Question

using the appropriate bond energies, calculate the heat of reaction δh for the following reaction: 2h−h + o=o → 2o−h (with h attached to o). you can find a table of bond energies by using the data button on the aleks toolbar. round your answer to the nearest kj/mol. note: for clarity, all lone pairs have been omitted from the molecular structures. unit: kj/mol

Explanation:

Step1: Identify Bond Energies

The reaction is \( 2\ce{H-H} + \ce{O=O}
ightarrow 2\ce{H-O-H} \) (or \( 2\ce{H_2} + \ce{O_2}
ightarrow 2\ce{H_2O} \)).
Bond energies (from standard tables):

• \( \ce{H-H} \): \( 436\ \text{kJ/mol} \)
• \( \ce{O=O} \): \( 498\ \text{kJ/mol} \)
• \( \ce{O-H} \): \( 463\ \text{kJ/mol} \)

Step2: Calculate Energy to Break Bonds

Bonds broken: \( 2\) moles of \( \ce{H-H} \) and \( 1\) mole of \( \ce{O=O} \).
Energy required (endothermic):
\( (2 \times 436) + 498 = 872 + 498 = 1370\ \text{kJ} \)

Step3: Calculate Energy to Form Bonds

Bonds formed: \( 4\) moles of \( \ce{O-H} \) (since \( 2\ce{H_2O} \) has \( 4\ce{O-H} \) bonds).
Energy released (exothermic):
\( 4 \times 463 = 1852\ \text{kJ} \)

Step4: Calculate \( \Delta H \)

\( \Delta H = \text{Energy to break} - \text{Energy to form} \)
\( \Delta H = 1370 - 1852 = -482\ \text{kJ/mol} \)

Answer:

\(-482\ \text{kJ/mol}\)