Question

The first ionization energy for $Li$ is $5.4{\text{eV}}$ and the electron affinity of $Cl$ is $3.61{\text{eV}}$ . The ${{\Delta H(in kj/mol)}}$ for the reaction $L{i_{(g)}} + C{l_{(g)}} \to L{i^ + }_{(g)} + C{l^ - }_{(g)}$ is:
(if the resulting ion do not combine with each other)
A.$70$
B.$100$
C.$172$
D.$200$

Answer 1

Hint: The minimum amount of energy required to remove the loosely bounded electron from the isolated gaseous atom is known as is ionization energy and the energy required to add an electron to isolated gaseous atom is known as its electron affinity.

Complete step by step answer:
The minimum amount of energy required to remove the outermost electron from the outermost shell of an isolated gaseous atom is known as its ionization energy. Ionisation potential is the another name for ionization energy, previously ionization energy was called ionization potential. The ionization energy is determined experimentally, but we can compare the ionization potential or ionization energy by the size and charge present on the elemental atom.
Electron affinity basically describes the change in energy of a neutral atom when an electron is added to the element to form an anion. It is simply opposite of the ionization energy.
The first ionization energy for lithium is $5.4{\text{eV}}$.
\[Li(g) \to L{i^ + } + {e^ - }\] ---------------------(1)
So the $\Delta H1$ for the reaction is $5.4{\text{eV/atom}}$
The electron affinity for chlorine atom is $3.61{\text{eV}}$.
\[C{l_{(g)}} + {e^ - } \to C{l^ - }\] ---------------------(2)
So the $\Delta H2$ for the above reaction is $3.61{\text{eV/atom}}$
When we add equation (1) and equation (2) we get our desired reaction.
Equation (1) + Equation (2)
$L{i_{(g)}} + C{l_{(g)}} \to L{i^ + }_{(g)} + C{l^ - }_{(g)}$
The electrons present on both the sides are cancelled out and we can see that gaseous atoms are converted into ions directly in this reaction.
The $\Delta H$ for this reaction will be the sum of enthalpy of the initial two reactions because these reactions are added to get the final output reaction.
$\begin{gathered}
   \Rightarrow \Delta H = \Delta {H_1} + \Delta {H_2} \\
   \Rightarrow \Delta H = 5.46 - 3.61 = 2.05eV \\
\end{gathered} $
Now we will convert the units into kilojoule.
$\Delta H = 2.05eV = 2.05 \times 1.6 \times {10^{ - 19}}j/atom$
And now we will convert atoms into moles.
$\begin{gathered}
  \Delta H = 2.05 \times 1.6 \times {10^{ - 19}} \times 6.022 \times {10^{23}} \times {10^{ - 3}}j/mol \\
  \Delta H = 172.4Kj/mol \\
\end{gathered} $
So, the ${{\Delta H( kj/mol)}}$ for the reaction $L{i_{(g)}} + C{l_{(g)}} \to L{i^ + }_{(g)} + C{l^ - }_{(g)}$ is $172{\text{Kj/mol}}$.

Hence option (C) is correct.

Note:
As we have seen that for reactions the enthalpy is written as $\Delta H$ not simply $H$ because it is the difference of the enthalpies of products and reactants. The standard enthalpy for the reaction is calculated by summing up the enthalpy changes of each step.