
The dissociation energy of ${H_2}$ is $430.53$$kJ/mol$ . if ${H_2}$ is exposed to radiant energy of wavelength $253.7nm$. What percentage of radiant energy will be converted into kinetic energy?
A.$91.3\% $
B.$8.7\% $
C.$9.5\% $
D.$90.5\% $
Answer
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Hint: Hydrogen molecules are made up of covalent bonds which are stable and strong. There is a requirement that a certain amount of energy to break the covalent bond between the hydrogen atom is expressed as bond dissociation energy.
Complete answer:
Amount of energy required to break the bond of ${H_2}$ is given in the question which is $430.53$$kJ/mol$.
Which means one mole of hydrogen require $430.53$$kJ$ of energy
As we know one mole of hydrogen contains $6.022 \times {10^{23}}$ number of molecules.
Hence, one molecule of hydrogen requires –
${{\rm E}_1} = \dfrac{{430.53 \times {{10}^3}}}{{6.022 \times {{10}^{23}}}}$J per molecule
After solving the above equation, we get
${E_1} = 7.15 \times {10^{ - 19}}$ J per molecule
Now calculate the amount of energy released by a hydrogen molecule when it is exposed to radiation of wavelength $253.7nm$.
$E = \dfrac{{hc}}{\lambda }$
${E_2} = \dfrac{{6.0626 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{253.7 \times {{10}^{ - 9}}}}$
After solving the above equation, we get
${E_2} = 7.83 \times {10^{ - 19}}J$
Energy which remain to the molecule after dissociation is available to convert into its kinetic energy
Hence, kinetic energy of molecule is $\left( {{E_2} - {E_1}} \right)$
kinetic energy of molecule is $\left( {7.83 - 7.15} \right) \times {10^{ - 19}}J$
After solving the above equation, we get
kinetic energy of molecule is $\left( {0.68 \times {{10}^{ - 19}}} \right)J$
Therefore, the percentage of energy converted into kinetic energy is
$\% KE$$ = \dfrac{{0.68 \times {{10}^{ - 19}}}}{{7.83 \times {{10}^{ - 19}}}} \times 100$
After solving the equation, we get
$\% KE$$ = 8.68\% $
Hence, $8.68\% $ of radiant energy will be converted into kinetic energy. Therefore the option C is the correct option.
Note:
Bond dissociation energy is also known as bond enthalpy and it is used to determine the strength of bond between the different atoms. Bond dissociation energy is always calculated for an atom in its gaseous state.
Complete answer:
Amount of energy required to break the bond of ${H_2}$ is given in the question which is $430.53$$kJ/mol$.
Which means one mole of hydrogen require $430.53$$kJ$ of energy
As we know one mole of hydrogen contains $6.022 \times {10^{23}}$ number of molecules.
Hence, one molecule of hydrogen requires –
${{\rm E}_1} = \dfrac{{430.53 \times {{10}^3}}}{{6.022 \times {{10}^{23}}}}$J per molecule
After solving the above equation, we get
${E_1} = 7.15 \times {10^{ - 19}}$ J per molecule
Now calculate the amount of energy released by a hydrogen molecule when it is exposed to radiation of wavelength $253.7nm$.
$E = \dfrac{{hc}}{\lambda }$
${E_2} = \dfrac{{6.0626 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{253.7 \times {{10}^{ - 9}}}}$
After solving the above equation, we get
${E_2} = 7.83 \times {10^{ - 19}}J$
Energy which remain to the molecule after dissociation is available to convert into its kinetic energy
Hence, kinetic energy of molecule is $\left( {{E_2} - {E_1}} \right)$
kinetic energy of molecule is $\left( {7.83 - 7.15} \right) \times {10^{ - 19}}J$
After solving the above equation, we get
kinetic energy of molecule is $\left( {0.68 \times {{10}^{ - 19}}} \right)J$
Therefore, the percentage of energy converted into kinetic energy is
$\% KE$$ = \dfrac{{0.68 \times {{10}^{ - 19}}}}{{7.83 \times {{10}^{ - 19}}}} \times 100$
After solving the equation, we get
$\% KE$$ = 8.68\% $
Hence, $8.68\% $ of radiant energy will be converted into kinetic energy. Therefore the option C is the correct option.
Note:
Bond dissociation energy is also known as bond enthalpy and it is used to determine the strength of bond between the different atoms. Bond dissociation energy is always calculated for an atom in its gaseous state.
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