Animated Solution for Chemistry - Chemical Thermodynamics: The ionisation enthalpy of ${\text{Na}}^{+}$ formation from $\text{Na}(g)$ is $495.8{\text{ kJ mol}}^{-1}$, while the electron gain enthalpy of Br is $-325.0{\text{ kJ mol}}^{-1}$. Given, the lattice enthalpy of NaBr is $-728.4{\text{ kJ mol}}^{-1}$. The energy for the formation of NaBr ionic solid is $(-)\dots \cdots \times 10^{-1}{\text{ kJ mol}}^{-1}$.

Imagine you are an atomic architect tasked with building a solid crystal of sodium bromide ($\text{NaBr}$) from scratch. You are handed a gaseous sodium atom and a gaseous bromine atom. How much energy will this entire construction process absorb or release?

To find out, we can't just smash them together and stick a thermometer in it. Instead, we rely on a beautiful principle of thermodynamics called Hess's Law. Hess's Law states that the total enthalpy change of a reaction is independent of the pathway taken. It only depends on the initial and final states. This allows us to break down our complex construction project into a series of simple, measurable steps, much like a financial ledger of energy.

Step 1: Paying the Toll (Ionization) First, we need to prepare our building blocks. We must turn our neutral sodium atom into a positively charged ion (${\text{Na}}^{+}$). Electrons don't just leave voluntarily; they are bound to the nucleus. We have to pay an energy toll to rip an electron away. This is the ionization enthalpy (${\Delta }_{i}H$). For sodium, this costs us $495.8{\text{ kJ mol}}^{-1}$. Because we are putting energy into the system, this value is positive.

Step 2: Getting a Refund (Electron Gain) Now we have a free electron wandering around. Our bromine atom, being a halogen, is desperate for an extra electron to complete its octet. When it grabs that electron to become a bromide ion (${\text{Br}}^{-}$), it drops to a lower, more stable energy state. The excess energy is released into the surroundings. This is the electron gain enthalpy (${\Delta }_{eg}H$), and it gives us a "refund" of $-325.0{\text{ kJ mol}}^{-1}$. The negative sign indicates energy leaving the system.

Step 3: The Grand Finale (Lattice Formation) Finally, we have our ${\text{Na}}^{+}$ and ${\text{Br}}^{-}$ ions. Opposite charges attract strongly. When these gaseous ions crash together to form a highly ordered, stable solid crystal lattice, a massive amount of electrostatic potential energy is released. This is the lattice enthalpy (${\Delta }_{\text{lattice}}H$), providing a huge energy payout of $-728.4{\text{ kJ mol}}^{-1}$.

Hess's Law

According to Hess's Law, the total energy change for the formation of the solid (${\Delta }_{f}H$) is simply the sum of the energy changes of these individual steps:

Let's plug in our ledger values:

Let's do the math carefully. It's often easier to sum the negative terms first:

Now, add the positive ionization energy:

So, the overall process releases $557.6\text{ kJ}$ of energy per mole. However, in competitive exams like JEE, the final boss is often the formatting of the answer. The question asks for the value in the specific format of $(-)\dots \cdots \times 10^{-1}{\text{ kJ mol}}^{-1}$.

We need to adjust our decimal point to match this scientific notation:

Therefore, the integer that fills the blank is $5576$. This problem is a perfect reminder that while understanding the deep physics of energy cycles is crucial, meticulous attention to algebraic signs and final formatting is what secures the marks!