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# If bond energy of ${H_2}$ , ${F_2}$, and HF are in the ratio $2:1:3$ and $\Delta {H_a}$ $({H_2})$ = 400 kJ/mol. Then $\Delta {H_f}$ (HF) is:A. 0 kJ/mol B. -600 kJ/mol C. -1200 kJ/mol D. None of these

Last updated date: 15th Sep 2024
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Hint: Enthalpy of formation can be calculated from the bond energy of its gaseous constituents. Knowing the reactions involved in the chemical process and using additional subtraction to get the final equation results in the formation of enthalpy change.

The standard enthalpy of formation can be defined as the change in enthalpy when one mole of a substance in the standard state is formed from its pure elements in similar conditions. The bond dissociation energy is the amount of energy required to break apart one mole of covalently bonded gases into a pair of ions.
Now the ratio of bond energy of ${H_2}$ , ${F_2}$ , and HF are $2:1:3$ . Let the bond energies are for ${H_2}$ 2x kJ/mol, for ${F_2}$ X kJ/mol, for HF 3x kJ/mol. Now the atomization energy of ${H_2}$ is given which is, $\Delta {H_a}$ $({H_2})$ = 400 kJ/mol.
Hydrogen is a diatomic gas. Therefore, the atomization energy should be half of its bond energy.
Therefore, the bond energy or hydrogen would be double of its automation enthalpy. Which is
$400 \times 2 = 800kJ/mol$ .
Now according to the ratio of the bond energy for ${H_2}$ is 2x kJ/mol. So, the value of the x is,
$2x = 800 \\ x = 400 \\$
So, from the ratio, the bond energy of HF and ${F_2}$ is 1200kJ/mol and 400kJ/mol respectively.
We can arrange the given bond energies as per their dissociated chemical equations:
${H_2}(g) \to 2{H^ + }(g),\,\,\Delta H = \,800kJ/mol$ …………..(I)
${F_2}(g) \to 2{F^ - }(g),\,\,\Delta H = \,400kJ/mol$ ……………..(II)
$HF(g) \to {H^ + }(g) + {F^ - }(g),\,\,\Delta H = \,1200\,kJ/mol$ ………..(III)
The enthalpy of formation of HF can be calculated as:
$\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}{F_2}(g) \to HF(g)$
On dividing equation (I) and (II) by 2 and adding them, we get
$\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to {H^ + }(g) + {F^ - }(g),\,\,\,\Delta H = 600\,kJ/mol$ ………..(IV)
On subtracting the enthalpy change of (III) equation from (IV), we obtain $\Delta H$ = -600kJ/mol
$\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}{F_2}(g) \to HF(g)\Delta H = - 600kJ/mol$ .

Therefore, the correct answer is B.

Note:Another method to solve it is without involving all these equations and just putting the given energy values in the formula of enthalpy of formation which is the subtraction of bond energy of products from the sum of bond energies of reactants.
$\Delta {H_f} = \sum {H_{bond\,energy}}(reac\tan t) - \sum {{H_{bond\,energy}}(product)}$ .