For the reaction, $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O},$: ΔH = – 571kJ ; Bond energy of (H–H) = 435 kJ and (O=O) =498kJ ,  the average bond energy of O–H bond using the above data is

Find average O–H bond energy from reaction data

Given:

Reaction: 2 H2 + O2 → 2 H2O

ΔH (reaction) = −571 kJ

Bond energies: H–H = 435 kJ, O=O = 498 kJ

Let average O–H bond energy = x (kJ)

Step 1 — Use bond-energy sign convention

For a reaction, approximate enthalpy change using bond energies:

ΔH ≈ (sum of bond energies of bonds broken) − (sum of bond energies of bonds formed)

Step 2 — Count bonds broken and formed

Bonds broken: 2 H–H (from 2 H2) and 1 O=O → total energy broken = 2×435 + 1×498

Bonds formed: 2 H2O has 4 O–H bonds → total energy formed = 4×x

Step 3 — Write the equation and solve for x

ΔH = (2×435 + 1×498) − (4×x)

Substitute ΔH = −571 kJ:

−571 = (870 + 498) − 4x

Simplify:

−571 = 1368 − 4x

Rearrange to find x:

4x = 1368 + 571 = 1939 ⇒  x = 1939 / 4 = 484.75 kJ

Final answer: Average O–H bond energy = 484.75 kJ.

Note: Bond-energy calculations give approximate values because actual ΔH depends on molecular environment and state. Here we used the common bond-energy approximation and sign convention (breaking bonds = energy input, forming bonds = energy release).