Calculate the energy (in KJ) 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` id 436 KJ `mol^(-1)`. Give your answer excluding decimal places.

Explanation

Amount of hydorgen gas is
`n_(1) = (PV)/(RT) = ((1atm)(1L))/((0.082L atmK^(-1)mol^(-1))(298K))`
`= 0.04092 mol`
Energy consumed in dissociating hydorgen molecule
`E_(1) = (0.04092mol) (436kJ mol^(-1)) = 17.84 kJ`
Since each hydorgen molecule given two hydorgne atoms, the amount of hydorgen atoms will be
`n_(2) = 2n_(1) =2 xx 0.04092 mol = 0.08184 mol`
Energy required to excite one mole of hydorgen atoms from the ground electronic state to the first excited state is
`DeltaE = N_(A)R_(H)hc ((1)/(1^(2))-(1)/(2^(2))) = (3)/(4) N_(A)R_(H)hc`
`= ((3)/(4)) (6.023 xx 10^(23) mol^(-1)) (109676 cm^(-1)) xx (6.626 xx 10^(-34)J) xx (3xx 10^(8)cms^(-1))`
`= 9848.5 J mol^(-1)`
For exciting `0.08184 mol` of hydrogen atoms, we will have
`E_(2) = (0.08184mol) (9.8485kJ mol^(-1)) = 0.806 kJ`
Hence, the total energy required will be
`E = E_(1)+E_(2) =(17.84 +0.806) kJ =18.65 kJ`