
A photon of \[4000\text{ }A{}^\circ \] is used to break the iodine molecule, then the \[%\] of energy converted to the K.E. of iodine atoms if bond dissociation energy of \[{{I}_{2}}\]molecule is \[246.5\text{ }kJ/mol\] is:
A.\[8%\]
B.\[12%\]
C.\[17%\]
D.\[25%\]
Answer
506.1k+ views
Hint: We know that the bond energy or bond enthalpy can be visualized as the average amount of energy required or produced when a bond in a chemical compound is broken or formed. Also consider the factors that affect bond dissociation enthalpy.
Complete answer:
Here, consider the factor of atomic radius as bond dissociation enthalpy is indirectly proportional to atomic radii. Before answering the question, let us understand the meaning of the term bond dissociation enthalpy. Bond dissociation enthalpy is basically the energy required for breaking a solid crystal into its corresponding ions. There are factors that affect the lattice energy of any substance. Firstly it is the charge on the ions on the ions. Higher is the charge on each ion, higher will be the lattice energy and the second factor is the radius. Lower the internuclear distances among the ions in an ionic crystal, higher will be their lattice energy i.e. closer the two ions are, higher is the force of attraction between them. Bond energy gives an idea of the strength of a chemical bond and therefore the stability of the compound. Hence the more stable the compound, the more energy will be needed to break the bond.
Energy of one photon\[=\dfrac{12400}{4000}=3.1ev=3.11\times 6\times {{10}^{-19}}\text{ }Joules.\]
Thus, energy supplied by one mole photon in \[KJ/mole=3.11\times 6\times {{10}^{-19}}6\times {{10}^{23}}{{10}^{-3}}=297 KJ/mol\]
Therefore, Percentage of energy converted to \[K.E.=\dfrac{297-246.5}{297}=17.\]
Therefore the correct answer is option C.
Note:
Remember that when the bond in a chemical compound is broken, then there is a need for energy hence the bond energy will have a positive value. And when the bond in a chemical compound is formed, then the energy will be released hence the bond energy will have a negative value.
Complete answer:
Here, consider the factor of atomic radius as bond dissociation enthalpy is indirectly proportional to atomic radii. Before answering the question, let us understand the meaning of the term bond dissociation enthalpy. Bond dissociation enthalpy is basically the energy required for breaking a solid crystal into its corresponding ions. There are factors that affect the lattice energy of any substance. Firstly it is the charge on the ions on the ions. Higher is the charge on each ion, higher will be the lattice energy and the second factor is the radius. Lower the internuclear distances among the ions in an ionic crystal, higher will be their lattice energy i.e. closer the two ions are, higher is the force of attraction between them. Bond energy gives an idea of the strength of a chemical bond and therefore the stability of the compound. Hence the more stable the compound, the more energy will be needed to break the bond.
Energy of one photon\[=\dfrac{12400}{4000}=3.1ev=3.11\times 6\times {{10}^{-19}}\text{ }Joules.\]
Thus, energy supplied by one mole photon in \[KJ/mole=3.11\times 6\times {{10}^{-19}}6\times {{10}^{23}}{{10}^{-3}}=297 KJ/mol\]
Therefore, Percentage of energy converted to \[K.E.=\dfrac{297-246.5}{297}=17.\]
Therefore the correct answer is option C.
Note:
Remember that when the bond in a chemical compound is broken, then there is a need for energy hence the bond energy will have a positive value. And when the bond in a chemical compound is formed, then the energy will be released hence the bond energy will have a negative value.
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