Question

Calculate the enthalpy of formation of water, given that the bond energies of $H - H$ , $O - O$ , $O - H$ bond are $433kJ/mol,492kJ/mol,464kJ/mol$ respectively.
A. $\Delta H = - 249kJ$
B. $\Delta H = + 249kJ$
C. $\Delta H = - 649kJ$
D. None of these

Answer 1

Hint: Enthalpy is defined as the sum of internal energy and the product of pressure and volume of the thermodynamic system. Bond energy is defined as the amount of energy required to break a bond.

Complete step by step answer:
Enthalpy is given by the formula as follows:
$H = E + PV$
Where,
$H = $ enthalpy
$E = $ Internal energy
$P = $ Pressure
$V = $ volume.
The reaction of formation of water is given below:
${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$
Hess’s law states that the change in enthalpy is independent of the pathway (it does not depend on temperature and pressure) between the initial and final states of the reaction.
It is given by the formula as follows:
Heat of formation$\left( {\Delta H} \right) = $ sum of standard enthalpies of reactants$ - $ sum of standard enthalpies of products.
Given data:
Bond energy of $H - H$bond$ = 433kJ/mol$
Bond energy of $O - O$bond$ = 492kJ/mol$
Bond energy of $O - H$bond$ = 464kJ/mol$
Using Hess’s law we will find the enthalpy of formation of water.
Heat of formation $\left( {\Delta H} \right) = $ sum of standard enthalpies of reactants $ - $ sum of standard enthalpies of products
Heat of formation $\left( {\Delta H} \right)$ = bond energy of H - H + bond energy of $\dfrac{1}{2}{O_2} - 2 \times $bond energy of $O - H$
Substituting the values of the bond energies in the equation we get,
$ = \left( {433 + \dfrac{1}{2} \times 492} \right) - \left( {2 \times 464} \right)$
$ = \left( {433 + 246} \right) - \left( {928} \right)$
$ = 679 - 928$
$ = - 249kJ$
So the enthalpy required for the formation of water is $ - 249kJ$ .

So, the correct answer is Option A.

Note: Since the heat of formation is negative then the reaction is exothermic. If the heat of formation is positive then the reaction will be endothermic.
Hess’s law is very useful to calculate the enthalpy changes that cannot be measured directly from the experiments.
Bond breaking is an endothermic process.