Posted by Jude on Monday, December 13, 2010 at 4:33pm.
I worked this for you earlier but I wasn't sure if the delta Hf was -336 or not. It is; therefore, the delta Hrxn is -672 for the reaction as written.
X2 + Y2 ==> 2XY
(sum B.E.X2 + B.E.Y2)-(sum B.E. 2XY) = deltaHrxn*2
(414+159)-(2*XY) = -672
Solve for XY and I get 622.5 kJ/mol for X-Y.
Check that this way:
(1/2)X2 + (1/2)Y2 = XY
(sum B.E.X2 + B.E.Y2)-(B.E.XY) = ??
(207+79.5)-(622.5) = ??
(286.5)-(622.5) = -336
Check my thinking.
Consider hypothetical elements X and Y. Suppose the enthalpy of formation of the compound XY is −336 kJ/mol, the bond energy for X2 is 421 kJ/mol, and the bond energy for Y2 is 152 kJ/mol. Estimate the XY bond energy in units of kJ/mol.
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