Question

The dissociation energy of $H_2$ is 430.53 kJ/mol. If $H_2$ is exposed to a radiation of wavelength of 253.7 nm, what percent of radiant energy will be converted into kinetic energy?

• A. 5.69%
• B. 13.59%
• C. 8.46%
• D. 18.56%

Answer 1

The correct option is C 8.46%

Dissociation energy per 1 mol is given as 430.53 kJ/mol which implies that for $N_A$ molecules the dissociation energy is 430.53 kJ/mol.

Hence, the dissociation energy for 1 molecule of $H_2$ = $\frac{430.53\times {10}^3}{6.023\times {10}^{23}}$ = $7.14\times {10}^{-19}$ J/molecule

According to Planck's constant, E = $\frac{hc}{\lambda }$
where, h = Planck's constant
c = Speed of light
$\lambda$ = Wavelength of radiant energy

Total energy E = $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{253.7\times {10}^{-9}}$ = $7.8\times {10}^{-19}J$

$7.14\times {10}^{-19}$ J energy is absorbed by 1 molecule of $H_2$ to dissociate

So Energy converted into KE = $7.8\times {10}^{-19}J-7.14\times {10}^{-19}$ J
= $0.66\times {10}^{-19}$ J

% energy converted to kinetic energy = $\frac{0.66\times {10}^{-19}}{7.8\times {10}^{-19}}\times 100$ = 8.46%