n^2}}}$$ kJ mol$$-$$1
$$\therefore$$ Energy required for excitation of 0.0818 moles of H atoms from ground state to first excited state
$$ = ({E_2} - {E_1}) \times 0.0818$$ kJ mol$$-$$1
$$ = - \left[ {\left( {{{ - 1312} \over {{2^2}}}} \right) - \left( {{{ - 1312} \over {{1^2}}}} \right)} \right] \times 0.0818$$ kJ
= 80.49 kJ
Total energy required for excitation of 1 litre H2 gas = 17.83 + 80.49 = 98.32 kJ
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