A) \[200\,kJ\,mo{{l}^{-1}}\]
B) \[100\,kJ\,mo{{l}^{-1}}\]
C) \[300\,kJ\,mo{{l}^{-1}}\]
D) \[400\,kJ\,mo{{l}^{-1}}\]
Correct Answer: D
Solution :
Let bond energy of \[{{A}_{2}}\] be x then bond energy of AB is also x and bond energy of \[{{B}_{2}}\] is \[x/2\]. Enthalpy of formation of AB is\[~-100\text{ }KJ/mole\]: \[{{A}_{2}}+{{B}_{2}}\to 2AB;\] \[\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\to AB;\,\Delta =-100KJ\] or\[-100=\left( \frac{x}{2}+\frac{x}{4} \right)-x\]\[\therefore \] \[-100=\frac{2x+x-4x}{4}\therefore x=400KJ\]You need to login to perform this action.
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