Question

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

Answer 1

Hint: 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 $\Delta {{H}_{C}}=-249\,kJ/mol$
The bond enthalpy of hydrogen – hydrogen bond is $433\,kJ/mol$
The bond enthalpy of oxygen – oxygen bond is $492\,kJ/mol$
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.
 ${{H}_{2(g)}}+\dfrac{1}{2}{{O}_{2(g)}}\to {{H}_{2}}{{O}_{(g)}}$
The enthalpy of combustion is calculated by the difference of bond enthalpies of reactants and products.
The formula can be written as:
 $\Delta {{H}_{C}}=B.{{E}_{R}}-B.{{E}_{P}}$
In this question, it is given that $\Delta {{H}_{C}}=-249\,kJ/mol$
Bond enthalpy of reactants = bond enthalpy of hydrogen – hydrogen bond, that is, $433\,kJ/mol$ and bond enthalpy of oxygen – oxygen bond, that is, $492\,kJ/mol$
As in the reaction, half ${{O}_{2}}$ is used, therefore, only half of the bond enthalpy of oxygen – oxygen will be considered.
Bond enthalpy of products = $2\times $ bond enthalpy of oxygen – hydrogen bond.
Bond enthalpy of oxygen – hydrogen bond is multiplied by two, because two $OH$ bonds are used in ${{H}_{2}}O$ structure.
 $\Delta {{H}_{C}}=\left( B.{{E}_{H-H}}+\dfrac{1}{2}B.{{E}_{O=O}} \right)-\left( 2\times B.{{E}_{O-H}} \right)$
Now, substituting the values in the above formula, we get
 $-249=\left( 433+\dfrac{1}{2}\times 492 \right)-\left( 2\times B.{{E}_{O-H}} \right)$
 $\Rightarrow -249=679-(2\times B.{{E}_{O-H}})$
 $\Rightarrow -928=-2\times B.{{E}_{O-H}}$
 $\Rightarrow B.{{E}_{O-H}}=464\,kJ/mol$
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 $(J)$ In this question, we have calculated the bond enthalpy of $OH$ with the help of the formula of enthalpy of combustion.