Grade 11 : Chemistry - Thermodynamics (Set 3)

Explanation: An intensive property is independent of the mass or size of the system. Enthalpy, volume, heat capacity, and internal energy depend on system size (extensive), whereas molar heat capacity (heat capacity per mole) is intensive.

Explanation: For an isothermal reversible expansion:
$$W = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$$
$$W = -2.303 \times 2.0 \times 8.314 \times 300 \times \log_{10}\left(\frac{20}{2}\right)$$
$$W = -11488.4\text{ J} = -11.49\text{ kJ}$$

Explanation: According to the First Law of Thermodynamics:
$$\Delta U = q + w$$
Heat absorbed by system $q = +500\text{ J}$. Work done by system $w = -200\text{ J}$.
$$\Delta U = +500 + (-200) = +300\text{ J}$$

Explanation: The standard enthalpy of formation is defined per mole of $\text{NH}_3$ produced.
$$\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$$
$$\Delta_r H^\circ = 2 \times (-46.1\text{ kJ}) - (0 + 0) = -92.2\text{ kJ}$$

Explanation: Relationship between $\Delta H$ and $\Delta U$:
$$\Delta H = \Delta U + \Delta n_g RT$$
$$\Delta n_g = n_{\text{products, gaseous}} - n_{\text{reactants, gaseous}} = 1 - (1 + 2) = -2$$
$$\Delta H^\circ = -885.0 + (-2) \times (8.314 \times 10^{-3}) \times 298 = -885.0 - 4.95 = -889.95\text{ kJ mol}^{-1}$$

Explanation: By definition, an adiabatic process is one in which no heat is transferred between the system and its surroundings, so $q = 0$.

Explanation: At equilibrium (phase change):
$$\Delta S_{\text{fusion}} = \frac{\Delta H_{\text{fusion}}}{T_m}$$
$$\Delta H_{\text{fusion}} = T_m \times \Delta S_{\text{fusion}} = 273 \times 22.0 = 6006\text{ J mol}^{-1} = 6.006\text{ kJ mol}^{-1} \approx 6.01\text{ kJ mol}^{-1}$$

Explanation: A reaction is spontaneous when $\Delta G < 0$.
$$\Delta G = \Delta H - T\Delta S = 0 \implies T = \frac{\Delta H}{\Delta S}$$
$$T = \frac{40000\text{ J mol}^{-1}}{100.0\text{ J K}^{-1}\text{ mol}^{-1}} = 400\text{ K}$$
Since $\Delta H > 0$ and $\Delta S > 0$, $\Delta G$ becomes negative for $T > 400\text{ K}$.

Explanation: Enthalpy ($H$) depends only on the current state of the system, not on the path taken. Heat ($q$) and work ($w$) are path functions.

Explanation: By standard thermodynamic convention, the enthalpy of formation of pure elements in their most stable standard state at 298 K and 1 bar is assigned a value of zero.

Explanation: $\Delta_r H = \sum \text{Bond Energy(reactants)} - \sum \text{Bond Energy(products)}$
$$-184 = [BE(\text{H}-\text{H}) + BE(\text{Cl}-\text{Cl})] - 2 \times BE(\text{H}-\text{Cl})$$
$$-184 = (436 + 242) - 2 \times BE(\text{H}-\text{Cl})$$
$$2 \times BE(\text{H}-\text{Cl}) = 678 + 184 = 862 \implies BE(\text{H}-\text{Cl}) = 431\text{ kJ mol}^{-1}$$

Explanation: Work done in irreversible expansion against constant external pressure:
$$w = -P_{\text{ext}} \Delta V = -2.0\text{ atm} \times (5.0 - 1.0)\text{ L} = -8.0\text{ L atm}$$
$$w = -8.0 \times 101.3\text{ J} = -810.4\text{ J}$$

Explanation: Neutralization of strong acid and strong base involves the net ionic reaction $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$, releasing a constant heat of approximately $-57.1\text{ kJ mol}^{-1}$ (or $-13.7\text{ kcal mol}^{-1}$).

Explanation: From $\Delta G = \Delta H - T\Delta S$, if $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in disorder), $\Delta G$ is negative at all absolute temperatures $T$, making the process always spontaneous.

Explanation: Adding the two reactions:
$$(A \rightarrow B) + (B \rightarrow C) \implies A \rightarrow C$$
$$\Delta H = \Delta H_1 + \Delta H_2 = (+50\text{ kJ}) + (-80\text{ kJ}) = -30\text{ kJ}$$

Explanation: Heat required:
$$q = m \cdot c \cdot \Delta T$$
$$q = 50.0\text{ g} \times 4.184\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times (45.0 - 25.0)^\circ\text{C}$$
$$q = 50.0 \times 4.184 \times 20.0 = 4184\text{ J} = 4.184\text{ kJ}$$

Explanation: For an ideal gas, internal energy $U$ depends only on temperature $T$. In an isothermal process, $\Delta T = 0$, so $\Delta U = 0$ and $\Delta H = 0$.

Explanation: Mayer's relation states that for 1 mole of an ideal gas, $C_p - C_v = R$, because expanding against constant pressure requires additional work ($P\Delta V = R\Delta T$).

Explanation: The equation relating $\Delta G^\circ$ and equilibrium constant $K$ is:
$$\Delta G^\circ = -RT \ln K$$
When $\Delta G^\circ = 0$, $-RT \ln K = 0 \implies \ln K = 0 \implies K = e^0 = 1$.

Explanation: At constant volume, $\Delta V = 0$, so boundary work $w = -P\Delta V = 0$. By the First Law, $\Delta U = q_v + w = q_v$. Thus, heat measured at constant volume is equal to internal energy change $\Delta U$.