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Determine the bond enthalpy of the \mathrm{O}-\mathrm{H} bond in water (\mathrm{H} _{2} \mathrm{O}) using the following data:

Bond dissociation enthalpy of H-H in hydrogen\mathrm{ (H2): 436 \mathrm{~kJ} / \mathrm{mol} }Bond dissociation enthalpy of \mathrm{O}-\mathrm{O} in oxygen (\mathrm{O} 2): 498 \mathrm{~kJ} / \mathrm{mol} Bond dissociation enthalpy of \mathrm{H}-\mathrm{O}-\mathrm{H} in hydrogen peroxide ( \mathrm{H} _{2} \mathrm{O} _{2}): 494 \mathrm{~kJ} / \mathrm{mol}

Option: 1

\mathrm{21 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{4 \mathrm{~kJ} / \mathrm{mol}}


Option: 3

\mathrm{6 \mathrm{~kJ} / \mathrm{mol}}


Option: 4

\mathrm{12 \mathrm{~kJ} / \mathrm{mol}}


Answers (1)

best_answer

1. Write the balanced reaction equation:

\mathrm{ \mathrm{H} 2 \mathrm{O} 2(g) \rightarrow \mathrm{H} 2 \mathrm{O}(g)+\mathrm{O} 2(g) }
2. Calculate the bonds broken in the reaction and their respective enthalpies:

Bond broken: \mathrm{O}-\mathrm{O} \, \, (in \, \, \mathrm{H} 2 \mathrm{O} 2 )

Enthalpy: 498 \mathrm{~kJ} / \mathrm{mol}

3. Calculate the bond formed in the reaction and its enthalpy:
Bond formed: \mathrm{O}-\mathrm{H}\, \, (in \, \, \mathrm{H} 2 \mathrm{O} )
Enthalpy: ?

4. Apply Hess's law:

\Delta H=\Sigma\left(\Delta H_{\text {bonds broken }}\right)-\Sigma\left(\Delta H_{\text {bonds formed }}\right)

5. Calculate \mathrm{\Delta H(O-H) :}\mathrm{ \Delta H(O-H)=498 \mathrm{~kJ} / \mathrm{mol}-\Delta H }

6. Rearrange the equation to solve for \Delta H:
 

\mathrm{ \Delta H=498 \mathrm{~kJ} / \mathrm{mol}-\Delta H(O-H) }

7. Substitute the value of the bond dissociation enthalpy of \mathrm{H}-\mathrm{O}-\mathrm{H} in hydrogen peroxide ( \mathrm{H} 2 \mathrm{O} 2):

\Delta H=498 \mathrm{~kJ} / \mathrm{mol}-494 \mathrm{~kJ} / \mathrm{mol}
8. Calculate the bond enthalpy of the \mathrm{O}-\mathrm{H} bond in water:

\Delta H=4 \mathrm{~kJ} / \mathrm{mol}

Answer: B) 4 \mathrm{~kJ} / \mathrm{mol}

Posted by

himanshu.meshram

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