Question

one method for estimating the enthalpy change for a reaction, δ h, is to calculate the energy difference between the bonds broken and the bonds formed. that is,
δ h = energy of bonds broken − energy of bonds formed

use this method, a value for the average o-o bond energy of 142 kj mol⁻¹, and the data in table 18.4 to estimate the enthalpy change for the following reaction:
h₂(g) + h₂o₂(g) → 2h₂o(g)

table 18.4 some average bond energies
bond bond energy (kj mol⁻¹) bond bond energy (kj mol⁻¹)
c—c 348 c—br 276
c═c 612 c—i 238
c≡c 960 h—h 436
c—h 412 h—f 565
c—n 305 h—cl 431
c═n 613 h—br 366
c≡n 890 h—i 299
c—o 360 h—n 388
o═o 743 h—o 463
c—f 484 h—s 338
c—cl 338 h—si 376

Explanation:

Step1: Identify bonds broken

In \( \ce{H2(g)} \), there is 1 \( \ce{H-H} \) bond. In \( \ce{H2O2(g)} \), there is 1 \( \ce{O-O} \) bond and 2 \( \ce{H-O} \) bonds. So bonds broken: 1 \( \ce{H-H} \) (436 kJ/mol), 1 \( \ce{O-O} \) (142 kJ/mol), 2 \( \ce{H-O} \) (but wait, no—wait, \( \ce{H2O2} \) structure is \( \ce{H-O-O-H} \), so bonds in \( \ce{H2O2} \): 2 \( \ce{H-O} \) and 1 \( \ce{O-O} \). Bonds in \( \ce{H2} \): 1 \( \ce{H-H} \). So total bonds broken: \( \ce{H-H} \) (1), \( \ce{O-O} \) (1), \( \ce{H-O} \) (2). Wait, no—wait, the reaction is \( \ce{H2 + H2O2 -> 2H2O} \). Let's list all bonds:

Reactants:

• \( \ce{H2} \): 1 \( \ce{H-H} \) bond (energy 436 kJ/mol)
• \( \ce{H2O2} \): 2 \( \ce{H-O} \) bonds (each 463 kJ/mol) and 1 \( \ce{O-O} \) bond (142 kJ/mol)

So energy of bonds broken: \( (1 \times 436) + (2 \times 463) + (1 \times 142) \)

Step2: Calculate bonds broken energy

Calculate: \( 436 + (2 \times 463) + 142 = 436 + 926 + 142 = 1504 \) kJ/mol

Step3: Identify bonds formed

Products: \( 2 \ce{H2O} \). Each \( \ce{H2O} \) has 2 \( \ce{H-O} \) bonds, so 2 molecules have \( 2 \times 2 = 4 \) \( \ce{H-O} \) bonds.

Energy of bonds formed: \( 4 \times 463 \) (since each \( \ce{H-O} \) is 463 kJ/mol)

Step4: Calculate bonds formed energy

\( 4 \times 463 = 1852 \) kJ/mol

Step5: Calculate \( \Delta H \)

\( \Delta H = \) bonds broken - bonds formed \( = 1504 - 1852 = -348 \) kJ/mol

Answer:

-348