
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
608.7k+ views
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

