Question

The ionization enthalpy of \[N{a^ + }\] formation from \[Na(g)\] is \[495.8\,kJ\,mo{l^{ - 1}}\] , while the electron gain enthalpy of \[Br\] is \[ - 325.0\,kJ\,mo{l^{ - 1}}\] . given the lattice enthalpy of \[NaBr\] is \[ - 728.4\,kJ\,mo{l^{ - 1}}\] . The energy for the formation of \[NaBr\] ionic solid is ( \[ - \] ) __________ \[ \times {10^{ - 1}}\,kJ\,mo{l^{ - 1}}\] .

Answer 1

Hint: Ionisation energy is defined as the effort or difficulty involved in removing an electron from an atom or ion. It can also be defined as an atom's or ion's tendency to give up an electron. The amount of energy released when an atom accepts an atom from any neutral isolated gaseous atom to form a negative gaseous ion (anion) in the process is referred to as the electron gain enthalpy. Here, in this question, we have to calculate the energy of formation by the summation of enthalpies.

Complete Step by Step Solution:
The change in Enthalpy is associated with the synthesis of one mole of an ionic compound from its gaseous ions, while all other parameters stay constant. It's the amount of energy necessary to convert one mole of an ionic substance into gaseous ions completely.

There is an ionisation enthalpy of the formation of \[N{a^ + }\] is given.
\[Na\xrightarrow{{}}N{a^ + } + {e^ - }\,\,\,\,\,\,\,\Delta H = 495.8\,kJ\,mo{l^{ - 1}}\]
Form the electron gain enthalpy of \[Br\] ,
\[\dfrac{1}{2}B{r_2} + {e^ - }\xrightarrow{{}}Br\,\,\,\,\,\,\Delta H = - 325\,kJ\,mo{l^{ - 1}}\]
From the lattice energy of \[NaBr\] ,
\[N{a^ + } + B{r^ - }\xrightarrow{{}}NaBr\,\,\,\,\,\Delta H = - 728.4\,kJ\,mo{l^{ - 1}}\]
Now, we know that the energy of formation of \[NaBr\] ionic solid is the summation of ionisation enthalpy of the formation of \[N{a^ + }\] , electron gain enthalpy of \[Br\] , and lattice energy of \[NaBr\] .

Let us calculate the energy of formation of \[NaBr\] as follows:
\[
  \Delta {H_{f(NaBr)}} = 495.8 - 325 - 728.4 \\
   \Rightarrow \Delta {H_{f(NaBr)}} = - 557.6\,kJ\,mo{l^{ - 1}} \\
   \Rightarrow \Delta {H_{f(NaBr)}} = - 5576 \times {10^{ - 1}}\,kJ\,mo{l^{ - 1}} \\
 \]
As a result, the energy of formation of \[NaBr\] is \[ - 5576 \times {10^{ - 1}}\,kJ\,mo{l^{ - 1}}\] .

Note: Hess's Law of Heat Summation is the theory that is used to compute the heat absorbed or heat evolved for a reaction. The law of Heat Summation is named after Henri Hess, who suggested it for the first time in Russia in 1840.