Question

The enthalpy change for the reaction is:
     \[Xe{{F}_{4}}\to X{{e}^{+}}+{{F}^{-}}+{{F}_{2}}+F\text{ ; }\Delta \text{H = ?}\]
The average Xe – F bond energy is 34kcal/mol. $I.{{E}_{1}}$ of Xe is 279kcal/mol and electron affinity of F is 85 kcal/mol and ${{e}_{F-F}}$ is 38 kcal/mol.
[A] 345 Kcal/mol
[B] 292 Kcal/mol
[C] 174 Kcal/mol
[D] None of these

Answer 1

Hint: To solve this you must know what bond energy, ionisation enthalpy and electron affinity is. Find the energy change for each of these processes. If energy is released then the energy change is negative and if energy is required, energy change is positive. Use this to find the solution.

Complete step by step answer:
In thermodynamics, we describe the energy change by the term enthalpy. Enthalpy is the change in internal energy of the reaction. Enthalpy is written in terms of internal energy, pressure and volume but considering only the internal energy part, we can write enthalpy as-
     \[\Delta H=U(product)-U(reactant)\]
Now, let us see the question given to us.
Here, the Xe – F bond energy is given to us as 34kcal/mol. In $Xe{{F}_{4}}$ we have 4 Xe – F bonds.
Therefore, total bond energy = 4 $\times $ 34kcal/mol = 136 Kcal/mol.
So, we can write that-
     \[Xe{{F}_{4}}\to 4F+Xe\text{ ; }\Delta \text{H = 136Kcal/mol}\]

Now, we know that ionization enthalpy is the amount of energy required for the ionisation of the atom i.e. removal of the loosely bound outermost electron from the atom.
So, we can write that: $Xe\to X{{e}^{+}}+{{e}^{-}}$ $I.{{E}_{1}}$ = +279kcal/mol
Then we have electron affinity. In simpler words, we can say that electron affinity is the affinity of an atom to gain an electron and is denoted as${{E}_{ea}}$. So, $F\to {{F}^{-}}-{{e}^{-}}$ , ${{E}_{ea}}$ = - 85 kcal/mol
And for ${{e}_{{{F}_{2}}}}$ , $2F\to {{F}_{2}}$ , energy released = -38 kcal/mol
So, the enthalpy change = $I.{{E}_{1}}$ + ${{E}_{ea}}$ + ${{e}_{{{F}_{2}}}}$ + 4 $B.{{E}_{Xe-F}}$
So, putting the values we will get, $\Delta H=279-85-38+136=292Kcal/mol$
We can see from the above calculation that enthalpy change for the reaction is 292 Kcal/mol.
So, the correct answer is “Option B”.

Note: We know that $\Delta H$ is the enthalpy of the system. Enthalpy is a state function and is dependent upon the change in internal energy, volume and pressure of the system. The enthalpy change of a reaction is also dependent upon the temperature change. Enthalpy is a measure of change in work done or heat released or required during the reaction.