Question

Calculate the energy required to excite $1\text{L}$ hydrogen gas at $1\text{atm}$ and $298\text{K}$ to first excited state of atomic hydrogen. The energy for the dissociation of $H-H$ bond is $450\text{kJ/mol}$. Take R = $\frac{1}{12}\text{Latm/mol},N_A=6\times {10}^{23},1eV=1.6\times {10}^{-19}\text{J}$

Answer 1

$PV=nRT$
$n=\frac{PV}{RT}$
$n=\frac{1\times 1}{1/12\times 300}=0.04$
$H_2\to 2H;\Delta H=\text{kJ/mol}$
Energy required to bring 0.04 moles of hydrogen gas to atomic state $=450\times 0.04=18\text{kJ}$
1 mole of $H_2$ contains = $6\times {10}^{23}$ molecules
0.04 mole of $H_2$ contains = $6\times {10}^{23}\times 0.04$ molecules
$\because$ 1 molecule of $H_2$ gas has 2 hydrogen atoms
$6\times {10}^{23}\times 0.04$ molecules of $H_2$ gas has $=2\times 6\times {10}^{23}\times 0.04=4.8\times {10}^{22}$ hydrogen atoms
$E=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})\text{eV}$
$E=13.6(\frac{1}{1}-\frac{1}{4})=13.6\times \frac{3}{4}=10.2\text{eV}=1.632\times {10}^{-21}\text{kJ}$
Energy required to excite $4.8\times {10}^{22}$ atoms = $1.632\times {10}^{-21}\times 4.8\times {10}^{22}=7.834\times 10=78.37\text{kJ}$
$\therefore$ Total energy required $=18+78.34=96.34\text{kJ}$