
If the bond dissociation energy of XY, ${X_2}$, ${Y_2}$ (all are gaseous diatomic molecules) are in the ratio of 1:1:0.5 and enthalpy for the formation of XY is -200kJ/mol. The bond dissociation energy of ${X_2}$ will be:
A. 200 KJ/mol
B. 100 KJ/mol
C. 800 KJ/mol
D. 300 KJ/mol
Answer
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Hint: In this question assume that a, a, a/2 are the dissociation enthalpy of X, Y, ${X_2}$ and ${Y_2}$ also make the equation using the given information and find out the correct option.
Complete answer:
By the given information
Let a, a, a/2 be the dissociation energies of X, Y, ${X_2}$ and ${Y_2}$.
$XY \to X\left( g \right) + Y\left( g \right);\vartriangle H = + 1a$ KJ/mol (equation 1)
\[{X_2} \to 2X;\vartriangle H = + 1a\] KJ/mole (equation 2)
${Y_2} \to 2Y;\vartriangle H = + 0.5a$ KJ/mole (equation 3)
Adding equation 2 and 3
\[{X_2} + {Y_2} \to 2X + 2Y\] \[ \Rightarrow \dfrac{1}{2}{X_2} + \dfrac{1}{2}{Y_2} \to X + Y\]
Substituting the value of X + Y in equation 1
$\dfrac{1}{2}{X_2} + \dfrac{1}{2}{Y_2} \to XY$
We have $\vartriangle {H_f}$= -200 kJ/mol
Using the formula
$\vartriangle {H_f}$ = Enthalpy of bond dissociation – Enthalpy of bond formed
Substituting the values in the formula
$ - 200 = \dfrac{1}{2}\left( a \right) + \dfrac{1}{2}\left( {\dfrac{a}{2}} \right) - a$
$ \Rightarrow \dfrac{{ - a}}{4} = - 200$
$ \Rightarrow $a = 800
Since a is the bond dissociation energy of ${X_2}$ = 800 KJ/mol
Hence, the correct option is C.
Note: In the above solution we have gone through the term Bond dissociation energy many times it can be explained as the energy needed to sever a chemical bond. This is one way of quantifying the strength of a chemical bond. Bond dissociation energy equals bond energy only for diatomic molecules. The strongest bond dissociation energy is for the Si-F bond.
Complete answer:
By the given information
Let a, a, a/2 be the dissociation energies of X, Y, ${X_2}$ and ${Y_2}$.
$XY \to X\left( g \right) + Y\left( g \right);\vartriangle H = + 1a$ KJ/mol (equation 1)
\[{X_2} \to 2X;\vartriangle H = + 1a\] KJ/mole (equation 2)
${Y_2} \to 2Y;\vartriangle H = + 0.5a$ KJ/mole (equation 3)
Adding equation 2 and 3
\[{X_2} + {Y_2} \to 2X + 2Y\] \[ \Rightarrow \dfrac{1}{2}{X_2} + \dfrac{1}{2}{Y_2} \to X + Y\]
Substituting the value of X + Y in equation 1
$\dfrac{1}{2}{X_2} + \dfrac{1}{2}{Y_2} \to XY$
We have $\vartriangle {H_f}$= -200 kJ/mol
Using the formula
$\vartriangle {H_f}$ = Enthalpy of bond dissociation – Enthalpy of bond formed
Substituting the values in the formula
$ - 200 = \dfrac{1}{2}\left( a \right) + \dfrac{1}{2}\left( {\dfrac{a}{2}} \right) - a$
$ \Rightarrow \dfrac{{ - a}}{4} = - 200$
$ \Rightarrow $a = 800
Since a is the bond dissociation energy of ${X_2}$ = 800 KJ/mol
Hence, the correct option is C.
Note: In the above solution we have gone through the term Bond dissociation energy many times it can be explained as the energy needed to sever a chemical bond. This is one way of quantifying the strength of a chemical bond. Bond dissociation energy equals bond energy only for diatomic molecules. The strongest bond dissociation energy is for the Si-F bond.
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