Respuesta :
Given question is incomplete. The complete question is as follows.
[tex]AB_{2}[/tex] is a molecule that reacts readily with water. Calculate the bond energy of the A–B bond using the standard enthalpy of reaction and the bond energy data provided. Enter a number in kJ to 0 decimal places.
[tex]2AB_{2}(g) + 2H_{2}O(g) \rightarrow O=O(g) + 4HB(g) + A_{2}[/tex] [tex]\Delta H^{o}[/tex] = –142 kJ
Bond: O–H O=O H–B [tex]A \rightarrow A^{+}[/tex]
Bond energy (kJ/mol): 467 498 450 321
Explanation:
The given reaction is as follows.
[tex]2AB_{2} + 2H_{2}O \rightarrow O_{2} + 4HB + A_{2}[/tex]
Now, we will calculate the enthalpy of reaction as follows.
[tex]\Delta H^{o}_{R}[/tex] = -142 kJ
Also, we know that
[tex]\Delta H^{o}_{R} = \Delta H_{reactants} - \Delta H_{products}[/tex]
= [tex][(2 \times 2 (A-B) + 2 \times 2 (O-H)] - [(O=O) + 4(H-B) + (A-A)][/tex]
-142 = [tex]4(A-B) + 4 \times 467 - 498 - 4(450) - 321[/tex]
[tex]4(A-B)[/tex] = -142 - 1868 + 498 + 1800 + 321
= 609
(A-B) = 152.25 kJ/mol
Thus, we can conclude that the bond energy of the A–B bond is 152.25 kJ/mol.
