Question

AB2 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. 2AB2(g) + 2H2O(g) ⟶ O=O(g) + 4HB(g) + A2 ΔH° = –142 kJ

Answer 1

Given question is incomplete. The complete question is as follows.

`AB_(2)` 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.

`2AB_(2)(g) + 2H_(2)O(g) \rightarrow O=O(g) + 4HB(g) + A_(2)`
`\Delta H^(o)` = –142 kJ

Bond: O–H O=O H–B
`A \rightarrow A^(+)`

Bond energy (kJ/mol): 467 498 450 321

Step-by-step explanation:

The given reaction is as follows.

`2AB_(2) + 2H_(2)O \rightarrow O_(2) + 4HB + A_(2)`

Now, we will calculate the enthalpy of reaction as follows.

`\Delta H^(o)_(R)` = -142 kJ

Also, we know that

`\Delta H^(o)_(R) = \Delta H_(reactants) - \Delta H_(products)`

=
`[(2 * 2 (A-B) + 2 * 2 (O-H)] - [(O=O) + 4(H-B) + (A-A)]`

-142 =
`4(A-B) + 4 * 467 - 498 - 4(450) - 321`

`4(A-B)` = -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.