By using the standard enthalpy of reaction, the bond energy of the (A-B) bond is equal to 152.25 kJ/mol.
[tex]AB_2[/tex] is a molecule that reacts readily with water ([tex]H_2O[/tex]) as represented by this properly balanced chemical equation:
[tex]2AB_2 +2H_2O \rightarrow O_2 +4HB+A_2\; \Delta H = -142\;kJ[/tex]
Bond: O–H O=O H–B [tex]A \rightarrow A^+[/tex]
Bond energy (kJ/mol): 467 498 450 321
Mathematically, the standard enthalpy of reaction can be calculated by using this formula:
[tex]\Delta H^{\circ}_R=\Delta H_{reactants}-\Delta H_{products}[/tex]
Substituting the given parameters into the formula, we have;
[tex]-142=[(2\times2(A-B)+(2\times2(O-H)-[(O=O)+4(H-B)+(A-A)]\\\\-142=4(A-B)+4(467)-498-4(450)-321\\\\-142=4(A-B)+1868-498-1800-321\\\\4(A-B)=-142-1868+498+1800+321\\\\4(A-B)=609\\\\(A-B)=\frac{609}{4}[/tex]
(A-B) = 152.25 kJ/mol.
Read more on bond energy here: brainly.com/question/5650115
