Use the following data to calculate $\Delta_{\text {lattice }} H^{\mathrm{s}}$ for $\mathrm{NaBr} . \Delta_{\text {sub }} H^{\mathrm{s}}$ for sodium metal $=108.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$, ionisation enthalpy of sodium $=496 \mathrm{~kJ} \mathrm{~mol}^{-1}$, electron gain enthalpy of bromine $=-325 \mathrm{~kJ} \mathrm{~mol}^{-1}$, bond dissociation enthalpy of bromine $=192 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_f H^{\mathrm{s}}$ for $\mathrm{NaBr}(s)=-360.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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,
$$\begin{aligned} \Delta_f H^{\mathrm{s}} & =\Delta_{\text {sub }} H^{\mathrm{s}}+I E+\Delta_{\text {diss }} H^{\mathrm{s}}+\Delta_{\text {eg }} H^{\mathrm{s}}+U \\ -360.1 & =108.4+496+96+(-325)-U \\ U & =+735.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$
Given that $\Delta H=0$ for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
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. Write the mathematical relation which relates these three parameters.
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
$$\begin{aligned} \mathrm{q}_{\mathrm{rev}} & =\text { heat of reversible reaction } \\ T & =\text { temperature } \end{aligned}$$
Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
Yes, the temperature of system and surroundings be the same when they are in thermal equilibrium.
At $298 \mathrm{~K}, K_p$ for reaction $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$ is 0.98 . Predict whether the reaction is spontaneous or not.
For the reaction, $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g), K_p=0.98$
As we know that $\quad \Delta_r G^{\mathrm{s}}=-2.303 R \mathrm{log} K_p$
Here, $K_p=0.98$ i.e., $K_p<1$ therefore, $\Delta_r G^{\circ}$ is positive, hence the reaction is non-spontaneous.