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Question: The enthalpy of combustion of \({{H}_{2(g)}}\) to give \({{H}_{2}}{{O}_{(g)}}\) is \(-249\,kJ\,mo{{l...

The enthalpy of combustion of H2(g){{H}_{2(g)}} to give H2O(g){{H}_{2}}{{O}_{(g)}} is −249 kJ mol−1-249\,kJ\,mo{{l}^{-1}} and bond enthalpies of H−HH-H and O=OO=O are 433 kJ mol−1433\,kJ\,mo{{l}^{-1}} and 492 kJ mol−1492\,kJ\,mo{{l}^{-1}} respectively. The bond enthalpy of O−HO-H is:
A. +464 kJ mol−1+464\,kJ\,mo{{l}^{-1}}
B. −464 kJ mol−1-464\,kJ\,mo{{l}^{-1}}
C. +232 kJ mol−1+232\,kJ\,mo{{l}^{-1}}
D. +232 kJ mol−1+232\,kJ\,mo{{l}^{-1}}

Explanation

Solution

When a substance reacts with oxygen rapidly, it releases heat in the environment. This process is known as combustion. In this question, we have talked about enthalpy, which is a thermodynamic property, that is used to calculate many measurements.

Complete solution:
Here, it is given that the enthalpy of combustion ΔHC=−249 kJ/mol\Delta {{H}_{C}}=-249\,kJ/mol
The bond enthalpy of hydrogen – hydrogen bond is 433 kJ/mol433\,kJ/mol
The bond enthalpy of oxygen – oxygen bond is 492 kJ/mol492\,kJ/mol
As we have discussed, combustion is the reaction of a substance with oxygen to give a compound and release heat.
Let us see a reaction based on this question.
H2(g)+12O2(g)→H2O(g){{H}_{2(g)}}+\dfrac{1}{2}{{O}_{2(g)}}\to {{H}_{2}}{{O}_{(g)}}
The enthalpy of combustion is calculated by the difference of bond enthalpies of reactants and products.
The formula can be written as:
ΔHC=B.ER−B.EP\Delta {{H}_{C}}=B.{{E}_{R}}-B.{{E}_{P}}
In this question, it is given that ΔHC=−249 kJ/mol\Delta {{H}_{C}}=-249\,kJ/mol
Bond enthalpy of reactants = bond enthalpy of hydrogen – hydrogen bond, that is, 433 kJ/mol433\,kJ/mol and bond enthalpy of oxygen – oxygen bond, that is, 492 kJ/mol492\,kJ/mol
As in the reaction, half O2{{O}_{2}} is used, therefore, only half of the bond enthalpy of oxygen – oxygen will be considered.
Bond enthalpy of products = 2×2\times bond enthalpy of oxygen – hydrogen bond.
Bond enthalpy of oxygen – hydrogen bond is multiplied by two, because two OHOH bonds are used in H2O{{H}_{2}}O structure.
ΔHC=(B.EH−H+12B.EO=O)−(2×B.EO−H)\Delta {{H}_{C}}=\left( B.{{E}_{H-H}}+\dfrac{1}{2}B.{{E}_{O=O}} \right)-\left( 2\times B.{{E}_{O-H}} \right)
Now, substituting the values in the above formula, we get
−249=(433+12×492)−(2×B.EO−H)-249=\left( 433+\dfrac{1}{2}\times 492 \right)-\left( 2\times B.{{E}_{O-H}} \right)
⇒−249=679−(2×B.EO−H)\Rightarrow -249=679-(2\times B.{{E}_{O-H}})
⇒−928=−2×B.EO−H\Rightarrow -928=-2\times B.{{E}_{O-H}}
⇒B.EO−H=464 kJ/mol\Rightarrow B.{{E}_{O-H}}=464\,kJ/mol
Therefore, the correct option is (A).

Note: Enthalpy is defined as a thermodynamic property which is calculated by the sum of internal energy and the product of pressure and volume of a system. The SI unit of the enthalpy is Joule (J)(J) In this question, we have calculated the bond enthalpy of OHOH with the help of the formula of enthalpy of combustion.