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AB,A(2) and B(2) are diatomic molecules....

`AB,A_(2)` and `B_(2)` are diatomic molecules. If the bond enthalpies of `A_(2), AB` and `B_(2)` are in the ratio `1:1:0.5` and the enthalpy of formation of `AB` from `A_(2)` and `B_(2)` is `-100kJ mol^(-1)` , what is the bond enthalpy of `A_(2)` ?

A

`400 kJ mol^(-1)`

B

`1650 kJ mol^(-1)`

C

`1200 kJ mol^(-1)`

D

`200 kJ mol^(-1)`

Text Solution

Verified by Experts

`(1)/(2)A_(2)+(1)/(2)B_(2)rarr AB, " " Delta H=-100`
`(1)/(2)BE_(A-A)+(1)/(2)BE_(B-B)-BE_(A-B)=-100`
Let A - A bond energy be equal to 'x', then `(x)/(2)+(x)/(4)-x=-100`
or `x=+400 kJ mol^(-1)`
`(1)/(2)A_(2)+(1)/(2)O_(2)rarr AO , Delta H_(C )=-1200 kJ mol^(-1)`
`(1)/(2)(400)+(1)/(2)BE_(O-O)-BE_(A-O)=-1200 kJ mol^(-1)`
`BE_(A-O)=200+250+1200=1650 kJ mol^(-1)`
The correct answer is
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