In $\mathrm{CH}_4$, there are four $\mathrm{C}-\mathrm{H}$ bonds. The enthalpy of atomisation of 1 mole of $\mathrm{CH}_4$ means dissociation of four moles of $\mathrm{C}-\mathrm{H}$ bond.

$$\begin{aligned} \therefore \quad \mathrm{C}-\mathrm{H} \text { bond energy per } \mathrm{mol} & =\frac{1665 \mathrm{~kJ}}{4 \mathrm{~mol}} \\ & =416.25 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$

Given that, $\Delta_{\text {sub }} H^5$ for Na metal $=108.4 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

IE of $\mathrm{Na}=496 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\text {eg }} H^{\mathrm{s}}$ of $\mathrm{Br}=-325 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\text {diss }} H^{\mathrm{s}}$ of $\mathrm{Br}=192 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_f H^{\mathrm{s}}$ for $\mathrm{NaBr}=-360.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Born-Haber cycle for the formation of NaBr is as

By applying Hess's law,

$$\begin{aligned} \Delta_f H^{\mathrm{s}} & =\Delta_{\text {sub }} H^{\mathrm{s}}+I E+\Delta_{\text {diss }} H^{\mathrm{s}}+\Delta_{\text {eg }} H^{\mathrm{s}}+U \\ -360.1 & =108.4+496+96+(-325)-U \\ U & =+735.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$

The mixing of two gases have $\Delta H$ equal to zero. Therefore, it is spontaneous process because energy factor has no role to play but randomness increases i.e., randomness factor favours the process.

Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is

$$\Delta S=\frac{q_{\mathrm{rev}}}{T}$$

Here, $\quad \Delta S=$ change in entropy

$$\begin{aligned} \mathrm{q}_{\mathrm{rev}} & =\text { heat of reversible reaction } \\ T & =\text { temperature } \end{aligned}$$

Yes, the temperature of system and surroundings be the same when they are in thermal equilibrium.