1. **State Function and Its Importance** State functions are properties whose values depend only on the current state of the system, not on the path taken to reach that state. Examples include internal energy, enthalpy, entropy, and pressure. They are important in thermodynamics because they allow us to describe the system completely at equilibrium and simplify calculations involving changes between states without needing details of the process path. 2. **Third Law of Thermodynamics and Microscopic Interpretation** The Third Law states that as the temperature of a perfect crystalline substance approaches absolute zero (0 K), its entropy approaches zero. Microscopically, this means that at 0 K, the system's particles occupy a unique ground state with no disorder or randomness, so the number of accessible microstates is one, leading to zero entropy. 3. **Sodium Dimerization at 1 bar: Reaction 2Na (g) ⇄ Na₂ (g)** (i) **Equilibrium Constant Kₚ at 1000 K and 1200 K** - The equilibrium constant at constant pressure is given by: $$K_p = \frac{p_{\mathrm{Na}_2}}{p_{\mathrm{Na}}^2}$$ - From statistical thermodynamics, relate $K_p$ to partition functions and dissociation energy: $$K_p = \left(\frac{q_{\mathrm{Na}_2} / V}{(q_{\mathrm{Na}}/V)^2}\right) \exp\left(-\frac{\Delta E}{k_B T}\right) \left(\frac{k_B T}{p^{\circ}}\right)^{\Delta n}$$ - Given dissociation energy $D_0=70\times10^3$ J/mol, use $\Delta E = -D_0$ for the association reaction. - Calculate $q_{\mathrm{Na}_2}$ and $q_{\mathrm{Na}}$ at each temperature: $\quad q_{\mathrm{Na}_2} = q_{\text{trans}} \times q_{\text{rot}} \times q_{\text{vib}}$ with: $\quad q_{\mathrm{trans}}/V = 1.508 \times 10^{19} T^{3/2}$ $\quad q_{\mathrm{rot}} = \frac{T}{\Theta_{\mathrm{rot}}} = \frac{T}{0.221}$ $\quad q_{\mathrm{vib}} = \frac{1}{1 - e^{-\Theta_{\mathrm{vib}}/T}} = \frac{1}{1 - e^{-229/T}}$ $\quad q_{\mathrm{Na}}/V = 7.543 \times 10^{18} T^{3/2}$ - Then, $$K_p = \frac{q_{\mathrm{Na}_2}/V}{(q_{\mathrm{Na}}/V)^2} \times e^{\frac{70\times10^3}{R T}} \times \left(\frac{k_B T}{p^{\circ}}\right)^{-1}$$ (accounting for $\Delta n = -1$) - Calculate numerically for $T=1000$ K and $1200$ K (using $R=8.314$ J/mol.K, $k_B$, and $p^{\circ}=1$ bar) (ii) **Fraction of Na atoms forming Na₂ dimers at 1000 K, $p_{total}=1$ bar** - Define $x$ as fraction of Na forming dimers: moles Na atoms initially $=N$, moles Na dimers $= (xN)/2$. - Expression for $K_p$ using partial pressures: $$K_p = \frac{p_{\mathrm{Na}_2}}{(p_{\mathrm{Na}})^2} = \frac{(x/2) p_{total}}{(1 - x)^2 p_{total}^2} = \frac{x/2}{(1-x)^2 p_{total}}$$ - Solve for $x$: $$x = \frac{2 K_p p_{total} (1-x)^2}{1}$$ - Rearrange and solve quadratic numerically to find $x$. (iii) **Estimate Enthalpy Change $\Delta_r H^\circ$** - Using van't Hoff equation: $$\frac{d \ln K_p}{d (1/T)} = -\frac{\Delta_r H^\circ}{R}$$ - Compute $\ln K_p$ at 1000 K and 1200 K from (i), calculate slope numerically: $$\Delta_r H^\circ \approx -R \frac{\ln K_p(1200) - \ln K_p(1000)}{(1/1200)-(1/1000)}$$ - Result provides enthalpy change estimate, compare to dissociation energy of 70 kJ/mol. **Summary:** - State functions depend solely on state variables. - Third Law: entropy approaches zero at 0 K due to unique microstate. - Use partition functions and Boltzmann factors to find $K_p$. - Calculate dimer fraction from equilibrium expressions. - Use van't Hoff relation to estimate enthalpy change. $q_count=3$