Question

Calculate the average bond enthalpy of $O - H$ bond in water at 298K using the data/information given below:-
1.${\Delta _f}{H^\circ }[{H_{(g)}}] = 218{\text{ kJ/mol}}$
2.${\Delta _f}{H^\circ }[{O_{(g)}}] = 249.2{\text{ kJ/mol}}$
3.\[{\Delta _f}{H^\circ }[{H_2}{O_{(g)}}] = - 241.8{\text{ kJ/mol}}\]
The average bond enthalpy of the –bond in water is defined as one-half of the enthalpy change for the reaction ${H_2}{O_{(g)}} \to 2{H_{(g)}} + {O_{(g)}}.$
Also, determine the $\Delta U$ of the $O - H$ bond in water at 298 K. Assume ideal gas behaviour.
A. $E(O - H) = 462{\text{ kJ/mol; }}\Delta {\text{U = 461 kJ/mol}}$
B. $E(O - H) = 463.5{\text{ kJ/mol; }}\Delta {\text{U = 463}}{\text{.5 kJ/mol}}$
C. $E(O - H) = 461{\text{ kJ/mol; }}\Delta {\text{U = 461 kJ/mol}}$
D. $E(O - H) = 463.5{\text{ kJ/mol; }}\Delta {\text{U = 461 kJ/mol}}$

Answer 1

Hint: Enthalpy of formation of a substance is defined as the heat change that takes place when one mole of a substance is formed from its element under a given condition of temperature and pressure. The value of enthalpy of formation of elements in a reaction is always considered as zero.

 Complete answer:
For calculating the average bond enthalpy of O-H bond in water, the bond energies of the reactants and the bond energies of the products from the following reaction are calculated.
\[2{H_{(g)}} + {\text{ }}{O_{(g)}} \to {H_2}O\;\]
Step 1: Calculating the total bond energies of reactant and product:
According to the question:
Enthalpy of formation of one mole of Hydrogen \[ = {\text{ }}218{\text{ }}kJ/mol\]
Enthalpy of formation for two moles of hydrogen \[2 \times 218{\text{ }} = 436{\text{ }}kJ/mol\]
Enthalpy of formation of one mole of Oxygen atom \[ = {\text{ }}249.2{\text{ }}kJ/mol\;\]
Therefore, the total enthalpy of reactants \[ = {\text{ }}436 + 249.2 = 685.2{\text{ }}kJ\]
Enthalpy of formation of one mole of water molecule \[ = - 241.8{\text{ }}kJ/mol\]
Therefore, the total bond enthalpy of product\[ = - 241.8{\text{ }}kJ/mol\]
Step 2: Calculating the enthalpy change of the reaction:
The enthalpy change of the reaction is equal to the bond energies of the reactant subtracted by the bond energies of the product. Thus, the enthalpy change of the above reaction is
\[\Delta {H_{reaction}} = 685.2 - \left( { - 241.80} \right)\]
\[\Delta {H_{reaction}} = 685.2 + 241.8\]
\[\Delta {H_{reaction}} = 927{\text{ k}}J/mol\]
Hence the enthalpy change of the reaction is equal to 927 kJ/mol.
Step 3: Calculating the average bond enthalpy of the O-H bond:
The average bond enthalpy of the O-H bond in water is defined as the one- half of the enthalpy change of the reaction.
\[\dfrac{{\Delta {H_{reaction}}}}{2} = \Delta {H_{O - H}}\]
$\dfrac{{927}}{2} = \Delta {H_{O - H}}$
$\Delta {H_{O - H}} = 463.5kJ/mol$
Hence, the average bond enthalpy of the O-H bond in water is 463.5 kJ/mol.
Step 4: Determining the change in internal energy:
For ideal gases the value change in internal energy can be calculated as
$\Delta H = \Delta U + nRT$
$\Delta U = \Delta {H_{O - H}} - nRT$
Where, n is equal to one in this case as there are two O-H bond in water and R is the universal gas constant equal to 8.314 $J{\text{ }}{{\text{k}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ and T is the temperature equal to 298K.$\Delta U = 463.5 - (\dfrac{{8.314 \times 298}}{{1000}})$
$\Delta U = 461{\text{ kJ/mol}}$
Therefore, the value of change in internal energy of the O-H bond in water at 298 K is equal to 461 KJ/mol.

 So, the correct answer is option D.

 Note:
The bond energy or bond enthalpy of a particular bond is either defined as the average amount of energy released when one mole of the bonds are formed from isolated gaseous atom or the amount of energy required when one mole of bonds are broken in order to obtain the separated gaseous atom.