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Question: Given that \(AB\) , \({A_2}\) and \({B_2}\) ​ are diatomic molecules. If the bond enthalpies of \(AB...

Given that ABAB , A2{A_2} and B2{B_2} ​ are diatomic molecules. If the bond enthalpies of ABAB , A2{A_2} and B2{B_2} ​ are in the ratio 1:1:0.51:1:0.5 and enthalpy of formation of ABAB from A2{A_2} and B2{B_2} ​ is 100kJmol1 - 100kJmo{l^{ - 1}}. What is the bond energy of A2{A_2} ​?
A. 200KJmol1200KJmo{l^{ - 1}}
B. 100KJmol1100KJmo{l^{ - 1}}
C. 300KJmol1300KJmo{l^{ - 1}}
D. 400KJmol1400KJmo{l^{ - 1}}

Explanation

Solution

As we know that write the common balanced equation of the reaction and make sure to make it for 1 mole of ABAB . You are given the enthalpy of reaction and a ratio of bond enthalpies, derive an equation from the reaction and put the values to get the value of bond enthalpy of A2{A_2} ..

Complete step by step answer:
Let’s write the initial equation which will be required to solve this question. The equation is given down below:
A2 + B22AB{{\text{A}}_2}{\text{ + }}{{\text{B}}_2} \to {\text{2AB}}
We are given that the bond enthalpies are in the ratio of 1:1:0.51:1:0.5 for A2:AB:B2{A_2}:AB:{B_2} , also the enthalpy of formation for AB is -100KJ/mol. So the formation of AB will take place if half of A2A_2 and half of B2B_2 reacts which means
12A2 + 12B2AB\dfrac{1}{2}{{\text{A}}_2}{\text{ + }}\dfrac{1}{2}{{\text{B}}_2} \to {\text{AB}}
Hence the enthalpy of this reaction is -100 KJ/mol. Now taking the bond enthalpy of A2A_2 as x we get the equation
12x + 12(x2) x = 100 KJmol1\dfrac{1}{2}x{\text{ + }}\dfrac{1}{2}(\dfrac{x}{2}) - {\text{ }}x{\text{ = }} - 100{\text{ KJmo}}{{\text{l}}^{ - 1}}
Solving this we will get the equation
x4=100 KJmol1\dfrac{x}{4} = 100{\text{ KJmo}}{{\text{l}}^{ - 1}}
Solving this we are getting x=400KJmol1x = 400KJmo{l^{ - 1}}

So, the correct answer is Option D.

Note: In many reactions it is necessary that we calculate the heat of the reaction and at this point the enthalpy of the reaction comes into the picture. Not only theoretically but practically also enthalpy values come handy. In calculating the refrigeration effect and the efficiency of the engine, enthalpy plays a huge role in calculating all these. As we know that the extensive properties like energy, volume and enthalpy depend on the mass of the system. For example, if the mass of the system gets doubled, the enthalpy of the system may also get doubled.