Question

Calculate the energy required to excite $2.46\text{L}$ of hydrogen gas at $1\text{atm}$ and $300\text{K}$ to the first excited state of atomic hydrogen. The energy of dissociation of $H-H$ bonds is $436{\text{kJ mol}}^{-1}$
(Given : $R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1}$, Energy of ground state of $H-$ atom $=-1312{\text{kJ mol}}^{-1}$)

Answer 1

From the given data, the number of moles of hydrogen $=n=\frac{PV}{RT}$
$n=\frac{1\text{atm}\times 2.46\text{L}}{0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1}\times 300K}=0.1\text{mole}$.
Energy required to dissociate $0.1$ mole of $H_2$
$=436{\text{kJ mol}}^{-1}\times 0.1\text{mol}=43.6\text{kJ}$
$\therefore$ Number of moles of hydrogen atoms
$=2\times 0.1=0.2$
We know that $E_n=\frac{-1312}{n^2}{\text{kj mol}}^{-1}$
($1$st excited state of H $=$ second shell n $=$$2$)
Thus, energy required for the exciation of $0.2$ moles of H-atoms $=(E_2-E_1)\times 0.2\text{kJ}=[-\frac{1312}{2^2}-(-\frac{1312}{1^2})]\times 0.2\text{kJ}=1312\times \frac{3}{4}\times 0.2=196.8\text{kJ}$
Thus, total energy required including energy of dissociation $=196.8+43.6=240.4\text{kJ}$