ΔE or ΔH can be calculated as:

$\overline{){\mathbf{\u2206}}{\mathbf{H}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{bond}}{\mathbf{}}{{\mathbf{energy}}}_{{\mathbf{reactants}}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathbf{bond}}{\mathbf{}}{{\mathbf{energy}}}_{{\mathbf{products}}}\mathbf{}}\phantom{\rule{0ex}{0ex}}\mathbf{\u2206}\mathbf{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{BE}\mathbf{}{\mathbf{A}}_{\mathbf{2}}\mathbf{+}\mathbf{BE}\mathbf{}{\mathbf{B}}_{\mathbf{2}}\mathbf{)}\mathbf{}\mathbf{-}\mathbf{2}\mathbf{(}\mathbf{BE}\mathbf{}\mathbf{AB}\mathbf{)}$

If A_{2} is half the amount of AB, we can set up the equation as:

Consider the following reaction:

A_{2} + B_{2 }→ 2AB ΔE = -285 kJ

The bond energy for A_{2} is one-half the amount of the AB bond energy. The bond energy of B _{2} = 432 kJ/mol. What is the bond energy of A_{2}?

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