chemistry
posted by chemistry on .
Consider the hypothetical elements X and Y. Suppose the enthalpy of formation for the compound XY is ƒ{336 kJ/mol, the bond energy for X2 is 414 kJ/mol, and the bond energy for Y2 is 159 kJ/mol. Estimate the XY bond energy in units of kJ/mol.

X2 + Y2 => 2XY delta H format = 2*336 kJ.
X2 = 414 = bond energy = B.E.
Y2 = 159
(sum B.E.reactantssum B.E. products) = delta H rxn.
(414+159)(2*B.E. products)= 672
NOTE: It isn't clear to me what the sign is for delta Hf for XY. I have assumed you intended to type a  sign; however, whatever the sign is for delta Hf that goes in for the 672. I have multiplied by 2 since there are two moles involved.
You can work out the B.E. products as the only unknown in the equation. You can check the number by the following:
[(414/2)+(150/2)](B.E. product you calculate) = delta Hf.