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.

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.

The net enthalpy change, $\Delta H$ for a cyclic process is zero as enthalpy change is a state function, i.e., $\Delta H$ (cycle) $=0$

The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$ is $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The solid form of $\mathrm{H}_2 \mathrm{O}$ is ice. In ice, molecules of $\mathrm{H}_2 \mathrm{O}$ are less random than in liquid water.

Thus, molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})<$ molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$. The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})$ is less than $70 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.